Exhibition Shows: Inverse Normal Distribution

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50 Solutions to Exercises

4 Solutions to Exercises

4.1 About these solutions

The solutions that follow were prepared by Darryl K. Nester. I occasionally pillaged or plagiarized solutions from the second edition (prepared by George McCabe), but I take full responsibility for any errors that may remain. Should you discover any errors or have any comments about these solutions (or the odd answers, in the back of the text), please report them to me:

Darryl Nester Bluffton College Bluffton, Ohio 45817 email: [email protected] WWW: http://www.bluffton.edu/~nesterd

4.2 Using the table of random digits

Grading SRSs chosen from the table of random digits is complicated by the fact that students can Þnd some creative ways to (mis)use the table. Some approaches are not mistakes, but may lead to different students having different ÒrightÓ answers. Correct answers will vary based on:

• The line in the table on which they begin (you may want to specify one if the text does not).

•Whether they start with, e.g., 00 or 01. •Whether or not they assign multiple labels to each unit. •Whether they assign labels across the rows or down the columns (nearly all lists in the text are alphabetized down the columns).

Some approaches can potentially lead to wrong answers. Mistakes to watch out for include:

• They may forget that all labels must be the same length, e.g., assigning labels like 0, 1, 2, . . . , 9, 10, . . . rather than 00, 01, 02, . . ..

• In assigning multiple labels, they may not give the same number of labels to all units. E.g., if there are 30 units, theymay try to use up all the two-digit numbers, thus assigning 4 labels to the Þrst ten units and only 3 to the remaining twenty.

4.3 Using statistical software

The use of computer software or a calculator is a must for all but the most cursory treatment of the material in this text. Be aware of the following considerations:

Acknowledgments 51

• Standard deviations: Students may easily get confused by software which gives both the so-called Òsample standard deviationÓ (the one used in the text) and the Òpopulation standard deviationÓ (dividing by n rather than n − 1). Symbolically, the former is usually given as ÒsÓ and the latter as ÒσÓ (sigma), but the distinction is not always clear. For example, many computer spreadsheets have a command such as ÒSTDEV(. . . )Ó to compute a standard deviation, but you may need to check the manual to Þnd out which kind it is. As a quick check: for the numbers 1, 2, 3, s = 1 while σ .= 0.8165. In general, if

two values are given, the larger one is s and the smaller is σ . If only one value is given, and it is the ÒwrongÓ one, use the relationship s = σ

√ n

n−1 .

• Quartiles and Þve-number summaries: Methods of computing quartiles vary between different packages. Some use the approach given in the text (that is, Q1 is the median of all the numbers below the location of the overall median, etc.), while others use a more complicated approach. For the numbers 1, 2, 3, 4, for example, we would have Q1 = 1.5 and Q3 = 2.5, but Minitab reports these as 1.25 and 2.75, respectively. Since I used Minitab for most of the analysis in these solutions, this was sometimes

a problem. However, I remedied the situation by writing a Minitab macro to compute quartiles the IPS way. (In effect, I was Òdumbing downÓ Minitab, since its method is more sophisticated.) This and other macros are available at my website.

• Boxplots: Some programs which draw boxplots use the convention that the ÒwhiskersÓ extend to the lower and upper deciles (the 10th and 90th percentiles) rather than to the minimum and maximum. (DeltaGraph, which I used for most of the graphs in these solutions, is one such program. It took some trickery on my part to convince it to make them as I wanted them.) While the decile method is merely different from that given in the text, some methods

are (in my opinion) just plain wrong. Some graphing calculators from Sharp draw Òbox charts,Ó which have a center line at the mean (not the median), and a box extending from x − σ to x + σ ! I know of no statistics text that uses that method.

4.4 Acknowledgments

I should mention the software I used in putting these solutions together:

• For typesetting: TEX Ñ speciÞcally, Textures, from Blue Sky Software. • For the graphs: DeltaGraph (SPSS), Adobe Illustrator, and PSMathGraphs II (MaryAnn Software).

• For statistical analysis: Minitab, G•Power, JMP IN, and GLMStatÑthe latter two mostly for the Chapters 14 and 15. George McCabe supplied output from SAS for Chapter 15. G•Power is available as freeware on the Internet, while GLMStat is share- ware. Additionally, I used the TI-82, TI-85, TI-86, and TI-92 calculators from Texas Instruments.

52 Chapter 1 Looking at Data Ñ Distributions

Chapter 1 Solutions

Section 1: Displaying Distributions with Graphs

1.1 (a) Categorical. (b) Quantitative. (c) Categorical. (d) Categorical. (e) Quantitative. (f) Quantitative.

1.2 Gender: categorical. Age: quantitative. Household income: quantitative. Voting Democratic/Republican: categorical.

1.3 The individuals are vehicles (or ÒcarsÓ). Variables: vehicle type (categorical), where made (categorical), city MPG (quantitative), and highway MPG (quantitative).

1.4 Possible answers (unit; instrument):

• number of pages (pages; eyes) • number of chapters (chapters; eyes) • number of words (words; eyes [likely bloodshot after all that counting]) • weight or mass (pounds/ounces or kilograms; scale or balance) • height and/or width and/or thickness (inches or centimeters; ruler or measuring tape) • volume (cubic inches or cubic centimeters; ruler ormeasuring tape [and a calculator])

Any one of the Þrst three could be used to estimate the time required to read the book; the last two would help determine how well the book would Þt into a book bag.

1.5 A tape measure (the measuring instrument) can be used to measure (in units of inches or centimeters) various lengths such as the longest single hair, length of hair on sides or back or front. Details on how to measure should be given. The case of a bald (or balding) person would make an interesting class discussion.

1.6 Possible answers (reasons should be given): unemployment rate, average (mean or median) income, quality/availability of public transportation, number of entertainment and cultural events, housing costs, crime statistics, population, population density, number of automobiles, various measures of air quality, commuting times (or other measures of trafÞc), parking availability, taxes, quality of schools.

1.7 For (a), the number of deaths would tend to rise with the increasing population, even if cancer treatments become more effective over time: Since there are more people, there are more potential cases of cancer. Even if treatment is more effective, the increasing cure rate may not be sufÞcient to overcome the rising number of cases. For (b), if treatments for other diseases are also improving, people who might have

died from other causes would instead live long enough to succumb to cancer.

Solutions 53

Even if treatments were becoming less effective, many forms of cancer are detected earlier as better tests are developed. In measuring Þve-year survival rates for (c), if we can detect cancer (say) one year earlier than was previously possible, then effectively, each patient lives one year longer after the cancer is detected, thus raising the Þve-year survival rate.

1.8 (a) 1988: 94924,800,000 .= 0.00003827 = 38.27 deaths per million riders. 1992: 90354,632,000

.= 0.00001653 = 16.53 deaths per million riders. Death rates are less than half what they were; bicycle riding is safer. (b) It seems unlikely that the number of riders more than doubled in a six-year period.

1.9 Using the proportion or percentage of repairs, BrandA ismore reliable: 294213,376 .= 0.22 =

22% for Brand A, and 192480 = 0.4 = 40% for Brand B.

1.10 (a) Student preferences may vary; be sure they give a reason. Method 1 is faster, but less accurateÑit will only give values that are multiples of 10. (b) In either method 1 or 2, fractions of a beat will be lostÑfor example, we cannot observe 7.3 beats in 6 seconds, only 7. The formula 60× 50÷ t , where t is the time needed for 50 beats, would give a more accurate rate since the inaccuracy is limited to the error in measuring t (which can be measured to the nearest second, or perhaps even more accurately).

1.11 Possible answers are total proÞts, number of employees, total value of stock, and total assets.

1.12 (a) Yes: The sum of the ethnic group counts is 12,261,000. (b) A bar graph or pie chart (not recom- mended) may be used. In order to see the contrast of the heights of the bars, the chart needs to be fairly tall.

American Indian

Asian non-Hispanic

Black

Hispanic

non-Hispanic white

Foreign

1000

2000

3000

4000

5000

6000

7000

8000

9000

N um

be r o

f s tu

de nt

s (th

ou sa

nd s)

54 Chapter 1 Looking at Data Ñ Distributions

1.13 (a) Shown at right. The bars are given in the same order as the data in the tableÑthe most obvious wayÑbut that is not necessary (since the vari- able is nominal, not ordinal). (b) A pie chart would not be appropriate, since the different entries in the ta- ble do not represent parts of a single whole.

Comp. Sci.

Life Sci.

Educ. Engin. Phys. Sci.

Psych. 0

Pe rc

en t o

f f em

al e

do ct

or at

1.14 (a) Below. For example, ÒMotor VehiclesÓ is 46% since 41,89390,523 = 0.4627 . . .. The ÒOther causesÓ category is needed so that the total is 100%. (b) Below. The bars may be in any order. (c) A pie chart could also be used, since the categories represent parts of a whole (all accidental deaths).

Cause Percent Motor vehicles 46 Falls 15 Drowning 4 Fires 4 Poisoning 8 Other causes 23

Motor Vehicle

Falls Drowning Fires Poison Other Causes

Pe rc

en t o

f a cc

id en

ta l d

ea th

1.15 Figure 1.10(a) is strongly skewed to the right with a peak at 0; Figure 1.10(b) is somewhat symmetric with a central peak at 4. The peak is at the lowest value in 1.10(a), and at a central value in 1.10(b).

1.16 The distribution is skewed to the right with a single peak. There are no gaps or outliers.

1.17 There are two peaks. Most of the ACT states are located in the upper portion of the distribution, since in such states, only the stronger students take the SAT.

1.18 The distribution is roughly symmetric. There are peaks at .230Ð.240 and .270Ð.290. The middle of the distribution is about .270. Ignoring the outlier, the range is about .345− .185 = .160 (or .350− .180 = .170).

1.19 Sketches will vary. The distribution of coin years would be left-skewed because newer coins are more common than older coins.

Solutions 55

1.20 (a) Among the women, 200 appears to an outlier. Among the men, the two high scores would probably not be consid- ered outliers. (b) The womenÕs median is 138.5; the range is 99 (101 to 200). The menÕs median is 114.5; the range is 117 (70 to 187). Generally, women have higher scores.

Men Women 50 7 8 8 21 9 984 10 139 5543 11 5

6 12 669 2 13 77 60 14 08 1 15 244 9 16 55

17 8 70 18

19 20 0

1.21 The back-to-back stemplot shown has split stems. There does not seem to be a substantial difference be- tween the two groups; this is supported by the fact that the medians are 111.5 (calcium) and 112 (placebo)Ñ almost identical. Before treatments, the two groups are very similar.

Calcium Placebo 9 8

2 10 2 77 10 9

2210 11 0224 11 79

3 12 3 9 12

13 0 6 13

1.22 If the Þrst two digits are used as stems, both distributions appear very spread out and one might conclude that there are outliers. This stemplot uses the hundreds digit for (split) stems and the tens digit for leaves. (The usual practice with stemplots is to truncateÑignore the ones digitÑrather than to round.) This display suggests that the experimental chicks had greater weight gain; the medians were 358 grams for the control chicks and 406.5 grams for the high-lysine group.

Control Experimental 87 2

44221 3 123 98866555 3 6799

310 4 00001222334 65 4 67

1.23 A histogram (using the classes 10Ð14, 15Ð19, 20Ð24, etc.) is essentially the same as the stemplot shown (with split stems). Preferences may vary; for example, some students Þnd stemplots easier to make, or prefer them because one can Þnd actual data values by looking at stemplots. The distribution is slightly left- skewed.

1 44 1 5899 2 2 2 55667789 3 13344 3 555589 4 0011234 4 5667789 5 1224

56 Chapter 1 Looking at Data Ñ Distributions

1.24 The stemplot gives more information than a histogram (since all the original numbers can be read off the stemplot), but both give the same impression. The distribution is roughly symmetric with one value that is somewhat low. The center of the distribution is between 5.4 and 5.5 (the median is 5.46).

48 8 49 50 7 51 0 52 6799 53 04469 54 2467 55 03578 56 12358 57 59 58 5

1.25 (a) Preferences will vary. The Þrst plot has the advantage of being compact, while the split stems suggest that there may be a second peak. (b) In either plot, the distri- bution is roughly symmetric, with center around 12.6 or 12.7 percent. Alaska and Florida appear to be outliers; Alaska is low presumably because of its less attractive climate, while Florida is high because many retirees move there.

4 9 5 6 7 8 8 9 10 0029 11 011344469 12 003445556666 13 0133445677999 14 23455 15 2379 16 17 18 6

4 9 5 5 6 6 7 7 8 8 8 9 9 10 002 10 9 11 0113444 11 69 12 00344 12 5556666 13 013344 13 5677999 14 234 14 55 15 23 15 79 16 16 17 17 18 18 6

1.26 (a) A stemplot is shown; a histogram would have a similar appearance. Percents are truncated, and stems were split to keep the branches from getting too long. (b) −26.6% is substantially lower than all other returns. With the outlier omitted, the distribution is fairly symmetric with center around 2% to 3%, spread from −14.7% to 19.2%. (c) The time plot (below) reveals no apparent pattern.

Solutions 57

−2 6 −2 −1 −1 44000 −0 98765 −0 44433332222222100 0 00000111122222233333334444444 0 556666777778899 1 0001133 1 789

-30 -25 -20 -15 -10 -5 0 5

10 15

1991 1992 1993 1994 1995 1996 1997

M on

th ly

re tu

rn ra

Date

1.27 A stemplot is shown; a histogram would also be appropriate. There are no outliers, although the distribution is clearly right- skewed. The split stems emphasize the skew- ness by showing the gaps.

0 44 0 5555566777888888888889999999 1 000000000001112222333444 1 56777899 2 1144 2 3 2 3 8 4 0 4 5 12 5 9

1.28 (a) There are four variables: GPA, IQ, and self-concept are quantitative, while gender is categorical. (OBS is not a variable, since it is not really a ÒcharacteristicÓ of a student.) (b) Below. (c) The distribution is skewed to the left, with center (median) around 7.8. GPAs are spread from 0.5 to 10.8, with only 15 below 6. (d) There is more variability among the boys; in fact, there seems to be a subset of boys with GPAs from 0.5 to 4.9. Ignoring that group, the two distributions have similar shapes.

0 5 1 8 2 4 3 4689 4 0679 5 1259 6 0112249 7 22333556666666788899 8 0000222223347899 9 002223344556668 10 01678

Female Male 0 5 1 8 2 4

4 3 689 7 4 069

952 5 1 4210 6 129

98866533 7 223566666789 997320 8 0002222348 65300 9 2223445668 710 10 68

58 Chapter 1 Looking at Data Ñ Distributions

1.29 Stemplot at right, with split stems. The distribution is fairly symmetricÑperhaps slightly left-skewedÑ with center around 110 (clearly above 100). IQs range from the low 70s to the high 130s, with a ÒgapÓ in the low 80s.

7 24 7 79 8 8 69 9 0133 9 6778 10 0022333344 10 555666777789 11 0000111122223334444 11 55688999 12 003344 12 677888 13 02 13 6

1.30 Stemplot at right, with split stems. The distribution is skewed to the left, with center around 59.5. Most self- concept scores are between 35 and 73, with a few below that, and one high score of 80 (but not really high enough to be an outlier).

2 01 2 8 3 0 3 5679 4 02344 4 6799 5 1111223344444 5 556668899 6 00001233344444 6 55666677777899 7 0000111223 7 8 0

1.31 (a) Table at right. (b) Histogram below. Children (under 10) represent the single largest group in the population; about one out of Þve Americans was under 10 in 1950. There is a slight dip in the 10Ð19 age bracket, then the percentages trail off gradually after that. (c) Histogram below. The pro- jections show a much greater proportion in the higher age bracketsÑthere is now a gradual rise in the pro- portion up to ages 40Ð49, followed by the expected decline in the proportion of Òsenior citizens.Ó

Age Group 1950 2075 0Ð9 19.4% 11.2% 10Ð19 14.4 11.5 20Ð29 15.9 11.8 30Ð39 15.1 12.3 40Ð49 12.8 12.2 50Ð59 10.3 12.1 60Ð69 7.3 11.1 70Ð79 3.6 8.8 80Ð89 1.1 6.1 90Ð99 0.1 2.5 100Ð109 0.0 0.5

0 20 40 60 80 100 Age

2075

0 20 40 60 80 100 0

Pe rc

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f p op

ul at

io n

Age

1950

Solutions 59

1.32 Use relative frequency histograms, since there are considerablymoremen thanwomen. The two histograms are both skewed to the right (as income distributions often are). WomenÕs salaries are generally lower than menÕs.

10 20 30 40 50 60 Salary ($1000)

M en

10 20 30 40 50 60 0

Pe rc

en t o

f w or

ke rs

Salary ($1000)

W omen

1.33 A class that is $20,000 wide should have bars one-fourth as tall as the bars for the $5,000-wide classes.

15 25 35 45 55 65 75

Pe rc

en t o

f g ra

du at

Salary ($1000)

1.34 (a) Right. (b) The plot shows a decreasing trendÑfewer distur- bances overall in the later yearsÑ and more importantly, there is an apparent cyclic behavior. Look- ing at the table, the spring and summer months (April through September) generally have the most disturbancesÑprobably for the simple reason that more people are outside during those periods.

1968 1969 1970 1971 1972 1973

N um

be r o

f D is

tu rb

an ce

Period

60 Chapter 1 Looking at Data Ñ Distributions

1.35 (a) Right. The death rate de- creases fairly steadily over time. (b) The drop from the mid-1970s to the mid-1980s appears to be part of the overall trend; there is no partic- ular suggestion in the plot that the decrease is any greater during that time, and thus no evidence that the lower speed limits saved lives (espe- cially since the decrease continues after the mid-1980s). (c) A histogram could be made, but it would probably not be very useful: The most important thing to study about these numbers is the change over time, not the number of times that, e.g., the death rate was between 4.5 and 5.0.

1960 1964 1968 1972 1976 1980 1984 1988 1992

M ot

or v

eh ic

le d

ea th

ra te

Year

1.36

1.5

2.5

3.5

4.5

5.5

1960 1968 1976 1984 1992

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Year

1960 1968 1976 1984 1992

M ot

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1.37 In his Þrst Þve years, Ruth had few home runs (pitchers donÕt play in as many games as outÞelders). After that, until the last years of his career, his home-run output ßuctuated but was consistently high (25 or more).

1915 1920 1925 1930 1935

N um

be r o

f h om

e ru

Year

Solutions 61

1.38 MenÕs times gradually decreased over time, with little change since the late 1970s. The times of women decreased quite rapidly from 1972 until the mid-1980s; since that time, they have been fairly consistent.

110

120

130

140

150

160

170

180

190

1960 1970 1980 1990

W in

ni ng

ti m

es (m

in ut

es )

Year

W omen

M en

1.39 (a)Weights are generally lower for toddlers with summer and late fall birthdays (JuneÐ November), and highest for toddlers with DecemberÐMarch birthdays. (b) Toddlers with summer birthdays appear to be slightly taller than those with winter birthdays (though there is more deviation from this pattern than there was for the weight pattern). (c) Both plots have extremes in the summer and winter, but they are opposite: When one is high, the other is low. As a crude summary, the two plots together suggest that summer two- year-olds are tall and skinny, while winter two-year-olds are short and heavy.

Jan Apr Jul Oct 10.6

10.8

11.0

11.2

11.4

W ei

gh t (

kg )

Month Jan Apr Jul Oct

79.2

79.4

79.6

79.8

80.0

80.2

H ei

gh t (

cm )

Month

62 Chapter 1 Looking at Data Ñ Distributions

1.40 (a) Diarrhea is the worst from April through August, especially April, May, and June. In other months the percentage is generally low (about 2.5% or less). (b) There is some hint of a second, smaller peak in October/November, and maybe even a third small peak in January (recall that this graph would theoretically wrap around from De- cember to January). However, these smaller peaks may be mere random ßuctuation. (c) The prevalence of diarrhea in April, May, and June may account for the low weights for children with birthdays from June through November.

Jan Apr Jul Oct 0

Pe rc

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f d ay

s w

ith d

ia rr

he a

Month

Section 2: Describing Distributions with Numbers

1.41 (a) Stemplot shown with stems split Þve ways. The mean is 516.3 revolutions; the median is 516.5 revolutions. These are similar because the distribution is fairly symmetric. (b) s = 44.2. Because of the symme- try, x and s are appropriate.

4 55 4 4 5 001 5 3 5 555 5 5 8

1.42 (a) Stemplot at right; it is relatively symmetric. (b) M = 50.7. (c) Q3 = 58.1; there were landslides in 1964, 1972, and 1984.

4 33 4 999 5 003 5 578 6 01

1.43 (a) See solution to Exercise 1.20. (b) & (c) The right skewness makes x > M in both cases. The IQR for the women is 28, so the outlier test gives an upper limit of 154+ 42 = 196Ñmaking the score of 200 an outlier.

x M Five-number summaries Women 141.06 138.5 101 126 138.5 154 200 Men 121.25 114.5 70 98 114.5 143 187

(d) All the displays and descriptions reveal that women generally score higher than men. The menÕs scores (IQR = 45) are more spread out than the womenÕs (even if we donÕt ignore the outlier); this is fairly clear from the boxplot but not so obvious from the stemplot.

Women Men 60

100

120

140

160

180

200

SS H

A s

co re

Solutions 63

1.44 With the outlier: x = 141.06 and M = 138.50. Without the outlier: x = 137.59 and M = 137. Both drop, but the removal of the outlier has a greater effect on the mean than the median.

1.45 (a) Control: x = 366.3 grams and s = 50.8 grams. Experimental: x = 402.95 grams and s = 42.73 grams. (b) Both distributions appear to be relatively symmetric, with no outliersÑwhich makes x and s appropriate descriptions.

1.46 For measurements in ounces, divide x and s by 28.35. Thus for the control group xnew = 12.92 oz and snew = 1.79 oz, and for the experimental group xnew = 14.21 oz and snew = 1.507 oz.

1.47 The distribution of wealth will be skewed to the right, so the median is less than the mean: M = $800, 000 and x = $2.2 million.

1.48 One would expect stock prices to be skewed to the right (many inexpensive stocks, with a few stocks having higher prices), so the median should be less than the mean.

1.49 x = $62, 500 and M = $25, 000. Seven of the eight employeesÑall but the ownerÑ earned less than the mean.

1.50 If three brothers earn $0, $0, and $20,000, the reported median is $20,000. If the two brothers with no income take jobs at $14,000 each, the median decreases to $14,000. The same thing can happen to the mean: In this example, the mean drops from $20,000 to $16,000.

1.51 The mean rises to $87,500, while the median is unchanged.

1.52 (a) x = 5.4479 and s = 0.2209. (b) The Þrst measurement corresponds to 5.50 × 62.43 = 343.365 pounds per cubic foot. To Þnd xnew and snew, we similarly multiply by 62.43: xnew

.= 340.11 and snew .= 13.79.

1.53 (a) The stemplot with split stems shows a peak in the high 80s, and suggests a slight skew to the right. (Without split stems, the skewness is not very apparent.) (b) x = 89.67, s2 = 61.3089, and s = 7.83. (c) M = 88.5, Q1 = 84.5, Q3 = 93, and IQR = 8.5. There are no outliers: no scores are less than Q1−1.5× IQR = 71.75 or greater than Q3 + 1.5× IQR = 105.75. (d) Answers may vary; the slight skewness suggests that the quartiles should be used.

7 9 8 13 8 6789 9 01 9 5 10 2 10 5

64 Chapter 1 Looking at Data Ñ Distributions

1.54 Details at right.

x = 11, 200 7 = 1600,

s2 = 214, 872 6

= 35, 812, and

s = √ 35, 812 .= 189.24.

xi xi − x (xi − x)2 1792 192 36864 1666 66 4356 1362 −238 56644 1614 14 196 1460 −140 19600 1867 267 71289 1439 −161 25921 11200 0 214872

1.55 (a) 1, 1, 1, 1 (no spread) is one answer. (b) 0, 0, 10, 10 (greatest spread) is the only answer. (c)Any collection of equal numbers has variance 0, so (b) has 11 correct answers. The answer to (b) is unique.

1.56 Answers will vary. Typical calculators will carry only about 12 to 15 digits. Minitab (at least the version used to prepare these answers) fails at 100,000,001 (nine digits).

1.57 See Exercise 1.25 for stemplots. There is a low outlier of 4.9% (Alaska) and a high outlier of 18.6% (Florida). Because of the outliers, the Þve-number summary is a good choice: Min = 4.9%, Q1 = 11.4%, M = 12.6%, Q3 = 13.9%, Max = 18.6%.

1.58 (a) x = 1.887% and s = 7.6%. In an average month, $100 would grow to $101.89. (b) The low outlier is−26.6%; this would change a $100 investment to $74.40. Without the outlier, x = 2.238% and s = 6.944%Ñrespectively higher and lower than the values with the outlier included. The median and quartiles would change very little, if at all, since they are resistant to outliers. [In fact, only Q3 changes, from 6.7 to 6.75.]

1.59 See Exercise 1.27 for the stemplot. The survival times are skewed to the right, so the Þve-number summary is a good choice: Min = 43, Q1 = 82.5, M = 102.5, Q3 = 151.5, Max = 598 days. Half the guinea pigs lived less than 102.5 days; typical lifetimes were 82.5 to 151.5 days. The longest-lived guinea pig died just short of 600 days, while one guinea pig lived only 43 days.

1.60 See Exercise 1.29 for a stemplot; the distribution is fairly symmetric. x = 108.92, M = 110, and s = 13.17; the mean and median are close. (Although the four low scores are not outliers, they Òdrag downÓ the mean.)

1.61 The logical number to choose as the 10th percentile is 10.55% (the average of 10.2% and 10.9%Ñconsistent with how we compute medians). Likewise, the 90th percentile is 14.85% (the average of 14.5% and 15.2%). The top 10%are Iowa (15.2%),WestVirginia (15.3%), Rhode Island (15.7%), Pennsyl-

vania (15.9%), and Florida (18.6%). The bottom 10% are Alaska (4.9%), Utah (8.8%), Colorado and Georgia (10%), and Texas (10.2%). [Regardless of how we choose the percentiles, these answers must be the same: they are the top and bottom Þve states.]

Solutions 65

1.62 Answers may vary slightly depending on the exact methods students use, but they should be similar to 101, 107, 112, 119. (Sort the numbers in order, then choose the numbers in about the 16th, 32nd, 48th, and 63rd locations.)

1.63 The Þve-number summaries for sodium content are below (all numbers in mg):

Type Min Q1 M Q3 Max Beef 253 320.5 380.5 478 645 Meat 144 379 405 501 545 Poultry 357 379 430 535 588

Overall, beef hot dogs have less sodium (except for the one with the most sodium: 645 mg). Even if we ignore the low outlier among meat hot dogs, meat holds a slight edge over poultry. It is difÞcult to make a general recommendation, but clearly, the best advice is to avoid poultry hot dogs; either buy beef (and hope that you donÕt get the worst one) or buy meat hot dogs (and hope that you get the best one).

Beef Meat Poultry 100

200

300

400

500

600

So di

um (m

1.64 (a) Before recentering, verbal scores were clearly lower than math scores. Both sets of scores were raised by the recentering, and the SATV scores ended up slightly higher than the SATM scores. (b) The two peaks (referred to Exercise 1.17) are not visible in the boxplots.

SATV 1990

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SATV 1996

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1.65 (a) xnew = 746x = 746 · 140 = 104, 440 watts (a = 0, b = 746). (b) xnew = x/0.62 = 65/0.62 .= 104.8 kph (a = 0, b = 1/0.62 .= 1.61). (c) xnew = x − 98.6 degrees (a = −98.6, b = 1). (d) xnew = 130x · 100% =

10 3 x% (a = 0, b = 10/3).

1.66 Min = $17, 500, Q1 = $27, 500, M = $32, 500, Q3 = $37, 500, Max = $70, 000. For example: 14% of salaries are below $25,000 and 35% are below $30,000, so Q1 (the 25th percentile) is $27,500.

1.67 Variance is changed by a factor of 2.542 = 6.4516; generally, for a transformation xnew = a + bx , the new variance is b2 times the old variance.

1.68 There are 72 survival times, so to Þnd the 10% trimmed mean, remove the highest and lowest 7 values (leaving 58). Remove the highest and lowest 14 values (leaving 44) for the 20% trimmed mean.

66 Chapter 1 Looking at Data Ñ Distributions

The mean and median for the full data set are x = 141.8 and M = 102.5. The 10% trimmed mean is x∗ = 118.16, and the 20% trimmed mean is x∗∗ = 111.68. Since the distribution is right-skewed, removing the extremes lowers the mean.

Section 3: The Normal Distributions

1.69 (a) The curve forms a 1 × 1 square, which has area 1. (b) P(X < 0.25) = 0.25. (c) P(0.1 < X < 0.9) = 0.8.

0.25 10 0.1 0.9

1.70 (a) The height should be 12 , since the area under the curve must be 1. The density curve is at right. (b) P(X ≤ 1) = 12 . (c) P(0.5 < X < 1.3) = 0.4. 20 1

1.71 The mean and median both equal 0.5; the quartiles are Q1 = 0.25 and Q3 = 0.75.

1.72 (a) Mean is C, median is B (right skew pulls the mean to the right). (b) Mean A, Median A. (c)Mean A, Median B (left skew pulls the mean to the left).

For 1.73.

1.85 2.25 2.651.050.650.25 1.45

For 1.74.

67.9 72.7 77.558.353.548.7 63.1

1.75 Using the 68Ð95Ð99.7 rule: 1.45± 2(0.40) = 1.45± 0.80, or 0.65 to 2.25 grams per mile. Using table values: 1.45 ± 1.96(0.40) = 1.45 ± 0.784, or 0.666 to 2.234 grams per mile.

1.76 The 68% interval is 63.1± 4.8 = 58.3 to 67.9 kg. 95%: 63.1± 2(4.8) = 53.5 to 72.7 kg. 99.7%: 63.1± 3(4.8) = 48.7 to 77.5 kg.

1.77 (a) 266± 2(16) = 266± 32, or 234 to 298 days. (b) Less than 234 days; longer than 298 days.

1.78 x = 108.92 and s = 13.17. About 70.5% (55/78) of the IQs are in the range x ± s = 95.75 to 122.09 (96Ð122). About 93.6% (73/78) of the IQs are in the range x±2s = 82.58 to 135.26 (83Ð135). All (100%) of the IQs are in the range x±3s = 69.41 to 148.43 (70Ð148).

Solutions 67

1.79 Eleanor: z = 680−500100 = 1.8. Gerald: z = 27−186 = 1.5. EleanorÕs score is higher.

1.80 The three stand close together, an astounding four standard deviations above the typical hitter. (Williams has a slight edge, but perhaps not large enough to declare him Òthe best.Ó)

Cobb .420−.266 .0371 = 4.15

Williams .406−.267 .0326 = 4.26

Brett .390−.261 .0317 = 4.07

1.81 (a) 0.9978. (b) 0.0022. (c) 0.9515. (d) 0.9515− 0.0022 = 0.9493.

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1.82 (a) 0.0122. (b) 0.9878. (c) 0.0384. (d) 0.9878− 0.0384 = 0.9494.

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1.83 (a) −0.67 or −0.68 (software: −0.6745). (b) 0.25 (software: 0.2533).

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1.84 (a) 0.84 (software: 0.8416). (b) 0.38 or 0.39 (software: 0.3853).

1.85 SAT scores of 800+ correspond to z scores above 3; this is 0.15% (using the 68Ð95Ð 99.7 rule).

1.86 (a) 12% ± 2(16.5%) = −21% to 45% (or 12% ± 1.96(16.5%) = −20.34% to 44.34%). (b) About 23%: R < 0% means Z < 0−1216.5

.= −0.7273; the table gives 0.2327 for Z < −0.73. (b)About 21.5%: R ≥ 25%means Z ≥ 25−1216.5

.= 0.7879; the table gives 0.2148 for Z ≥ 0.79.

1.87 (a) X > 700 means Z > 700−544103 .= 1.5146; the table gives 0.0655 for Z > 1.51.

(b) X < 500 means Z < 500−544103 .= −0.4272; the table gives 0.3336 for Z < −0.43.

0.3336 = 0.6600.

1.88 (a) About 5.21%: P(X < 240) = P(Z < 240−26616 ) = P(Z < −1.625) = 0.0521. This software value is also halfway between the two table values 0.0516 (for −1.63) and 0.0526 (for −1.62). (b) About 54.7%: P(240 < X < 270) = P(−1.625 < Z < 270−266 16 ) = P(−1.625 < Z < 0.25) = 0.5987 − 0.0521 = 0.5466. (c) 279 days or

68 Chapter 1 Looking at Data Ñ Distributions

longer: The 80th percentile for a standard normal distribution is 0.8416 (or 0.84 from the table), so take 266+ 0.8416(16).

1.89 About 6.68%: If X is her measured potassium level, then X < 3.5 meq/l means Z < 3.5−3.80.2 = −1.5, for which Table A gives 0.0668.

1.90 The standard score for 1.7 is z = −1.625, and for 2.1 it is z = −1.125. P(X < 1.7) = 0.0521; this software value is also halfway between the two table values 0.0516 (for −1.63) and 0.0526 (for −1.62). P(1.7 < X < 2.1) = 0.1303− 0.0521 = 0.0782; 0.1303 is halfway between the two table values 0.1292 (for −1.13) and 0.1314 (for −1.12).

1.91 SarahÕs z score is 135−11025 = 1, while her motherÕs z score is 120−9025 = 1.2, so SarahÕs mother scored relatively higher. But Sarah had the higher raw score, so she does stand higher in the variable measured. Sarah scored at the 84th percentile (0.8413). Her mother scored at the 88.5th percentile

(0.8849).

1.92 To score among 30% who are most Anglo/English: about 3.42 or more. To score among 30% who are most Mexican/Spanish: about 2.58 or less. For the Þrst answer, the 70th percentile for a standard normal distribution is 0.5244 (or

0.52 from the table), so take 3 + 0.5244(0.8). For the second answer, use the 30th per- centile for a N (0, 1) distribution, which is−0.5244 (or−0.52), and take 3−0.5244(0.8).

1.93 (a) 50%: P(W < 100) = P(Z < 0) = 0.5. (b)W < 80means Z < 80−10015 .= −1.33;

the table gives 0.0918, or 9.18%. (c) W > 140 means Z > 140−10015 .= −2.67; the table

gives 0.38%. (d) 100 < W < 120 means 0 < Z < 120−10015 .= 1.33; the table gives

40.82%.

1.94 The top 5% is about 125 or higher: The 95th percentile for a N (0, 1) distribution is 1.645, so take 100+ 1.645(15) = 124.675. The top 1% is about 135 or higher: The 99th percentile for a N (0, 1) distribution is

2.326, so take 100+ 2.326(15) = 134.89.

1.95 (a) The area should be 25%, so Q1 .= −0.67. For the third quartile, the area should

be 75%, so Q3 .= 0.67. (A more accurate value is ±0.675). (b) Q1 = 100 − 15 ×

0.67 = 100 − 10.05 = 89.95 (89.875 using 0.675), and Q3 = 110.05 (or 110.125). (c) IQR = Q3− Q1 = 1.34 (or 1.35). (d) 1.5× IQR = 2.01 (or 2.025), so the suspected outliers are below Q1−1.5× IQR = −2.68 (or−2.7), and above Q3+1.5× IQR = 2.68 (or 2.7). This percentage is 2× 0.0037 = 0.74% (or 2× 0.0035 = 0.70%).

Solutions 69

1.96 (a) Software gives 1.2816 for the 90th percentile and−1.2816 for the 10th percentile. Using Table A, we would choose ±1.28. (b) About 245.5 and 286.5 days: Take 266 ± (1.2816)(16).

1.97 The plot does not suggest any major deviations from normality, except that the tails are less extreme than would be expected. This means extremely high and extremely low scores are Òstacked upÓÑno one scored below 14 or above 54.

1.98 The right skewness is shown by the sharp rise at the right end; it indicates that the longest survival times are higher than what one would expect from a normal distribution.

1.99 The plot is reasonably close to a line, apart from the stair-step appearance produced by granularityÑpresumably due to limited accuracy of the measuring instrument.

1.100 The plot (below, left) suggests nomajor deviations from normality, although the three lowest measurements donÕt quite fall in line with the other points.

For 1.100.

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1.101 The plot (above, right) suggests that the distribution is normal (except for the low point, which was a suspected outlierÑsee Exercise 1.26).

1.102 See also Exercise 1.30. The left-skewness shows up as a slight curve in the normal probability plot. There are no particular outliers. The mean score is x = 56.96, and the Þve-number summary is Min = 20, Q1 = 51, M = 59.5, Q3 = 66, Max = 80.

2 01 2 8 3 0 3 5679 4 02344 4 6799 5 1111223344444 5 556668899 6 00001233344444 6 55666677777899 7 0000111223 7 8 0

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70 Chapter 1 Looking at Data Ñ Distributions

1.103 The boysÕ distribution seems to have two peaks, one in the low 50s and another in the high 60s/low 70s. Since both distributions are slightly skewed to the left, the Þve-number summaries may be more appropriate. The stemplot and Þve-number summaries suggest

that girlsÕ scores do not seem to extend as high as the highest boysÕ scores.

x Min Q1 M Q3 Max Boys 57.91 20 51 59 67 80 Girls 55.52 21 49 60 64 72

Girls Boys 1 2 0 8 2

3 0 975 3 6 4 4 0234 96 4 79 444 5 1111223344 8665 5 56899

444320000 6 13344 99765 6 566677778

20 7 00011123 7 8 0

1.104 A stemplot from one sample is shown. Histograms will vary slightly, but should suggest a bell curve. The normal probability plot (below, left) shows something fairly close to a line, but illustrates that even for actual normal data, the tails may deviate slightly from a line.

−2 6 −2 −1 9998877655 −1 4433332111111 −0 9998888776666555 −0 43333222211110000 0 001222223333444 0 555566666778889 1 1123444 1 5 2 0033 2 5

For 1.104.

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1.105 A stemplot from one sample is shown. Histograms will vary slightly, but should suggest the density curve of Figure 1.33 (but with more variation than students might expect). The normal probability plot (above, right) shows that, compared to a normal distribution, the uniform dis- tribution does not extend as low or as high (not surprising, since all observations are between 0 and 1).

0 0123446677778899 1 001123445556 2 12337789 3 0139 4 0023446689 5 12356667889 6 001224568 7 0234447788 8 00799 9 011225577999999

Solutions 71

Exercises

1.106 (a) Car makes: a bar chart or pie chart. Car age: a histogram or stemplot (or a boxplot). (b) Study time: a histogram or stemplot (or a boxplot). Change in study hours: a time plot (average hours studied vs. time). (c) A bar chart or pie chart. (d) A normal probability plot.

1.107 (a) Since a person cannot choose the day on which he or she has a heart attack, one would expect that all days are Òequally likelyÓÑno day is favored over any other. While there is some day-to-day variation, this expectation does seem to be supported by the chart. (b) Monday through Thursday are fairly similar, but there is a pronounced peak on Friday, and lows on Saturday and Sunday. Patients do have some choice about when they leave the hospital, and many probably choose to leave on Friday, perhaps so that they can spend the weekend with the family. Additionally, many hospitals cut back on stafÞng over the weekend, and they may wish to discharge any patients who are ready to leave before then.

1.108 No, and no: It is easy to imagine examples of many different data sets with mean 0 and standard deviation 1Ñe.g., {−1,0,1} and {−2,0,0,0,0,0,0,0,2}. Likewise, for any given Þve numbers a ≤ b ≤ c ≤ d ≤ e (not all the same), we

can create many data sets with that Þve number summary, simply by taking those Þve numbers and adding some additional numbers in between them, e.g. (in increasing order): 10, , 20, , , 30, , , 40, , 50. As long as the number in the Þrst blank is between 10 and 20, etc., the Þve-number summary will be 10, 20, 30, 40, 50.

1.109 The 1940 distribution is skewed to the left, while the 1980 distribution is fairly symmetric and considerably less spread out than the 1940 distribution. There are few low percentages in the 1980s, reßecting increased voting by blacks.

1.110 (a) The stemplot below (with stems split two ways) looks fairly symmetric, but observe that the lowest observation is considerably less than the others, and the twohighest are also somewhat set apart. (This is even more apparent if, e.g., we split stems Þve ways. This also makes the stemplot looks less symmetric.) (b) The lowest observation (6.75 min) and the highest two (9.75 and 10.17min) are these unusual situations. Without them, we Þnd x .= 8.36 min and s .= 0.4645 min. In addition (or in place of) these numbers, we can Þnd the Þve-number summary: 7.42, 7.92, 8.42, 8.67, 9.17. (c) Based on a normal probability plot (not shown), the distribution is reasonably normal. (The split-Þve-ways stemplot does not look too promising; such impressions can be misleading.) (d) Plot below. There is no strong indication of a trend, but the last ten days (starting a bit after Thanksgiving) are all above average.

72 Chapter 1 Looking at Data Ñ Distributions

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1.111 Either a bar chart or a pie chart would be appropriate; both are shown below. The pie chart labels might also show the actual percents. An ÒOther methodsÓ category (with 7.9%) is needed so that the total is 100%.

Hand- guns

Other Guns

Knives Body Part

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1.112 Salary distributions are right-skewed, so the mean will be higher than the median: x = $1, 160, 000 and M = $490, 000.

1.113 (a) xnew = −50+2x : b = 2will change the standard deviation to 20; it alsomultiplies the mean by 2, so use a = 100 − 2(75) = −50. (b) xnew = −49.09 + 2011x . b = 2011 changes the standard deviation to 20; a = 100− 2011(82) = −49.09 makes the mean 100. (c)David: xnew = −50+2(78) = 106. Nancy: xnew = −49.09+ 2011(78) = 92.72. David scored relatively higher. (d) Using either 78 from a N (75, 10) distribution or 106 from N (100, 20) distribution, DavidÕs score is z = 0.3 standard deviations above the mean, so about 61.79% of third graders score below 78. For Nancy, z = −0.36, so about 35.94% (or 35.81% using software) of sixth graders score below 78.

1.114 (a) P(S < 20) = P(Z < −1) = 0.1587 (or Òabout 16%,Ó using the 68Ð95Ð99.7 rule). (b) P(S < 10) = P(Z < −3) = 0.0013 (or Òabout 0.15%,Ó using the 68Ð95Ð99.7 rule). (c) About 28.4: The 75th percentile for a standard normal distribution is 0.6745 (or 0.67 from the table), so take 25+ 0.6745(5).

Solutions 73

1.115 A WISC score of 135 is z = 73 = 2.3 .= 2.33 standard deviations above the mean,

so about 0.99% score above 135. This is about 12 or 13 (12.87) of the 1300 students.

1.116 (a) xnew = 4x : b = 4 multiplies both the mean and standard deviation by 4, leaving them at 100 and 20, as desired. (b) xnew = 4(30) = 120. (c) The quartiles for a standard normal distribution are ±0.6745 (or ±0.67 from the table), so take 100± 0.6745(20) = 86.51 and 113.49 (or 86.6 and 113.4).

1.117 The normal quantile plot indicates that the data are approximately normally dis- tributed; the mean and standard deviation are good measures for normal distributions. The mean is 35.09, and the standard deviation is 11.19.

1.118 See also the stemplots in the solution to Exercise 1.22. Both normal plots appear reasonably linear, so the mean and standard deviation should be useful. For the control group: xc = 366.3 g and sc .= 50.81 g. For the experimental group: xe = 402.95 g and se .= 42.73 g.

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1.119 (a) Five-number summaries and boxplots below. Note in particular that the OL boxplot looks odd since M = Q3 for that position. (b) The heaviest players are on the offensive line, followed by defensive linemen and tight ends. The lightest players overall are the kickers, followed by wide receivers and defensive backs. (c) The (1.5 × IQR) outlier test reveals outliers in the OL and WR positions. SpeciÞcally, the outliers are the lightest (235 lb) and heaviest (335 lb) offensive linemen and the lightest (155 lb) wide receiver. Note that the outlier test can be applied ÒvisuallyÓ to the boxplots: Take the length of

the box (which is the IQR) and multiply its length by 1.5. If the boxesÕ ÒwhiskersÓ extend more than this distance from the box, this indicates that there are outliers. With this in mind, we can easily see that only the WR and OL positions need to be examined.

74 Chapter 1 Looking at Data Ñ Distributions

Min Q1 M Q3 Max QB 180 185 202 207.5 210 RB 170 190 211 225 230 OL 235 275 295 295 335 WR 155 179 182.5 189 202 TE 230 235 242.5 256 260 K 160 167.5 175 184 193 DB 170 176 190 193 195 LB 205 215 220 230 237 DL 220 240 245 265 285

(All numbers in lbs)

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1.120 Results will vary. One set of 20 samples gave the results at the right (normal probability plots are not shown). Theoretically, x will have a N (20, 1)

distributionÑso that about 99.7% of the time, one should Þnd x between 17 and 23. Mean- while, the theoretical distribution of s is nearly normal (slightly skewed) with mean .= 4.9482 and standard deviation .= 0.7178; about 99.7% of the time, s will be between 2.795 and 7.102. Note that Òon the average,Ó s underestimates σ (that is, 4.9482 < 5). Unlike the mean x , s is not an unbiased estimator of σ ; in fact, for a sample of size n, the mean of s/σ is

√ 2 0(n/2)√

n−1 0(n/2−1/2) . (This factor approaches 1 as n approaches inÞnity.) The proof of this fact is left as an exerciseÑfor the instructor, not for the average student!

Means 18 589 19 00124 19 7789 20 1333 20 21 223 21 5

Standard deviations 3 8 4 01 4 22 4 44455 4 66 4 9 5 000 5 22 5 45

1.121 The distribution is strongly right-skewed, so the Þve- number summary is appropriate:

Min $109,000 Q1 $205,000 M $1,250,290 Q3 $2,300,000 Max $9,237,500

The highest salary is deÞnitely an outlier.

0 1111111122568899 1 1444577 2 0233 3 0 4 066 5 03 6 7 8 9 2

Solutions 75

1.122 (a) Multiply by 1.29 franc/dollar. (b) The stemplot (or a histogram) looks similar to that of the previous exercise. (c) The values for francs can be found by multiplying the dollar values by 1.29. (d) Once again, multiply the dollar values by 1.29 to get the franc values.

0 111111122378 1 011248889 2 226899 3 8 4 5 19 6 048 7 8 9 10 11 9

x Q1 M Q3 s IQR Dollars 1,803,761 205,000 1,250,290 2,300,000 2,028,071 2,095,000 Francs 2,326,852 264,450 1,612,873 2,967,000 2,616,212 2,702,550

1.123 The stemplot shown does not include Los Angeles countyÑan ex- tremely high outlier. The distribution is strongly right-skewed, so we use the Þve-number summary:

Min 1,232 Q1 44,193 M 150,816 Q3 533,392 Max 9,127,751

With IQR = 489, 199, the outlier test says that any population over 533, 392 + 733, 798.5 = 1, 267, 190.5 is an outlier. This would give seven outliers, but viewing the stemplot, one is inclined to take only the three largest counties as outliers. One division would be to include all of the top 25% of the counties (i.e., the

counties with population over Q3 = 533, 392), some of the middle half (those with population between Q1 and Q3), and a small fraction of the bottom 25%.

0 000000000000000000000001111111111 0 22223333 0 445 0 66777 0 8 1 1 1 3 1 455 1 1 2 2 2 2 66

1.124 Stemplots for both variables are at the right. H2S is slightly right-skewed, while LACTIC is more spread out and symmetric. Normal probability plots (not shown) reßect that observation: LACTIC produces a fairly straight line, while H2S is slightly curved on the ends. From the summary statistics below, x and s are appropriate for LACTIC, while the Þve-number summary is better for H2S.

x s Five-number summary H2S 5.942 2.127 2.996 3.912 5.329 7.601 10.199 Lactic 1.4420 0.3035 0.860 1.250 1.450 1.680 2.010

H2S 2 9 3 1268899 4 17799 5 024 6 11679 7 4699 8 7 9 025 10 1

LACTIC 8 6 9 9 10 689 11 56 12 5599 13 013 14 469 15 2378 16 38 17 248 18 1 19 09 20 1

76 Chapter 1 Looking at Data Ñ Distributions

1.125 Men seem to have higher SATM scores than women; each number in the Þve-number summary is about 40 to 50 points higher than the corresponding number for women. Women generally have higher GPAs than men, but the difference is less striking; in fact, the menÕs median is slightly higher. All four normal probability plots (not

shown) look fairly linear, so all four data sets might be judged normal. How- ever, both GPA setsÑespecially the male GPAÑare somewhat left-skewed; there is some evidence of this in the long bottom tails of the GPA boxplots. Statistical tests indicate that the male GPA numbers would not be likely to come from a normal distribution.

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Min Q1 M Q3 Max Male GPA 0.12 2.135 2.75 3.19 4.00 Female GPA 0.39 2.250 2.72 3.33 4.00 Male SATM 400 550 620 670 800 Female SATM 300 510 570 630 740

Solutions 77

Chapter 2 Solutions

Section 1: Scatter plots

2.1 (a) Time spent studying is explanatory; the grade is the response variable. (b) Explore the relationship; there is no reason to view one or the other as explanatory. (c) Rainfall is explanatory; crop yield is the response variable. (d) Explore the relationship. (e) The fatherÕs class is explanatory; the sonÕs class is the response variable.

2.2 Height at age six is explanatory, and height at age 16 is the response. Both variables are quantitative.

2.3 (a) The two variables are negatively related; the plot shows a clear curve, with an outlier (one car with high nitrogen oxides). (b) No: High carbon monoxide is associated with low nitrogen oxides, and vice versa.

2.4 (a) City: 11 mpg. Highway: 16 mpg. (b) The plot shows a fairly strong positive linear relationship. We would expect that cars which are fuel efÞcient (or not) in one setting would also be efÞcient (or not) in the other.

2.5 (a) At right. Alcohol from wine should be on the horizontal axis. (b) There is a fairly strong linear relationship. (c) The association is negative: Countries with high wine consumption have fewer heart disease deaths, while low wine consumption tends to go with more deaths from heart disease. This does not prove causation; there may be some other reason for the link.

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2.6 (a) At right. First-round score should be on the horizontal axis; horizontal and vertical scales should be the same. (b) There is a fairly strong positive association; since the scores are those of the same golfers on two rounds, this association is expected. (c) The player with 105 on the Þrst round and 89 on the second lies outside the generally linear pattern. (The extreme point at (102, 107) lies in the pattern, so should not be considered an outlier.) We canÕt tell which round was unusual for the outlying player.

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78 Chapter 2 Looking at Data Ñ Relationships

2.7 (a) At right. Flow rate is explanatory. (b) As the ßow rate increases, the amount of eroded soil in- creases. Yes, the pattern is approximately linear; the association is positive.

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2.8 (a) At right; speed is explanatory. (b) The relationship is curvedÑlow in the middle, higher at the extremes. Since low ÒmileageÓ is actually good (it means that we use less fuel to travel 100 km), this makes sense: moderate speeds yield the best performance. Note that 60 km/hr is about 37 mph. (c) Above-average values of ÒmileageÓ are found with both low and high values of Òspeed.Ó (d) The relationship is very strongÑthere is little scatter around the curve, and it is very useful for prediction.

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u se

d (li

te rs

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Speed (km/hr)

2.9 (a) Franklin is marked with a + (in the lower left corner). (b) There is a mod- erately strong positive linear association. (It turns out that r 2 = 87.0%.) There are no really extreme observations, though Bank 9 did rather well (its point lies slightly above the pattern of the rest), and the Þrst three banks had high values for both variables (but Þt with the over- all pattern). Franklin does not look out of place.

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Solutions 79

2.10 (a) Body mass is the explanatory variable. Women are marked with solid circles, men with open circles. (b) There is a moderately strong, linear, positive association. The amount of scatter appears to increase with body mass. The relationship is basically the same for both genders, but males typically have larger values for both variables.

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2.11 (a) Shown. Fatal cases are marked with solid circles; those who survived are marked with open circles. (b) There is no clear relationship. (c) Generally, those with short incubation periods are more likely to die. (d) Person 6Ñthe 17-year- old with a short incubation (20 hours) who survivedÑmerits extra attention. He or she is also the youngest in the group by far. Among the other survivors, one (person 17) had an incubation period of 28 hours, and the rest had incubation periods of 43 hours or more.

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2.12 The scatterplot shows a weak positive association. Student 55, a male, has an IQ of 103 and a GPA of 0.530. Student 58, a female, has an IQ of 72 with a 7.295 GPA.

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G PA

2.13 (a) x = $177, 330. The distribution is left-skewed, so the Þve-number summary is more appropriate: Min = $113, 000, Q1 = $149, 000, M = $174, 900, Q3 = $188, 000, Max = $327, 500. (b) There is a weak positive relationship. (c) The Þve most expensive houses: The prices are outliers, and their points on the scatterplot lie

80 Chapter 2 Looking at Data Ñ Relationships

above the general pattern.

1 11 1 2222233 1 44444455555 1 6666677777777777 1 888888888999999 2 000 2 2 2 7 2 89 3 1 3 2

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Square feet

2.14 (a) Below, left. A strong relationshipÑa sort of negative association, but ÒangularÓ rather than linear. (The strength of the relationship is somewhat hard to judge because the points are so tightly packed together vertically [for the horizontal row of points] and horizontally [for the vertical column of points].) (b) The other scatterplot shows a reasonably linear negative relationship. (If common logarithms are used instead of natural logs, the plot will look the same, except the vertical and horizontal scales will be different.)

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Seed count

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2.15 (a) Means: 10.65, 10.43, 5.60, and 5.45. (b) There is little difference in the growth when comparing 0 and 1000 nematodes, or 5000 and 10,000 nematodesÑbut the growth drops sub- stantially between 1000 and 5000 nema- todes.

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Solutions 81

2.16 (a) Plot at right. The means are (in the order given) 47.16, 15.6, 31.5, and 14.83. (b) Yellow seems to be the most attractive, and green is second. White and blue boards are poor attractors. (c) Positive or negative association make no sense here because color is a categorical variable (what is an Òabove- averageÓ color?).

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Board color Yellow White Green Blue

2.17 (a) The means (by pecking order) are 1520, 1707, 1540, and 1816 g. These are connected in the scatterplot. (b) Against: Pecking order 1 had the lowest mean weight, while 4 was the heaviest on the average.

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Section 2: Correlation

2.18 x = 58.2 cm and sx .= 13.20 cm (for the femur measurements); y = 66 cm and sy .= 15.89 cm (for the humerus). The standardized values are at the right; the correlation is r = 3.97659/4 = 0.994.

zx zy zx zy −1.53048 −1.57329 2.40789 −0.16669 −0.18880 0.03147 0.06061 0.25173 0.01526 0.43944 0.37759 0.16593 1.19711 1.13277 1.35605

3.97659

2.19 (a) See the solution to Exercise 2.10 for the plot. It appears that the correlation for men will be slightly smaller, since the menÕs points are more scattered. (b) Women: r = 0.876. Men: r = 0.592. (c)Women: x = 43.03 kg. Men: x = 53.10 kg. This has no effect on the correlation. (d) The correlations would remain the same.

82 Chapter 2 Looking at Data Ñ Relationships

2.20 (a) Either variable may be on the horizontal axis; both axes should have the same scale. The scatter- plot suggests a positive correlation, not too close to 1. (b) r = 0.565. (c) r would be the same (since it is based only on the standard scores, which are un- changed if we decrease all menÕs heights by 6 inches). The correlation gives no information about who is taller. (d) Changing the units of measurement does not affect standard scores, and so does not change r . (e) r = 1 (this is a perfect straight line).

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2.21 See 2.6 for the scatterplot. r = 0.550; without player 7, r∗ = 0.661. Without player 7, the pattern of the scatterplot is more linear.

2.22 See 2.8 for the scatterplot. r = −0.172Ñit is close to zero because the relationship is a curve rather than a line.

2.23 (a) The solid circles in the plot. (b) The open circles. (c) r = r∗ = 0.253. The correlations are equal, since the scale (units) of x and y does not change standard scores.

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2.24 (a) Shown. (b) With the exception of Northern Ireland (in the upper left corner), there is a moderate positive association. (c) r = 0.224; without Northern Ireland, r∗ = 0.784. Removing Northern Ireland makes the pattern of the scatterplot more linear.

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To ba

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Alcohol spending (£)

2.25 The plot shows a relatively strong negative association, hence r is negative and large (close to−1). r does not describe the curve of the plot, or the different patterns observed within the ACT and SAT states.

2.26 (a) Standard deviations measure variability; we can see that the Equity Income Fund is less variable (ÒvolatileÓ) than the Science & Technology Fund. Put another way, the Equity Income Fund tends to be more consistent. (Note: This does not indicate which gives higher yields.) (b) TheMagellan Fund, with the higher correlation, tends to rise and

Solutions 83

fall with the S&P index. The Small Cap Stock Fund also generally rises and falls with the S&P index, but is not tied as closely (i.e., there are more exceptions to this ÒruleÓ).

2.27 (a) The new speed and fuel consumption (respectively) values are x∗ = x ÷ 1.609 and y∗ = y × 1.609 ÷ 100 ÷ 3.785 .= 0.004251y. (The factor of 1/100 is needed since we were measuring fuel consumption in liters/100 km.) The transformed data has the same correlation as the originalÑ r = −0.172Ñsince a linear transforma- tion does not alter the correlation. The scatterplot of the transformed data is not shown here; it resembles (except for scale) the plot of 2.8. (b) The new correlation is r∗ = −0.043; the new plot is even less linear than the Þrst.

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2.28 See 2.14 for the scatterplots. For the original data, r = −0.470, reßecting the negative association, as well as the marked nonlinearity of the scatterplot. After taking logarithms, r∗ = −0.929; the plot of the transformed data is much more linear.

2.29 The plot shows a weak positive asso- ciation; it is fairly linear. The correlation is r = 0.542; there is some tendency for GPAs and self-concept scores to be high (or low) together.

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Self-concept score

2.30 If the husbandÕs age is y and the wifeÕs x , the linear relationship y = x + 2 would hold, and hence r = 1.

2.31 The person who wrote the article interpreted a correlation close to 0 as if it were a correlation close to−1 (implying a negative association between teaching ability and re- search productivity). Professor McDanielÕs Þndings mean there is little linear association between research and teachingÑfor example, knowing a professor is a good researcher gives little information about whether she is a good or bad teacher.

2.32 (a) Since sex has a nominal scale, we cannot compute the correlation between sex and anything. [There is a strong association between sex and income. Some writers use ÒcorrelationÓ as a synonym for Òassociation.Ó It is much better to retain the more

84 Chapter 2 Looking at Data Ñ Relationships

speciÞc meaning.] (b) A correlation r = 1.09 is impossible, since −1 ≤ r ≤ 1 always. (c) Correlation has no units, so r = 0.23 bushels is incorrect.

Section 3: Least-Squares Regression

2.33 (a) Below, left. The range of values on the horizontal axis may vary. (b) When x = 20, y = 2500 dollars. (c) y = 500 + 200x . (The slope is his rate of savings, in dollars per year).

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2.34 y = 1500x . It might be worthwhile to point out that this is simply the familiar formula Òdistance equals velocity times time,Ó and that (meters/second) times (seconds) equals meters.

2.35 (a) Weight y = 100 + 40x g; the slope is 40 g/week. (b) Above, right. (c) When x = 104, y = 4260 grams, or about 9.4 poundsÑa rather frightening prospect. The regression line is only reliable for ÒyoungÓ rats; like humans, rats do not grow at a constant rate throughout their entire life.

2.36 Plot below, left. For analog service, the monthly bill is y1 = 19.99 + 0.85x . For digital service, the monthly bill is y2 = 24.99 + 0.60x . Digital service is cheaper for both 30 minutes ($42.99 vs. $45.49) and one hour ($60.99 vs. $70.99). In fact, digital service is cheaper for anything over 20 minutes.

Solutions 85

For 2.36.

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2.37 (a) Above, right. (b) The initial pH was 5.4247; the Þnal pH was 4.6350. (c) The slope is −0.0053; the pH decreased by 0.0053 units per week (on the average).

2.38 (a) Ideally, the scales should be the same on both axes. (b) For every additional unit of strength after 7 days, the concrete has an additional 0.96 units of strength after 28 days. (c) y = 1389 + (0.96)(3300) = 4557 pounds per square inch.

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2.39 (a) The plot shows a moderately strong positive linear relationship. (b) r = 0.941; about r 2 = 88.6% of variation in manatee deaths is explained by powerboat registrations, so predictions are reasonably accurate. (c) ŷ .= −41.4 + 0.125x ; when x = 716, y .= 48 dead manatees are predicted. (d) When x = 2000, y .= 208; extrapolation (in number of boats, as well as time) makes this prediction unreliable. (e) The additional points are shown as open circles. Two of the points (those for 1992 and 1993) lie below the overall pattern (i.e., there were fewer actual manatee deaths than we might expect), but otherwise there is no strong indication that the measures succeeded. (f) The mean for those years was 42Ñless than our predicted mean of 48 (which might suggest that the measures taken showed some results).

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Boats (thousands)

86 Chapter 2 Looking at Data Ñ Relationships

2.40 (a) At right. (b) ŷ .= 123+20.2x . For each additional degree-day per day, gas consumption increases by about 20.2 ft3

per day. (c) We predict y .= 931 ft3 of gas/day when x = 40 degree-days/day (carrying out more decimal places in the equation gives ŷ = 932.1 ft3). JoanÕs actual usage (870 ft3) is lower, so the insulation seems to be effective.

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Degree days

2.41 (a) At right. (b) r = 0.507 and r 2 = 0.257 = 25.7%. There is a positive association between U.S. and over- seas returns, but it is not very strong: Knowing the U.S. return accounts for only about 26% of the variation in overseas returns. (c) The regres- sion equation is ŷ = 5.64 + 0.692x . (d) ŷ = 12.6%; the residual (prediction error) is 32.9% − 12.6% = 20.3%. Since the correlation is so low, the predictions will not be very reliable.

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2.42 For degree-days: x = 21.54 and sx = 13.42. For gas consumption: y = 558.9 and sy = 274.4. The correlation is r = 0.989. The slope is therefore b = (0.989)(274.4)/13.42 .= 20.2 and the intercept is a =

558.9− (20.2)(21.54) .= 123 (there may be slight differences due to rounding).

2.43 (a) b = r · sy/sx = 0.16; a = y− bx = 30.2. (b) JulieÕs predicted score is ŷ = 78.2. (c) r 2 = 0.36; only 36% of the variability in y is accounted for by the regression, so the estimate ŷ = 78.2 could be quite different from the real score.

2.44 r = √0.16 = 0.40 (high attendance goes with high grades, so the correlation must be positive).

2.45 The correlation is r = 0.9670, so r 2 = 93.5% of the variation in erosion is explained by the relationship between ßow rate and erosion.

Solutions 87

2.46 WomenÕs heights are the x values; menÕs are the y values. The slope is b = (0.5)(2.7)/2.5 = 0.54 and the intercept is a = 68.5− (0.54)(64.5) = 33.67. The regression equation is ŷ = 33.67 + 0.54x .

Ideally, the scales should be the same on both axes. For a 67-inch tall wife, we predict the husbandÕs height will be about 69.85 inches.

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Wife's height

2.47 (a)Male height on female height: ŷ1 = 24+ 0.6818x . Female height on male height: ŷ2 = 33.66+ 0.4688x . (Note that x and y mean opposite things in these two equations.) The two slopes multiply to give r 2 = 0.3196, since the standard deviations cancel out. Put another way, the slopes are reciprocalsÑexcept for the factor of r attached to each. In general, the two slopes must have the same sign, since r determines whether they are positive or negative. (b) Since regression lines always pass through (x, y), they intersect at (66, 69)Ñthe

Þrst coordinate is the mean female height, while the second is the mean male height. When graphing, remember to plot the female vs. male line Òsideways.Ó That is, choose a value on the vertical axis as ÒxÓ (male height), then compute the corresponding ÒyÓ (female height) and Þnd this location on the horizontal axis. Alternatively, write the second equation as x = 33.66 + 0.4688y (which uses x and y in the same way as the Þrst equation) and solve to get y .= 2.133x − 71.8. (c) Since the slope is a ratio of heights, the conversion from inches to centimeters would

have no effect (the factor of 1/2.54 cancels out in the numerator and denominator).

2.48 Lean body mass: m = 46.74, and sm = 8.28 kg. Metabolic rate: r = 1369.5, and sr = 257.5 cal/day. The correlation is r = 0.865. For predicting metabolic rate from body mass, the slope is b1 = r · sr/sm .= 26.9 cal/day per kg. For predicting body mass from metabolic rate, the slope is b2 = r · sm/sr .= 0.0278 kg per cal/day.

2.49 (a) ŷ = 113 + 26.9x . For every 1 kg increase in lean body mass, the metabolic rises by about 26.9 cal/day. (b) x = 46.74 kg, sx = 8.28 kg; y = 1369.5 cal/day, sy = 257.5 cal/day; r = 0.865 (no units); b = 26.9 cal/day per kg, and a = 113 cal/day. (c) x = 102.83 lb, sx = 18.23 lb; y, sy, r , and a are unchanged; b = 12.2 cal/day per lb; ŷ = 113+ 12.2x .

2.50 The correlation of IQ with GPA is r1 = 0.634; for self-concept and GPA, r2 = 0.542. IQ does a slightly better job; it explains about r 21 = 40.2% of the variation in GPA, while self-concept explains about r 22 = 29.4% of the variation.

88 Chapter 2 Looking at Data Ñ Relationships

Section 4: Cautions about Correlation and Regression

2.51 (a) Below, left. (b) No: The pattern is not linear. (c) The sum is 0.01. The Þrst two and last four residuals are negative, and those in the middle are positive. Plot below, right.

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2.52 (a) The Eat Slim point is set apart from the others, but Þts in reasonably well with the pattern of the rest of the plot. (b) With all observations, ŷ = −91.2 + 3.21x (line 1 in the plot); without Eat Slim, ŷ = 46.9 + 2.40x (line 2). Eat Slim is inßuential; it moves the line quite a bit. (c) Use the second equation: We estimate ŷ .= 407 mg of sodium for a hot dog with 150 cal.

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2.53 (a) At right. Ideally, the scales should be the same on both axes. (b) The Þrst omits the outlier; it lies closer to the pattern of the other points (and farther from the omitted point). ¥

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2.54 (a) Below, left. (b) The regression equation is ŷ = −14.4+ 46.6x . (c) Below, right. The residuals for x = 0.25 and x = 20.0 are almost all positive; all those for the middle two x values are negative.

Solutions 89

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2.55 (a)Below, left. The relationship seems linear. (b)Regression line: ŷ = 1.77+0.0803x (y is stride rate, x is speed). (c) The residuals (reported by Minitab, then rounded to 3 decimal places) are 0.011,−0.001,−0.001,−0.011,−0.009, 0.003, 0.009. These add to 0.001. Results will vary with rounding, and also with the number of decimal places used in the regression equation. (d) Residuals are positive for low and high speeds, negative for moderate speeds; this suggests that a curve (like a parabola) may be a better Þt. We cannot plot residuals vs. time of observation since we do not have that information.

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2.56 (a) Ideally, the scales should be the same on both axes. (b) The regression equation is ŷ = 1.03+0.902x . (c) The predicted venous measurements are 6.44, 11.85, and 17.27 ml/minute; all these are within 10% of the respective microspheres measurements.

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2.57 (a) In 1986, the overseas return was 69.4%Ñmuch higher than would be expected. The residual is 50.9%. The original equation was ŷ = 5.64+ 0.692x ; without this point, it is ŷ = 4.13 + 0.653x . This is not much of a change; the point is not inßuential.

90 Chapter 2 Looking at Data Ñ Relationships

(b) There is no obvious pattern to the residual plot (below). (The residuals shown are for the regression with all the points.)

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2.58 (a) The plot (below, left) suggests a weak positive association between price and consumptionÑthe opposite of what we expected. (b) The regression equation is ŷ = 44.9+ 9.50x ; regression explains r 2 = 35.8% of the variation in consumption. (c) The residual plot vs. time (below, right) shows a pattern of rising and falling, rather than the ÒrandomÓ ßuctuations we expect.

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Solutions 91

2.59 (a) Shown, with three Þtted lines on the graph (the Þrst and third are nearly identical). Equations are given above the graph. (b) Sea scallops are relatively expensive; the point for scallops lies far away from the rest of the points, though it does not deviate greatly from the pattern. The Þtted line changes slightly (it becomes less steep) without that point. Lobsters might also be seen as outliers, though they are not as separated from the pack. Note that if we remove both scallops and lobsters, the resulting line is almost the same as the line for all the data. (c) r = 0.967; r 2 = 0.935 = 93.5% of the variation in 1980 prices is explained by 1970 prices. (d) Without scallops, r∗ = 0.940; without scallops and lobsters, r = 0.954. The correlation drops slightly since, in the absence of the outlier(s), the scatter of the data is less, so the scatter about a line is (relatively) greater. (e) Yes: The plot suggests a linear relationship.

All points: ŷ = −1.2+ 2.70x Minus scallops: ŷ = 11.0+ 2.25x

Minus scallops & lobster: ŷ = 0.31+ 2.72x

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1970 price

(maybe)

2.60 (a) To three decimal places, the correlations are all approximately 0.816 (rD actually rounds to 0.817), and the regression lines are all approximately ŷ = 3.000 + 0.500x . For all four sets, we predict ŷ .= 8 when x = 10. (b) Below. (c) For Set A, the use of the regression line seems to be reasonableÑthe data do seem to have a moderate linear association (albeit with a fair amount of scatter). For Set B, there is an obvious nonlinear relationship; we should Þt a parabola or other curve. For Set C, the point (13, 12.74) deviates from the (highly linear) pattern of the other points; if we can exclude it, regression would be would very useful for prediction. For Set D, the data point with x = 19 is a very inßuential pointÑthe other points alone give no indication of slope for the line. Seeing how widely scattered the y-coordinates of the other points are, we cannot place too much faith in the y-coordinate of the inßuential point; thus we cannot depend on the slope of the line, and so we cannot depend on the estimate when x = 10.

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2.61 (a) The regression line is ŷ = 11.06−0.01466x , but the plot does not suggest a linear relationship. Moral: Check to see if a line is an appropriate model for the data. (b) The regression line is ŷ = 2.72+ 0.166x , but there is an inßuential point: Northern Ireland,

92 Chapter 2 Looking at Data Ñ Relationships

which had much higher tobacco expenditures than one would suspect from its alcohol spending. Moral: Look for outliers and inßuential points.

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2.62 (a) There is some support for decreased mortality with higher volume (the line de- creases from left to right), but the relationship is not very strong, and the wide scatter makes it difÞcult to judge. Hospitals with more cases tend to be less variable in mortality rate; in particular, of hospitals with more than 200 cases, almost none had mortality rates over 0.2. However, part of this decreased scatter may be due to something else: If all hospitals

had the same mortality rate, we would expect more variation in deaths among hospitals with fewer cases, for the same reason that a mean or proportion from a small sample has more variation than the same quantity from a large sample. (For example, note that a hospital with only one case would have a mortality rate of either 0 or 100%!) (b) Above 150Ð200 cases, there does not seem to be strong evidence of a difference,

but below that, the wide scatter at least suggests that some hospitals are better than others. For this reason, it does seem advisable to avoid hospitals with fewer than 150 cases.

2.63 (a) Table below, plot at right. Min Q1 M Q3 Max

U.S. −26.4% 5.1% 17.5% 23.6% 37.6% Overseas −23.2% 2.5% 12.0% 29.6% 69.4%

(b) Either answer is defensible: One-fourth of the time, overseas stocks did better than 29.6% (vs. 23.6% for U.S. stocks). On the other hand, half the time, U.S. stocks returned 17.5% or more (vs. 12% for overseas stocks). (c) Overseas stocks are more volatileÑQ3 − Q1 = 27.1% for overseas stocks, about 50% larger than the U.S. IQR of 18.5%. Also, the boxplot shows a lot more spread, and the low U.S. return (−26.4%) is an outlier; not so with the overseas stocks.

US Overseas -30 -20 -10

0 10 20 30 40 50 60 70

R et

ur n

(% )

Solutions 93

2.64 (a) The plot (right) shows a strong positive linear relationship; ideally, the scales should be the same on both axes. Only one observationÑ(51, 69)Ñ deviates from the pattern slightly. The regression line ŷ = −2.580 + 1.0935x explains r 2 = 95.1% of the variation in the data. (b) Plots below. There is no striking relationship with x (rural reading); there may be an increasing spread over time. The large positive residual stands out. (c) The point (108, 123) is a potentially inßuential observation (although it does not seem to deviate from the pattern of the other points). It has the second-highest residual. (d) When x = 88, we predict ŷ = 93.65. (e) The quantile plot (below) shows that the residuals are right-skewed.

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2.65 (a) Right-hand points are Þlled circles; left-hand points are open circles. (b) The right-hand points lie below the left-hand points. (This means the right-hand times are shorter, so the subject is right-handed.) There is no striking pattern for the left-hand points; the pattern for right-hand points is obscured since they are squeezed at the bottom of the plot. (c) Right hand: ŷ = 99.4 + 0.0283x (r = 0.305, r 2 = 9.3%). Left hand: ŷ = 172 + 0.262x (r = 0.318, r 2 = 10.1%). The left-hand regression is slightly better, but neither is very good: distance accounts for only 9.3% (right) and 10.1% (left) of the variation in time. (d) Neither plot shows a systematic pattern. (Plots not shown.)

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94 Chapter 2 Looking at Data Ñ Relationships

2.66 The plot shown is a very simpliÞed (and not very realistic) exampleÑÞlled circles are economists in business; open circles are teaching economists. The plot should show positive association when either set of circles is viewed separately, and should show a large number of bachelorÕs degree economists in business and graduate degree economists in academia.

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ua l i

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10 00

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Years of post-HS education

2.67 r = 0.999. With individual runners, the correlation would be smaller (closer to 0), since using data from individual runners would increase the ÒscatterÓ on the scatterplot, thus decreasing the strength of the relationship.

2.68 (a) There is clearly higher scatter for higher predicted values; the regression more accurately predicts low salaries than high salaries. (b) The residual plot is curved, similar to Figure 2.19(b). Salaries are typically overestimated for players who are new to the majors, and for those who have been in the majors for 15 or more years (these resid- uals are mostly negative). Those in for eight years will generally have their salaries underestimated; these residuals are mostly positive.

Section 5: An Application: Exponential Growth and World Oil Production

2.69 1 hour (four 15-minute periods): 24 = 16. 5 hours (20 15-minute periods): 220 = 1, 048, 576.

2.70 (a) At right. (b) Below, left (the curve is y = 2x−1). (c) The 64th square should have 263 .= 9.22 × 1018 grains of rice. (d) Below, right. Logarithms given in the table. (e) y .= −0.30103 + 0.30103x (number of decimals in the answer may vary). This predicts y .= 18.86 for the logarithm of the number of grains on the 64th squareÑthe same as log(263).

Square Grains Logarithm 1 1 0 2 2 0.30103 3 4 0.60206 4 8 0.90309 5 16 1.20412 6 32 1.50515 7 64 1.80618 8 128 2.10721 9 256 2.40824 10 512 2.70927

Solutions 95

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2.71 (a) At right. The bond value after x years is $500(1.075)x ; all are rounded to 2 decimal places. (b) Below, left (the curve is y = 500(1.075)x ). (c) Below, right. Logarithms given in the table.

Years Bond Value Logarithm 1 $537.50 2.73038 2 577.81 2.76179 3 621.15 2.79320 4 667.73 2.82460 5 717.81 2.85601 6 771.65 2.88742 7 829.52 2.91883 8 891.74 2.95024 9 958.62 2.98165 10 1030.52 3.01305

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2.72 FredÕs balance after x years is $500+ $100x ; AliceÕs balance is $500(1.075)x . After 25 years, Alice has more money: $3049.17 vs. $3000.00.

2.73 (a) If the investment was made at the beginning of 1970: 1000(1.1134)26 .= $16, 327.95. If the investment was made at the end of 1970: 1000(1.1134)25 .= $14, 664.94. (b) 1000(1.0562)25 = $3, 923.32.

2.74 (a) Below, left. (b) The ratios are 3.6, 907,075226,260 .= 4.0, and 2,826,095907,075

.= 3.1. (c) Below, right. (d) The regression equation is ŷ = −1095 + 0.556x (or −38.5 + 0.556x , if we code the years as 78, 79, etc.). The predicted value of y is 7.0302, which means we predict about 107.03 .= 10.7 million acres defoliated.

96 Chapter 2 Looking at Data Ñ Relationships

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2.75 (a) There is a slight curve to the graph (below, left), suggesting exponential growth. (b) After taking logarithms, regression gives ŷ = −98.753 + 0.052501x (plot below, right). Growth was faster from about 1965 to 1985, when the points in the scatterplot rise faster than the line. (c) log 495, 710 = 5.695, which is less than ŷ .= 5.829, so the actual spending was less than predicted. Note: The problem asks students to predict 1992 spending, but comparing the loga-

rithms may be easier. For students who can follow the switch from Òlog(Spending)Ó back to ÒSpending,Ó we can observe the following: For 1992, we estimate log(Spending) = ŷ .= 5.829, so Spending .= 105.829 .= $674, 528. These answers are very sensitive to rounding; using full accuracy from software, the

predicted value is $673,585. In any case, the actual value is considerably less than the prediction.

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2.76 (a) Below, left. (b) Below, right. (c) The plot of log(population) is not linear, so growth was not exponentialÑor at least the rate of growth was not constant. The plot seems to be made up of two linear pieces, one for 1400Ð1750, the other (steeper) line from 1750Ð1950. The population grew more quickly after 1750.

Solutions 97

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2.77 (a)Below, left. (b)Below, right. The logarithm plot has a greater slopeÑrepresenting a faster growth rateÑup to 1880 than after 1880. (c) ŷ = −15.3815+0.0090210x . When x = 1997, ŷ = 2.633, which corresponds to a population of 429.5 million. (Computation with ÒexactÓ values gives 429.9 million.) The actual value is much smaller than the predicted value.

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2.78 (a) Below. (b) Growth from 1950Ð1980 was more exponential than linear (since the log plot is fairly linear over that range). (c) The points on the log plot from 1980Ð1995 appear to lie on a straight line (with a lower slope), suggesting exponential growth at a lower rate. (On the other hand, the points for 1980Ð1995 on the Þrst plot also seem to lie on a straight line, suggesting linear growth over that period. It is hard to spot minor deviations from linearity with only four points to consider.) (d) Vehicle registrations dropped during World War II.

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98 Chapter 2 Looking at Data Ñ Relationships

2.79 For 1945, log y = 1.60865, so ŷ = 101.60865 = 40.61 million vehicles. For 1995, log y = 2.43715, so ŷ = 102.43715 = 273.6 million vehicles.

2.80 Plots below. The log plot suggests exponential growth at one rate up to 1965, then at a slightly lower rate. The productivity index for 1985 is higher than the value suggested by the overall pattern.

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Section 6: Relations in Categorical Data

The answers for 2.81 through 2.85 are summarized in the following table, based on Table 2.15 of the text. The second entry in each cell is the column percent = cell entry/column total (e.g., 1.8% .= 362017 ), and the third entry is the row percent = cell entry/row total (e.g., 14.6% .= 36246 ). [Thus, except for round-off error, the second entries add to 100% down the columns; the third entries add to 100% across the rows.]

2-year 2-year 4-year 4-year Age full-time part-time full-time part-time

Under 18 36 98 75 37 246

1.8% 2.8% 1.2% 1.4% 1.8% 14.6% 39.8% 30.5% 15.0%

18Ð21 1126 711 3270 223 5330 55.8% 20.5% 54.5% 8.7% 37.9% 21.1% 13.3% 61.4% 4.2%

22Ð34 634 1575 2267 1380 5856

31.4% 45.3% 37.8% 54.0% 41.7% 10.8% 26.9% 38.7% 23.6%

35 and up 221 1092 390 915 2618

11.0% 31.4% 6.5% 35.8% 18.6% 8.4% 41.7% 14.9% 35.0% 2017 3476 6002 2555 14,050 Ñ Ñ Ñ Ñ

14.4% 24.7% 42.7% 18.2%

Solutions 99

2.81 (a) Adding across the bottom (total) row: 14,050 thousand, or 14,050,000. (b) At the right end of the second row of the table: 5330 14,050

.= 37.9%. (c) Reading across the second entry of the second row above: 11262017

.= 55.8%, 7113476

.= 20.5%, 32706002 .= 54.5%, 2232555

.= 8.7%. (d) 18- to 21-year-olds constitute the majority of full-time students at both 2- and 4- year institutions, but make up much a smaller proportion of part-time students.

2-year full-time

2-year part-time

4-year full-time

4-year part-time

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2.82 (a) There are 3476 thousand 2-year part-time students; 45.3% .= 15753476 are 22 to 34 years old. (b) There are 5856 thousand 22- to-34-year-old students; 26.9% .= 15755856 are enrolled part-time at 2-year colleges.

2.83 (a) These are in the right-hand ÒmarginÓ of the table: Adding across the rows, we Þnd 246 (thousand), 5330, 5856, and 2618, respectively. Dividing by 14,050 gives 1.8%, 37.9%, 41.7%, and 18.6%. (b) From the Ò2- year part-timeÓ column, we divide 98, 711, 1575, and 1092 by 3476 to get 2.8%, 20.5%, 45.3%, and 31.4%. (c) Two-year part-time students are more likely to be older (over 22, and even moreso over 35) than undergraduates in general. They are also slightly more likely to be under 18, and considerably less likely to be 18 to 21.

Under 18 18 to 21 22 to 34 35 and up 0

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2.84 For older students, take 221, 1092, 390, and 915, and divide by 2618 to get 8.4%, 41.7%, 14.9%, and 35.0%. We might then compare these with the same percentages for the whole population: From the bottom row of the table, divide 2017, 3476, 6002, and 2555 by 14,050 to get 14.4%, 24.7%, 42.7%, and 18.2%. From these percentages, and the bar chart at

right, we can see that older students are more likely to be part-time than students in general.

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100 Chapter 2 Looking at Data Ñ Relationships

2.85 For these distributions, divide the numbers in each cell by the total at the bottom of each column. (The columns in this table add to 100%.) The biggest difference is that full-time stu- dents tend to be youngerÑboth 2- and 4-year full-time students have similar age distributions. Meanwhile, part-time students have similar distributions at both types of institutions, and are pre- dominantly over 21, with about one-third over 35.

2-year 2-year 4-year 4-year full-time part-time full-time part-time

<18 1.8% 2.8% 1.2% 1.4% 18Ð21 55.8% 20.5% 54.5% 8.7% 22Ð34 31.4% 45.3% 37.8% 54.0% 35+ 11.0% 31.4% 6.5% 35.8%

Under 18 18 to 21 22 to 34 35 and up 0

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4-yr PT

2.86 Two examples are shown at right. In general, any number from 10 to 50 can be put in the upper left corner, and then all the other entries can be determined.

30 20 50 0 30 20 10 40

2.87 (a) 75+119+160600 = 59% did not respond. (b) 75200 = 37.5% of small businesses, 119 200 = 59.5% of medium-sized businesses, and 160200 = 80% of large businesses did not respond. Generally, the larger the business, the less likely it is to respond. (c) At right. (d) Small: 125246

.= 50.8%. Medium: 81246

.= 32.9%. Large: 40246 .=

16.3%. (e) No: Over half of respondents were small businesses, while less than 1/6 of responses come from large businesses.

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2.88 (a) Use column percents, e.g., 68225 .= 30.2% of females are in administration, etc. See

table and graph below. The biggest difference between women and men is in Administra- tion: a higher percentage of women chose this major. Meanwhile, a greater proportion of men chose other Þelds, especially Finance. (b) There were 386 responses; 336722

.= 46.5% did not respond.

Solutions 101

Female Male Overall Accting. 30.2% 34.8% 32.1% Admin. 40.4% 24.8% 33.9% Econ. 2.2% 3.7% 2.8% Fin. 27.1% 36.6% 31.1%

Accounting Administration Economics Finance 0 5

10 15 20 25 30 35 40

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Male

Overall

2.89 58.3% of desipramine users did not have a relapse, while 25.0% of lithium users and 16.7% of those who received placebos succeeded in breaking their addictions. Desipramine seems to be effective.

Desipramine Lithium Placebo 0

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ith ou

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2.90 Compute column percents, e.g., 61,941 355,265

.= 17.4% of those U.S. degrees considered in this table are in engineering, etc. See table and graph at right. We observe that there are considerably more social science degrees, and fewer engineering degrees, in the U.S. The Western Europe and Asia distributions are similar.

United Western Field States Europe Asia Overall Eng. 17.4% 38.3% 37.0% 32.8% Nat. sci. 31.3% 33.7% 32.0% 32.3% Soc. sci. 51.3% 28.0% 31.1% 34.9%

US Western Europe

Asia Overall 0

Pe rc

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2.91 (a) The sum is 58,929; the difference is due to roundoff error. (b)Divide each column total by 99,588 to obtain the percents in the bottom margin of the table below. Bar graph below, left. (c) For 18 to 24 years, divide all numbers in that row by 12,613; the percentages are on the third line of the top row of the table below (73.6%, 24.1%, 0.2%, 2.1%). For 40 to 64 years, divide by 36,713 to obtain the third line of the third row of the table (6.3%, 72.7%, 6.0%, 15.0%). Among the younger women, almost three-fourths have not yet married, and those who are married have had little time to become widowed or divorced. Most of the older group are or have been marriedÑonly about 6% are still single. (d) 48.1% of never-married women are 18Ð24, 36.0% are 25Ð39, 11.9% are 40Ð

102 Chapter 2 Looking at Data Ñ Relationships

64, and 4.0% are 65 or older. The bar chart is below, right; see also the Þrst column of the table. The target ages should be under 39.

Never married

Married Widowed Divorced 0

Pe rc

en t o

f a ll

w om

18Ð24 25Ð39 40Ð64 ³ 65 0

Pe rc

en t o

f s in

gl e

w om

Never Age Married Married Widowed Divorced

18Ð24 9,289 3,046 19 260 12,613 48.1% 5.2% 0.2% 2.5% 12.7% 73.6% 24.1% 0.2% 2.1%

25Ð39 6,948 21,437 206 3,408 32,000 36.0% 36.4% 1.9% 33.2% 32.1% 21.7% 67.0% 0.6% 10.7%

40Ð64 2,307 26,679 2,219 5,508 36,713 11.9% 45.3% 20.0% 53.6% 36.9% 6.3% 72.7% 6.0% 15.0%

≥ 65 768 7,767 8,636 1,091 18,264 4.0% 13.2% 77.9% 10.6% 18.3% 4.2% 42.5% 47.3% 6.0% 19,312 58,931 11,080 10,266 99,588

Ñ Ñ Ñ Ñ 19.4% 59.2% 11.1% 10.3%

2.92 Percents and bar graph below; for example, 64.5% .= 16,38125,415 . Both genders use Þrearms more than any other method, but they are considerably more common with men (64.5% of male suicides, but only 42.0% of female suicides, used Þrearms). Women are more likely to use poison (34.6% vs. 14.0% for men).

Male Female Firearms 64.5% 42.0% Poison 14.0% 34.6% Hanging 15.0% 13.2% Other 6.5% 10.2%

Firearms Poison Hanging Other 0

Pe rc

en t o

f s ui

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Men

Women

Solutions 103

2.93 (a) At right. (b) 490700 = 70% of male applicants are admitted, while only 280500 = 56% of females are admitted. (c) 80% of male business school applicants are admitted, compared with 90% of females; in the law school, 10% of males are admitted, compared with 33.3% of females. (d) A majority (6/7) of male applicants apply to the business school, which admits 83% of all applicants. Meanwhile, a majority (3/5) of women apply to the law school, which only admits 27.5% of its applicants.

Admit Deny Male 490 210 Female 280 220

2.94 (a) Alaska Airlines: 5013274+501 .= 13.3%.

America West: 7876438+787 .= 10.9%. (b) See

the table at the right. (c) Both airlines do best at Phoenix, where America West has 72.7% of its ßights, and Alaska Airlines has only 6.2% of its ßights. Seattle is the worst city for both; Alaska West has 56.8% of its ßights there, compared with 3.6% for America West. The large percentage of ÒgoodÓ (Phoenix) ßights for America West, and the large percentage of ÒbadÓ (Seattle) ßights for Alaska Airlines, makes America West look better.

Percent Delayed Alaska America Airlines West

Los Angeles 11.1% 14.4% Phoenix 5.2 7.9 San Diego 8.6 14.5 San Francisco 16.9 28.7 Seattle 14.2 23.3

2.95 Examples will vary, of course; here is one very simplistic possibility (the two-way table is at the right; the three-way table is below). The key is to be sure that the three-way table has a lower percentage of overweight people among the smokers than among the nonsmokers.

Early Death Yes No

Overweight 4 6 Not overweight 5 5

Smoker Early Death Nonsmoker Early Death Yes No Yes No

Overweight 1 0 Overweight 3 6 Not overweight 4 2 Not overweight 1 3

2.96 (a) At right. (b) Overall, 11.9% of white defendants and 10.2% of black defendants get the death penalty. However, for white victims, the percentages are 12.6% and 17.5% (respec- tively); when the victim is black, they are 0% and 5.8%. (c) In cases involving white victims, 14% of defendants got the death penalty; when the victim was black, only 5.4% of defendants were sentenced to death. White defendants killed whites 94.3% of the timeÑbut are less likely to get the death penalty than blacks who killed whites.

Death Penalty? Yes No

White defendant 19 141 Black defendant 17 149

104 Chapter 2 Looking at Data Ñ Relationships

Section 7: The Question of Causation

2.97 Both reading ability and shoe size tend to increase with ageÑthe lurking variable z. Diagram below.

For 2.97.

Age

S hoe

size

R eading

ability

For 2.98.

F lu

cases

W eather

Ice cream

demand

2.98 Flu tends to increase, and ice cream sales to decrease, during the winter months. Common response to weather. Diagram above.

2.99 No: The high death rate for C may occur because C is the anesthetic of choice in serious operations or for patients in poor condition.

2.100 The diagram below illustrates the confounding between exposure to chemicals and standing up.

For 2.100.

E xposure to

chemica ls

T ime

standing up

M iscarriages?

For 2.101.

H ospita l

size

Length

of stay

S eriousness

of illness

2.101 Patients suffering frommore serious illnesses are more likely to go to larger hospitals (which may have more or better facilities) for treatment. They are also likely to require more time to recuperate afterwards.

2.102 Spending more time watching TV means that less time is spent on other activities; these may suggest lurking variables. For example, perhaps the parents of heavy TV watchers do not spend as much time at home as other parents. Also, heavy TV watchers would typically not get as much exercise.

2.103 In this case, there may be a causative effect, but in the direction opposite to the one suggested: people who are overweight are more likely to be on diets, and so choose

Solutions 105

artiÞcial sweeteners over sugar. [Also, heavier persons are at a higher risk to develop diabetes; if they do, they are likely to switch to artiÞcial sweeteners.]

2.104 The explanatory and response variableswere Òconsumption of herbal teaÓ and Òcheer- fulness.Ó The most important lurking variable is social interactionÑmany of the nursing home residents may have been lonely before the students started visiting.

2.105 The explanatory variable is whether or not a student has taken at least two years of foreign language, and the score on the test is the response. The lurking variable is the studentsÕ English skills before taking (or not taking) the foreign language: Students who have a good command of English early in their high school career are more likely to choose (or be advised to choose) to take a foreign language.

2.106 We might want to know, for example, information about proximity to power lines, tracking in our study some children who do not live near power lines or other electromag- netic Þeld sources. It may also be useful to know family history for those who develop leukemia.

2.107 We need information on the type of surgery, and on the age, sex, and condition of the patient.

Exercises

2.108 (a) Correlation measures the strength and direction of the linear association between actual and recalled consumption; it will be high (close to 1) if there is a good match be- tween actual and recalled consumption. (ÒA good matchÓ does not necessarily mean that actual and recalled consumption are nearly the same; it only means that the subjects re- member high consumption for high-quantity foods and low consumption for low-quantity foods.) The second aim of the study was to make predictions about actual consumption, so regression is the appropriate tool. (b) A correlation of 0.217 indicates a rather weak association. This might mean, for example, that among subjects who remembered eating a lot of beef, some really did eat a lot of beef, but others ate average or below-average quantities. (c) The value of r 2 is the fraction of variation in age-30 food intake accounted for by predicting with each of the other two variables (recalled intake and current intake). The higher r 2 is, the more reliable the prediction.

2.109 (a) Yes: The two lines appear to Þt the data well. There do not appear to be any outliers or inßuential points. (b)Compare the slopes: beforeÑ0.189; afterÑ0.157. (The units for these slopes are 100 ft3 per degree-day/day; for students who are comfortable with units, 18.9 ft3 vs. 15.7 ft3 would be a better answer.) (c) Before: ŷ = 1.089 + 0.189(35) = 7.704. After: ŷ = 0.853 + 0.157(35) = 6.348. This amounts to an additional ($0.75)(7.704− 6.348) = $1.017 per day, or $31.53 for the month.

106 Chapter 2 Looking at Data Ñ Relationships

2.110 (a) b = r · sy/sx .= 1.1694; a = y − bx .= 0.3531. The regression equation is ŷ = 0.3531 + 1.1694x ; it explains r 2 .= 27.6% of the volatility in Philip Morris stock. (b) On the average, for every percentage-point rise in the S&P monthly return, Philip Morris stock returns rise about 1.17 percentage points. (And similarly, Philip Morris returns fall 1.17% for each 1% drop in the S&P index return.) (c) When the market is rising, the investor would like to earn money faster than the prevailing rate, and so prefers beta > 1. When the market falls, returns on stocks with beta < 1 will drop more slowly than the prevailing rate.

2.111 (a) Explanatory: weeds per meter (wpm). Response: corn yield. (b) The stemplots (below) give some evidence that yield decreases when there are more lambÕs-quarter plants. (c) Scatterplot below. The regression equation is ŷ = 166 − 1.10x . Each additional lambÕs-quarter per meter decreases yield by about 1.1 bushels/acre. (d) ŷ = 166− 1.10(6) = 159.4 bushels/acre.

0 wpm 14 14 15 15 16 16 57 17 2 17 7

1 wpm 14 14 15 15 7 16 1 16 67 17 17

3 wpm 14 14 15 3 15 69 16 16 17 17 6

9 wpm 14 2 14 15 15 16 233 16 17 17

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0 1 2 3 4 5 6 7 8 9

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n yi

el d

(b us

he ls

/a cr

Weeds per meter

2.112 (a) Below, left. (b) The regression equation is ŷ = 1.71+0.0795x . (c) Below, right. The points for the residuals, like those of the original data, are split with women above the line (zero), and men below. (Men are taller on the average, so they have longer legs, and therefore longer strides. Thus, they need fewer steps per second to run at a given speed.)

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15 16 17 18 19 20 21 22

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R es

id ua

Speed (feet/second)

Solutions 107

2.113 (a) At right. (b) The plot shows a negative association (longer beams are less strong), with no outliers. (c) The regression equation is ŷ = 488 − 20.7x ; it is not a good match because the scatterplot does not suggest a straight line. (d) Length 5 to 9 inches: ŷ = 668 − 46.9x . Length 9 to 14 inches: ŷ = 283 − 3.37x . These two lines together describe the data fairly well. One might ask why strength at Þrst decreases so rapidly with increasing length, then almost levels off.

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4 6 8 10 12 14

St re

ng th

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in 2 )

Length (inches)

2.114 (a) The Þrst two graphs below. (b) From 0 to 6 hours (the Þrst phase), the growth is ÒßatÓÑthe colony size does not change much. From 6 to 24 hours, the log(mean colony size) plot looks like a positively sloped line, suggesting a period of exponential growth. At some point between 24 and 36 hours, the growth rate drops off (the 36-hour point is considerably below the linear pattern of the 6- to-24-hour points). (c) The regression equation is ŷ = −0.594+ 0.0851x ; the prediction for x = 10 hours is ŷ .= 0.257, so the predicted mean colony size is about 100.257 = 1.81. (d) For hours 6Ð24 and log(mean colony size), r = 0.9915. For time and log(individual colony size), r∗ = 9846. This is smaller because individual measurements have more scatter (see scatterplot below, right); the points do not cluster as tightly around a line.

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120

0 6 12 18 24 30 36

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2.115 (a) 1120 = 55.0%, 6891 .= 74.7%, and

3 8 = 37.5%. Some (but not too much) time spent in extracurricular activities seems to be beneÞcial. (b) No: There may be a lurking variable that affects bothÑe.g., a personality trait that ÒcausesÓ students to do well, and also to participate in extracurricular activities in moderation.

< 2 hours 2 to 12 hours 12+ hours 0

Pe rc

en t w

ith A

, B , o

r C

Activity hours/week

108 Chapter 2 Looking at Data Ñ Relationships

2.116 The upper part of the table shown gives the percentages of all homicides and all sui- cides committed with each type of Þrearm. This table supports the hypothesis that long guns are used more often for suicides than for homicides: We observe that handguns accounted for about 89% of homicides but only about 71% of suicides. It is also possible to compute the percent-

age of all handgun deaths which were homicides, etc. (the lower part of the table) and observe that considerably higher percentages of shotgun and riße deaths were suicides. Either of these sets of numbers may be used to construct a bar chart (not shown

here). Note that the evidence in these tables says nothing about the accuracy of our explanation of this difference. Also, students may misinterpret the hypothesis as saying that we expect to see long guns used more often than handguns for suicides. Be sure they answer the right question!

Homicides Suicides Handgun 89.3% 70.9% Shotgun 5.3% 12.6% Riße 2.9% 13.7% Not speciÞed 2.5% 2.9%

Handgun 79.1% 20.9% Shotgun 56.0% 44.0% Riße 38.5% 61.5% Not speciÞed 72.2% 27.8%

2.117 Some departments pay higher salaries than others; if women are concentrated in the lower-paying disciplines, their overall median salary will be lower than that of men even if all salaries in each department are identical.

2.118 Number of ÞreÞghters and amount of damage are common responses to the serious- ness of the Þre.

2.119 (a) 68,838109,672 .= 62.8%, and similarly we

get 61.9%, 60.9%, 55.2%, 53.5%, 52.8%, 53.3%, 50.3%, 55.1%, and 49.1%. There is a fairly steady decline in participation, with a noticeably large drop after the 1960s. (b) More college students became eligible to vote after 1970, and that group may be more likely to miss an election, either because of apathy or because they are away from home when elections occur.

1956 1964 1972 1980 1988 1996

Pe rc

en t v

ot in

Year

Solutions 109

2.120 (a) At right. (b) HC and CO are positively associated; NOX is negatively associated with both HC and CO. (c) The HC/CO plot has no particular outliers; the two or three points in the upper right corners are nicely in line with the pattern of the rest of the plot. In the HC/NOX plot, four points (possibly more) lie above the line suggested by the rest of the points. In the NOX/CO plot, three points deviate from the overall pattern. (Note: All these an- swers may vary. Some students may consider more points to be outliers; some might circle fewer points.) (d) rHC,CO = 0.9008, rHC,NOX = −0.5588, and rNOX,CO = −0.6851. Without the outliers circled, r∗

HC,NOX = −0.6418 and

r∗ NOX,CO

= −0.7406. These answers will vary with what students considered to be outliers in (c). (e) The regression equations are HC = 0.322+ 0.0288 CO HC = 0.810− 0.194 NOX NOX = 1.83− 0.0631 CO

(f) Without engines 11, 22, 24, and 32: HC = 0.774− 0.191 NOX

Without engines 22, 24, and 32: NOX = 1.85− 0.0724 CO

(g) The best relationship for prediction is HC/CO; the other two relationships are less linear and not as strong (the correlations are smaller, even after omitting out- liers). The NOX/CO relationship might be good for prediction using a nonlinear function, if we omit engine 32.

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110 Chapter 2 Looking at Data Ñ Relationships

2.121 (a) Both distributions are skewed to the right. Five-number summaries at right; stemplots below. There are no outliers for teachersÕ pay. Spending over $6.52+1.5($6.52−$4.65) = $8.875 thousand per student qualiÞes as an outlier; these states are New York, New Jersey, and Alaska (which also had high [non- outlier] pay values). (b) There is a moderate positive association. This makes sense since money spent for teacher salaries is part of the education budget; more money spent per pupil would typically translate to more money spent overall. (c) Regression equation: ŷ = 14.1 + 3.53x . For each additional $1000 spent per student, teacher salaries increase by about $3,530. Regression of pay on spending explains about 62.9% of the variation in spending. (d) The residuals for the three states are small (their points are close to the line). Without those states, the regression line is ŷ = 12.5 + 3.82x , which has a slightly greater slope than beforeÑso the three points are somewhat inßuential (although we see below that the line does not change much).

Min Q1 M Q3 Max Pay 26.0 30.8 32.6 39.1 50.0 Spending 3.67 4.95 5.66 6.52 9.93

Pay 2 6666 2 8888999 3 000111111 3 222222 3 444455 3 6667 3 8899 4 001 4 23 4 4 4 677 4 8 5 0

Spending 3 6 4 12234 4 56778999 5 1123444 5 556667788 6 00011124 6 579 7 01233 7 8 2 8 5 9 4 9 89

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3 4 5 6 7 8 9 10

Te ac

he rs

' p ay

($ 10

00 )

Spending per pupil ($1000)

2.122 (a) & (b) The Þve-number summaries are

TeachersÕ Pay ($1000) Spending per Pupil ($1000) Min Q1 M Q3 Max Min Q1 M Q3 Max

Coastal 29.0 36.20 40.90 46.10 50.0 4.73 6.16 6.805 8.50 9.93 South 26.5 29.60 31.55 32.60 40.7 4.12 4.50 5.175 5.69 7.17 Midwest 26.0 31.05 35.30 37.25 47.4 4.60 5.04 5.470 5.88 7.00

The boxplots are below. Only teacherÕs salaries in the south and midwest states have outliers: In the south, those above $32.6 + 1.5($3) = $37.1 are Delaware ($39.1) and Maryland ($40.7). In the midwest, Michigan ($47.4) is above $37.25 + 1.5($6.2) = $46.55. (c) The coastal states are clearly higher in both salaries and spending; the midwest is slightly higher than the south in salaries, but not very different in spending per pupil. (d) The residuals for the south are considerably less variable, and more than three- quarters are negative. There is no striking difference between the coastal and midwest residuals.

Solutions 111

Residuals Min Q1 M Q3 Max

Coastal −6.69841 −2.75212 1.14705 5.02827 10.3386 South −5.17373 −1.73138 −0.95191 −0.12244 2.9210 Midwest −5.14927 −2.17094 −0.05066 3.34934 8.8804

Coastal South Midwest 4

Sp en

di ng

p er

p up

il ($

10 00

)

Coastal South Midwest -8 -6 -4 -2 0 2 4 6 8

R es

id ua

Coastal South Midwest 25

Te ac

he rs

' s al

ar ie

s ($

10 00

) MI

MD Del.

2.123 The scatterplot is not very promis- ing. The regression equation is ŷ = 1.28 + 0.00227x ; the correlation is r = 0.252, and the regression explains r 2 = 6.3% of the variation in GPA. By itself, SATM does not give reliable predictions of GPA.

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G PA

SATM

2.124 (a) There is a strong positive linear relationship, as we would expect since the two measurements should be nearly equal. Ideally, the scales should be the same on both axes. (b) r = 0.9965; the process is quite reliable. (c) The regression equation is ŷ = −0.0333 + 1.02x . With x = 1.6298, sx = 0.1694, y = 1.6252, and sy = 0.1730, we compute the same slope from b = r sy/sx .

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1.4

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1.2 1.4 1.6 1.8 2

112 Chapter 2 Looking at Data Ñ Relationships

2.125 (a) T2 vs. T1 at right; residuals below. (b) The distributions of T1 and T2 (stemplots below) do not appear to have any outliers. (c) With the new point, ŷ = −0.017+1.011x ; without it, ŷ = −0.033+1.018x . The two lines are very similar, so the point is not inßuential. (d) With the new point, r = 0.966; without it, r = 0.996.

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T1 12 4 12 13 334 13 8 14 12 14 579 15 11234 15 678899 16 0003334 16 578889 17 02234 17 569 18 013 18 6789 19 0 19 20 2

T2 12 5 13 233 13 7 14 024 14 59 15 011344 15 5667889 16 224 16 5567789 17 00344 17 559 18 113 18 6788 19 01 19 20 4

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2.126 (a) The stemplot of T2 is nearly the same as the one above; just remove the Ò0Ó leaf from the 19 stem, and add a stem 12 and leaf 0 at the top. The stemplot of T1 is at the right; 2.2 is high (it is the leaf 0 at the bottom). (b) The summary measures (and original values) are x = 1.6410 (1.6298), sx = 0.1858 (0.1694), y = 1.6169 (1.6252), sy = 0.1813 (0.1730). These values only change slightly. (c) The new point appears to be, and is, inßuential. (d) r = 0.709, compared to 0.996 without the inßuential pointÑa big change.

12 4 13 3348 14 12579 15 11234678899 16 003334578889 17 02234569 18 0136789 19 0 20 2 21 22 0

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Solutions 113

Chapter 3 Solutions

Section 1: First Steps

3.1 One observation could have many explanations. Were the sunßowers in the sun and the okra in the shade? Were the sunßowers upwind from the okra? Repeated trials in controlled conditions are needed.

3.2 The anecdote describes a single unusual event. We would like data on deaths and injuries for occupants wearing/not wearing restraints for many accidents.

3.3 It is an observational study: information is gathered without imposing any treatment. A voterÕs gender is the explanatory variable, and political party is the response variable.

3.4 (a) This is an experiment: a treatment is imposed. (b) The explanatory variable is the teachingmethod (computer assisted or standard), and the response variable is the increase in reading ability based on the pre- and posttests.

3.5 (a)Which surgery was performed is the explanatory variable, while survival time is the response. (b) This study uses available data; it is not an experiment because the study itself imposes no treatment on the subjects. (c) Any conclusions drawn from this study would have to be viewed with suspicion, because doctors may recommend treatment based on the patientÕs condition. Perhaps some doctors are more likely to suggest one treatment for more advanced cases; those patients would have a poorer prognosis than the patients for whom the doctors suggest the other treatment.

3.6 It was not an experiment, since we observe variables without imposing any treatments. The explanatory variable is whether or not a family had been accepted in public housing, and the response variable is Òfamily stabilityÓ (and Òother variablesÓ).

3.7 This was an experiment; the treatment was walking briskly on the treadmill. The fact that eating was not recorded limits the conclusions that can be drawn. The explanatory variable was time after exercise, and the response variable was the metabolic rate.

3.8 (a) The anesthetic used (the ÒtreatmentÓ) was not imposed, but rather was chosen by the doctors caring for each patient. The nature and seriousness of the illness and the patientÕs overall physical condition may inßuence the choice of anesthetic and also inßuence the death rate. (b) The high death rate for C may occur because C is the anesthetic of choice in serious operations or for patients in poor condition. We should get information on the type of surgery, and on the age, sex, and condition of the patient.

114 Chapter 3 Producing Data

Section 2: Design of Experiments

3.9 Subjects: 300 sickle cell patients. Factor: drug given. Treatments: hydroxyurea and placebo. Response variable: number of pain episodes.

3.10 Experimental units: pairs of pieces of package liner. Factor: temperature of jaws. Treatments: 250◦F, 275◦F, 300◦F, 325◦F. Response variable: peel strength of the seal.

3.11 Subjects: students. Factors: length of commercial, and number of repetitions. Treat- ments: 30 seconds repeated 1, 3, or 5 times, and 90 seconds repeated 1, 3, or 5 times. Response variables: recollection of ad, attitude about camera, and intention to buy cam- era.

3.12 Experimental units: chicks. Factors: corn variety and protein level. Treatments: standard at 12%, 16%, or 20% protein; opaque-2 at 12%, 16%, or 20% protein; and ßoury-2 at 12%, 16%, or 20% protein. Response variables: weight gain.

3.13 (a) Below. (b) A placebo allows researchers to control for the relief subjects might experience due to the psychological effect of taking a drug.

Random Assignment

��* Group 1

150 patients - Treatment 1

Hydroxyurea HHHj

HHHj Group 2 150 patients

- Treatment 2 Placebo

��* Observe

pain episodes

3.14 (a) Measure the blood pressure for all subjects, then randomly select half to get a calcium supplement, with the other half getting a placebo.

Random Assignment

��* Group 1 20 subjects

- Treatment 1 Calcium HHHj

HHHj Group 2 20 subjects

- Treatment 2 Placebo

��* Observe

change in BP

(b) If we assign labels 01 to 40 (down the columns), then choose two digits at a time from line 131, we give calcium to the subjects listed in the table below. (They are chosen in the order given, reading down the columns.) See note on page 50 about using Table B.

05ÐChen 29ÐOÕBrian 31ÐPlochman 02ÐAsihiro 32ÐRodriguez 20ÐImrani 18ÐHoward 36ÐTownsend 19ÐHruska 16ÐGuillen 07ÐCranston 23ÐKrushchev 04ÐBikalis 37ÐTullock 13ÐFratianna 27ÐMarsden 25ÐLiang 39ÐWillis 33ÐRosen 35ÐTompkins

Solutions 115

3.15 (a) The response variable is the company chosen.

Random Assignment

��* Group 1 20 women

- Brochure B with child care HHHj

HHHj Group 2 20 women

- Brochure B with no child care

��*

Observe which company is chosen

(b) If we assign labels 01 to 40 (down the columns), then choose two digits at a time beginning on line 121, we choose the subjects listed in the table below for the child-care brochure. (They are chosen in the order given, reading down the columns.) See note on page 50 about using Table B.

29ÐNg 25ÐLippman 09ÐDanielson 28ÐMorse 07ÐCortez 13ÐGarcia 08ÐCurzakis 18ÐHoward 34ÐSugiwara 38ÐUllmann 27ÐMcNeill 03ÐAÞÞ 22ÐKaplan 15ÐGreen 23ÐKim 01ÐAbrams 10ÐDurr 05ÐCansico 30ÐQuinones 36ÐTravers

3.16 Diagram below. Choose two digits at a time beginning on line 120. Group 1 will be 16, 04, 19, 07, and 10; Group 2 is 13, 15, 05, 09, and 08; Group 3 is 18, 03, 01, 06, and 11. The others are in Group 4. See note on page 50 about using Table B.

Random Assignment

�

Group 1 5 pairs

- Treatment 1 250◦F

J J J Ĵ��

�*

Group 2 5 pairs

- Treatment 2 275◦F HHHj

HHHj Group 3 5 pairs

- Treatment 3 300◦F

�� *

J J J Ĵ Group 4

5 pairs - Treatment 4

325◦F

�

Measure peel strength

3.17 Diagram below. Assign labels 01 to 20, then choose two digits at a time beginning on line 145. Use method A in plots 19, 06, 09, 10, 16, 01, 08, 20, 02, and 07. See note on page 50 about using Table B.

01 Ð A 02 Ð A 03 Ð B 04 Ð B 05 Ð B 06 Ð A 07 Ð A 08 Ð A 09 Ð A 10 Ð A 11 Ð B 12 Ð B 13 Ð B 14 Ð B 15 Ð B 16 Ð A 17 Ð B 18 Ð B 19 Ð A 20 Ð A

Random Assignment

��* Group 1 10 plots

- Treatment 1 Method A HHHj

HHHj Group 2 10 plots

- Treatment 2 Method B

��*

Measure tubers per plant and fresh weight

116 Chapter 3 Producing Data

3.18 In the Þrst designÑan observational studyÑthe men who exercise (and those who choose not to) may have other characteristics (lurking variables) which might affect their risk of having a heart attack. Since treatments are assigned to the subjects in the second design, the randomization should Òwash outÓ these factors.

3.19 If this year is considerably different in some way from last year, we cannot compare electricity consumption over the two years. For example, if this summer is warmer, the customers may run their air conditioners more. The possible differences between the two years would confound the effects of the treatments.

3.20 (a) An experiment is not possible, since the explanatory variable (gender) cannot be ÒimposedÓ on the subjects. (b) An experiment is possible, but there may be some ethical difÞculties in randomly assigning a surgical treatment to cancer patients (especially if the attending physician recommends the other treatment to the patient).

3.21 Diagram below. Assign labels 01 to 16, then choose two digits at a time beginning on line 115: use blue on poles 04, 09, 14, and 03; green on 10, 06, 11, and 16; white on 02, 07, 13, and 15; and yellow on the rest.

01 02 03 04 yellow white blue blue 05 06 07 08

yellow green white yellow 09 10 11 12 blue green green yellow 13 14 15 16 white blue white green

Random Assignment

�

Group 1 4 poles

- Treatment 1 Blue

J J J Ĵ��

�*

Group 2 4 poles

- Treatment 2 Green HHHj

HHHj Group 3 4 poles

- Treatment 3 White

�� *

J J J Ĵ Group 4

4 poles - Treatment 4

Yellow

�

Count Beetles

3.22 (a) The factors are question location (two levels: before or after text passage) and question type (three levels: simple fact, computation, word problem). This gives six treatments: before/simple fact, after/simple fact, before/computation, after/computation, before/word problem, after/word problem. (b) We start with 12 classes, and randomly split them into six groups of two each; see diagram below. Randomization will vary with starting line. See note on page 50 about using Table B.

Solutions 117

Random Assignment

� � � � � � � � ��

Group 1 2 classes

- before/ simple fact

B B B B B B B BBN

�

Group 2 2 classes

- after/ simple fact

J J J Ĵ��

�*

Group 3 2 classes

- before/ computation HHHjHHHj Group 4

2 classes - after/

computation �� *J

J J JĴ

Group 5 2 classes

- before/ word problem

�B

B B B B B B B BN Group 6

2 classes - after/

word problem � � � � � � � ��

Observe mathematics test score

3.23 Since there are 6 treatments, we can assign 16 students to each treatment and have 4 left over. These 4 can be ignored, or each can be randomly assigned to one of the six groups.

Random Assignment

� � � � � � � � ��

Group 1 16 students

- 30 seconds 1 time

B B B B B B B BBN

�

Group 2 16 students

- 30 seconds 3 times

J J J Ĵ��

�*

Group 3 16 students

- 30 seconds 5 times HHHjHHHj Group 4

16 students - 90 seconds

1 time �� *J

J J JĴ

Group 5 16 students

- 90 seconds 3 times

�B

B B B B B B B BN Group 6 16 students

- 90 seconds 5 times

� � � � � � � ��

Measure recollection,

etc.

3.24 (a) Lack of control means that the speciÞc effects of the meditation technique cannot be distinguished from the effect of investing a month in any activity with the expectation that it will reduce your anxiety. (b)The experimenter expects meditation to lower anxiety, and probably hopes to show that it does. This will unconsciously inßuence the diagnosis. (c) The control group might receive no treatment other than the before and after interview (which itself may affect anxiety) ormight receive an alternative treatment such as physical exercise. Ideally the interviewer should not know the treatment received by an individual,

118 Chapter 3 Producing Data

but this is difÞcult in practice. An ÒobjectiveÓ test of anxiety avoids this problem. If an interviewer is used, he or she should be an outside party with no stake in the experiment.

3.25 For each person, ßip the coin to decide which hand they should use Þrst (heads: right hand Þrst; tails: left hand Þrst).

3.26 The randomization will vary with the starting line in Table B. Completely randomized design: Randomly assign 10 students to ÒGroup 1Ó (which

has the trend-highlighting software) and the other 10 to ÒGroup 2Ó (which does not). Compare the performance of Group 1 with that of Group 2. Matched pairs design: Each student does the activity twice, once with the software

and once without. Randomly decide (for each student) whether they have the software the Þrst or second time. Compare performance with the software and without it. (This randomization can be done by ßipping a coin 20 times, or by picking 20 digits from Table B and using the software Þrst if the digit is even, etc.) Alternate matched pairs design: Again, all students do the activity twice. Randomly

assign 10 students to Group 1 and 10 to Group 2. Group 1 uses the software the Þrst time; Group 2 uses the software the second time.

3.27 (a) Ordered by increasing weight, the Þve blocks are

(1) Williams 22 Festinger 24 Hernandez 25 Moses 25 (2) Santiago 27 Kendall 28 Mann 28 Smith 29 (3) Brunk 30 Obrach 30 Rodriguez 30 Loren 32 (4) Jackson 33 Stall 33 Brown 34 Dixon 34 (5) Birnbaum 35 Tran 35 Nevesky 39 Wilansky 42

(b) The exact randomization will vary with the starting line in Table B. Different methods are possible; perhaps the simplest is to number from 1 to 4 within each block, then assign the members of block 1 to a weight-loss treatment, then assign block 2, etc. For example, starting on line 133, we assign 4ÐMoses to treatment A, 1ÐWilliams to B, and 3ÐHernandez to C (so that 2ÐFestinger gets treatment D), then carry on for block 2, etc. (either continuing on the same line, or starting over somewhere else).

3.28 In each Þeld, have two boards for each color. The diagram is below. One method of randomization would be to assign labels 1Ð8 (ignore 0 and 9) to each pole in Þeld 1, then select from line 105: 5 and 2 for blue; 4 and 7 for green; 6 and 1 for white; and the other two (3 and 8) for yellow. Proceeding on from there, in the second Þeld we assign 1 and 4 for blue; 8 and 6 for green; 7 and 5 for white; and the other two (2 and 3) for yellow. See note on page 50 about using Table B.

1 2 3 4 5 6 7 8 W B Y G B W G Y 1 2 3 4 5 6 7 8 B Y Y B W G W G

Solutions 119

Experimental units (boards)

� �3 Field 1

- Random Assignment

��3 ��* HHjQQs

Group 1

Group 2

Group 3

Group 4

Blue

Green

White

Yellow

QQs HHj ��*

��3

Count Beetles

Q Qs

Field 2 - Random Assignment

��3 ��* HHjQQs

Group 1

Group 2

Group 3

Group 4

Blue

Green

White

Yellow

QQs HHj ��*

��3

Count Beetles

3.29 (a) Below. (b) Practically, it may take a long time until enough claims have been Þled to have the information we need. Ethically, this outline suggests that we assign subjects to an insurance plan; some might object to that. Other answers are possible.

Random Assignment

�

Group 1 n subjects

- Treatment 1 100% of costs

J J J Ĵ��

�*

Group 2 n subjects

- Treatment 2 75% of costs HHHj

HHHj Group 3 n subjects

- Treatment 3 50% of costs

�� *

J J J Ĵ Group 4

n subjects - Treatment 4

0% of costs

�

Observe amount spent on health care

3.30 Use a block design, separately assigning the men and the women to the six treatment groups. The diagram would be quite large, but it would be modeled after Figure 3.4.

3.31 (a) False. Such regularity holds only in the long run. If it were true, you could look at the Þrst 39 digits and know whether or not the 40th was a 0. (b) True. All pairs of digits (there are 100, from 00 to 99) are equally likely. (c) False. Four random digits have chance 1/10000 to be 0000, so this sequence will occasionally occur. 0000 is no more or less random than 1234 or 2718, or any other four-digit sequence.

3.32 The mean IQ of the whole group is µ = 108.92. Logically, when we select 39 from this group, half the time the mean x of this smaller group will be more than 108.92, and half the time it will be less. The theoretical distribution of x is too difÞcult to Þnd exactly, but based on 1000 simulated samples, it is approximately normal with mean 108.92 (the same as the ÒpopulationÓ mean) and standard deviation sx

.= 1.54. (Therefore, x will almost always be between 104.3 and 113.5.)

120 Chapter 3 Producing Data

Section 3: Sampling Design

3.33 Population: Employed adult women. Sample: The 48 women who return the ques- tionnaires. 52% did not respond.

3.34 Population: All words in Tom ClancyÕs novels. Sample: The 250 words recorded. Variable: Number of letters in a word.

3.35 (a) Adult U.S. residents. (b) U.S. households. (c) All regulators from the supplier, or the regulators in the last shipment.

3.36 Variable: Approval of presidentÕs job performance. Population: Adult citizens of the U.S., or perhaps just registered voters. Sample: The 1210 adults interviewed. Possible sources of bias: Only adults with phones were contacted. Alaska and Hawaii were omitted.

3.37 Beginning with Agarwal and going down the columns, label the people with the numbers 01 to 28. From line 139 we select

04ÐBowman, 10ÐFrank, 17ÐLiang, 19ÐNaber, 12ÐGoel, 13ÐGupta See note on page 50 about using Table B.

3.38 Labels: 000 to 439 (or 001 to 440, or two labels each). With either starting label, the Þrst Þve districts from line 117 are (for one label each)

381, 262, 183, 322, 341 With two labels each (starting with either 000/440 or 001/441), the list is

381, 679 (= 239), 853 (= 413), 262, 183 See note on page 50 about using Table B.

3.39 Taking three-digit numbers beginning on line 125 gives the following sample: 214, 313, 409, 306, 511

Note we can only use the numbers 101Ð114, 201Ð215, 301Ð317, 401Ð410, and 501Ð513. Alternatively, we might assign 2-digit labels 00 to 72 (or 01 to 73), rather than use the

3-digit block numbers as labels. When this is done in some order (say, numerical order of block numbers), line 125 gives

96 (ignore), 74 (ignore), 61, 21, 49, 37, 82 (ignore), 37 (repeatÑignore), 18 See note on page 50 about using Table B.

3.40 This defeats the purpose of randomization; if we always start on the same line, our choices are no longer random.

3.41 (a)We will choose one of the Þrst 40 at random and then the addresses 40, 80, 120, and 160 places down the list from it. Beginning on line 120, the addresses selected are 35, 75, 115, 155, 195. (Only the Þrst number is chosen from the table.) (b)All addresses are equally likelyÑeach has chance 1/40 of being selected. To see this, note that each

Solutions 121

of the Þrst 40 has chance 1/40 since one is chosen at random. But each address in the second 40 is chosen exactly when the corresponding address in the Þrst 40 is, so each of the second 40 also has chance 1/40. And so on. This is not an SRS because the only possible samples have exactly one address from

the Þrst 25, one address from the second 25, and so on. An SRS could contain any Þve of the 200 addresses in the population. Note that this view of systematic sampling assumes that the number in the population is a multiple of the sample size.

3.42 Label the students 00, . . . , 24 and use Table B. Then label the faculty 0, . . . , 9 and use the table again. Students may try somemethod of choosing both samples simultaneously. We simply want to choose two separate SRSs, one from the students and one from the faculty. See note on page 50 about using Table B.

3.43 Give each name on the alphabetized lists a number: 001 to 500 for females and 0001 to 2000 for males. From line 122 of Table B, the Þrst Þve females selected are 138, 159, 052, 087, and 359. Continuing on from where we left off, the Þrst Þve men are 1369, 0815, 0727, 1025, and 1868.

3.44 It is not an SRS, because it is impossible to choose a sample with anything but 50 women and 200 men.

3.45 (a)Households without telephones or with unlisted numbers. Such households would likely be made up of poor individuals (who cannot afford a phone), those who choose not to have phones, and those who do not wish to have their phone number published. (b)Thosewith unlisted numberswould be included in the sampling framewhen a random- digit dialer is used.

3.46 The higher no-answer was probably the second periodÑmore families are likely to be gone for vacations, etc. Nonresponse of this type might underrepresent those who are more afßuent (and are able to travel).

3.47 Voluntary response is the big reason. Opponents of gun control usually feel more strongly than supporters, and so are more likely to call. The sampling method also reduces response from poorer people by requiring a phone and willingness to pay for the call.

3.48 Call-in polls, and Òvoluntary responseÓ polls in general, tend to attract responses from those who have strong opinions on the subject, and therefore are often not representative of the population as a whole. On the other hand, there is no reason to believe that the 500 randomly chosen adults overrepresent any particular group, so the 72% ÒyesÓ from that poll is more reliable as an estimate of the true population proportion.

3.49 Form A would draw the higher response favoring the ban. It is phrased to produce a negative reaction: Ògiving huge sums of moneyÓ versus Òcontributing,Ó and giving Òto

122 Chapter 3 Producing Data

candidatesÓ rather than Òto campaigns.Ó Also, form B presents both sides of the issue, allowing for special interest groups to have Òa right to contribute.Ó

3.50 (a) The question is clear, and not particularly slanted, but somemay be embarrassed to say ÒyesÓ to this. (b) This question is likely to elicit more responses against gun control (that is, more people will choose 2). The two options presented are too extreme; no middle position on gun control is allowed. (c) This is clearly slanted in favor of national health insurance. (d) The wording is too technical for many people to understandÑand for those who do understand it, it is slanted because it suggests reasons why one should support recycling. It could be rewritten to something like ÒDo you support economic incentives to promote recycling?Ó

Section 4: Toward Statistical Inference

3.51 6.2% is a statistic.

3.52 2.503 cm is a parameter; 2.515 cm is a statistic.

3.53 43 is a statistic; 52% is a parameter.

3.54 Both 335 g and 289 g are statistics.

3.55 (a) High variability, high bias (wide scatter, many are low). (b) Low variability, low bias (little scatter, close to parameter). (c)High variability, low bias (wide scatter, neither too low nor too high). (d) Low variability, high bias (little scatter, but too high). Make sure that students understand that Òhigh biasÓ means that the values are far from the parameter, not that they are too high.

3.56 The larger sample will give more precise resultsÑthat is, the results are more likely to be close to the population truth (if bias is small).

3.57 For this exercise we assume that the population proportions in all states are about the same. The effect of the population proportion on the variability will be studied further later. Additionally, we ignore Þnite-population corrections in this course. (a) No: The precision of an SRS of size 2000 is the same no matter what the population size (as long as the population is about 10 times the size of the sample or larger). (b) Yes: The sample sizes will vary from 32,000 (California) to 485 (Wyoming), so the precision will also vary (larger samples are less variable).

3.58 The variability would be practically the same for either population. (This makes the [certainly correct] assumption that the pollÕs sample size was less than 800,000Ñ10% of the population of New Jersey.)

Solutions 123

3.59 (a) Answers will vary. If, for example, 8 heads are observed, then p̂ = 820 = 0.4 = 40%. (b) Note that all the leaves in the stemplot should be either 0 or 5, since all possible p̂-values end in 0 or 5. For comparison, here is the sampling distribution (assuming p really is 0.5). An individual studentÕs stemplot will probably not resemble this much, but pooled efforts may be fairly close.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

3.60 (a) The scores will vary depending on the starting row. Note that the smallest possible mean is 61.75 (from the sample 58, 62, 62, 65) and the largest is 77.25 (from 73, 74, 80, 82). (b) Answers will vary; shown below are two views of the sampling distribution. The Þrst shows all possible values of the experiment (so the Þrst rectangle is for 61.75, the next is for 62.00, etc.); the other shows values grouped from 61 to 61.75, 62 to 62.75, etc. (which makes the histogram less bumpy). The tallest rectangle in the Þrst picture is 8 units; in the second, the tallest is 28 units. Technical note: These histograms were found by considering all

( 10 4

) = 210 of the

possible samples. It happens that half (105) of those samples yield a mean smaller than 69.4, and half yield a greater mean. In Exercise 3.32, it was also the case that half of the samples gave means higher than µ, and half lower. In this exercise, it just happens to work out that way; in 3.32, it had to (because we were sampling half of the population).

69.461 7769.461.75 77.25

3.61 (a) We let the digits 0 and 1 represent the presence of eggs, while the other digits represent the absence of eggs. Use ten digits in each sample (one for each square yard). Answers will vary with the line chosen from Table B. (b) To make the stemplot, view each p̂ value as having a 0 in the second place after the decimalÑe.g., p̂ = 0.20 rather than just p̂ = 0.2Ñand use 0 for the leaf. For comparison, here is the sampling distribution. An individual studentÕs stemplot will probably not resemble this much, but pooled efforts may be fairly close.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

3.62 (a) We let the digits 0Ð3 represent ÒyesÓ responses, while the other digits represent Òno.Ó Use 20 digits in each sample. Answers will vary with the line chosen from Table B. (b) The mean of 10 proportions from samples of size 20 is the same as the proportion

124 Chapter 3 Producing Data

from a single sample of size 200. Almost always (99.7% of the time), this value will be in the range 0.4± 3

√ (0.4)(0.6) 200

.= 0.296 and 0.504.

3.63 (a) p = 2795 .= 0.2842. (b) Assign labels

01 through 95 to the players, then take digits two at a time. (In fact, it is easier to simply say that 01Ð27 correspond to offensive backs, and pay no attention to the table.) Answers will vary with the starting line in Table B; if the sample contains, say, 3 offensive backs, then p̂ = 315 = 0.2. (c) The distribution should look roughly normal, centered at p. For comparison, the sampling distribution of p̂ is shown. (d) The mean should be fairly close to p; the lack of bias is (should be) illustrated in that the histogram is clustered around p. [The number of offensive backs has a hypergeometric distribution with parameters N = 95, r = 27, n = 15. The sampling distribution shown has standard deviation 0.1074; the average of 20 p̂ values would be approximately N (p, 0.024), so that about 99.7% of the time, p should be between 0.212 and 0.356.]

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

3.64 (a) Below is the population stemplot (which gives the same information as a histogram). The (population) mean GPA is µ .= 2.6352 and the standard deviation is σ .= 0.7794. [Technically, we should take σ .= 0.7777, which comes from dividing by n rather than n − 1, but few (if any) students would know this.] (b) Ð (e) These histograms are not shown; results will vary with starting line in Table B. The theoretical distribution of x is too difÞcult to Þnd exactly, but based on 1000 simulated samples, it is approximately normal with mean 2.6352 (the same as µ) and standard deviation sx

.= 0.167. (Therefore, x will almost always be between 2.13 and 3.14.) The histogram shown is based on these samples. Note that it is slightly left-

skewed, but less than the population distribution. Also note that the sx , the standard deviation of the sampling distribution, is smaller than σ/

√ 20 .= 0.174, since we are

sampling without replacement.

2 2.1 2.2 2.3 2.4 2.5 2.6

2.7 2.8 2.9 3 3.1

0 134 0 567889 1 0011233444 1 5566667888888888999999 2 000000000111111111222222222333333333444444444 2 5555555555555666666667777777777777788888888888888999999 3 0000000000000011111111112222222223333333333333333444444444 3 556666666677777788889 4 0000

Solutions 125

Exercises

3.65 It is an observational studyÑno treatment was imposed (clearly, there is no ethical way to impose a treatment for this kind of study).

3.66 It is an observational studyÑno treatment was imposed. Results of this study might establish a link between Þtness and personality, but could not establish causation.

3.67 For each taster ßip a coin. If heads, taste Pepsi Þrst, then Coke. If tails, taste Coke Þrst, then Pepsi.

3.68 The factors are whether or not the letter has a ZIP code (2 levels: yes or no) and the time of day the letter is mailed. The number of levels for the second factor may vary. To deal with lurking variables, all letters should be the same size and should be sent to

the same city, and the day on which a letter is sent should be randomly selected. Because most post ofÞces have shorter hours on Saturdays, one may wish to give that day some sort of Òspecial treatmentÓ (it might even be a good idea to have the day of the week be a third factor in this experiment).

3.69 Answers will vary. An example: You want to compare how long it takes to walk to class by two different routes. The experiment will take 20 days. The days are labeled from 01 to 20. Using Table B, the Þrst 10 numbers between 01 and 20 will be assigned to route A; the others will be assigned to route B. Take the designated route on each day and record the time to get to class. Note that this experiment is not blind; you know the route you take on each day.

3.70 (a) Each subject takes both tests; the order in which the tests are taken is randomly chosen. (b) Take 22 digits from Table B. If the Þrst digit is even, subject 1 takes the BI Þrst; if it is odd, he or she takes the ARSMA Þrst. (Or, administer the BI Þrst if the Þrst digit is 0Ð4, the ARSMA Þrst if it is 5Ð9).

3.71 (a) Below. (b) The patients are numbered from 01 to 30. Using line 125, those receiving the beta blockers are

21, 18, 23, 19, 10, 08, 03, 25, 06, 11, 15, 27, 13, 24, 28 See note on page 50 about using Table B.

Random Assignment

��* Group 1 15 patients

- Treatment 1 Beta blocker HHHj

HHHj Group 2 15 patients

- Treatment 2 Placebo

��* Observe pulse rate

126 Chapter 3 Producing Data

3.72 (a) Label the students from 0001 to 3478. (b) Taking four digits at a time beginning on line 105 gives 2940, 0769, 1481, 2975, and 1315. See note on page 50 about using Table B.

3.73 A stratiÞed random sample would be useful here; one could select 50 faculty members from each level. Alternatively, select 25 (or 50) institutions of each size, then choose 2 (or 1) faculty members at each institution. If a large proportion of faculty in your state works at a particular class of institution, it

may be useful to stratify unevenly. If, for example, about 50% teach at Class I institutions, you may want half your sample to come from Class I institutions.

3.74 (a) One possible population: all full-time undergraduate students in the fall term on a list provided by the Registrar. (b)A stratiÞed sample with 125 students from each year is one possibility. (c)Mailed questionnaires might have high nonresponse rates. Telephone interviews exclude those without phones, and may mean repeated calling for those that are not home. Face-to-face interviews might be more costly than your funding will allow.

3.75 (a) Use a block design:

Measure BP for all subjects

��1 Black Men

- Random Assignment

��

@@R

Group 1

Group 2

Calcium

Placebo

@@R

��

Observe change in BP

PPq White Men

- Random Assignment

��

@@R

Group 1

Group 2

Calcium

Placebo

@@R

��

Observe change in BP

(b)A larger group givesmore informationÑwhenmore subjects are involved, the random differences between individuals have less inßuence, and we can expect the average of our sample to be a better representation of the whole population.

3.76 (a) There are two factors (temperature and stirring rate) and six treatments (temperature-stirring rate combinations). Twelve batches are needed. (b) Below. (c) From line 128, the Þrst 10 numbers (between 01 and 12) are 06, 09, 03, 05, 04, 07, 02, 08, 10, and 11. So the 6th and 9th batches will receive treatment 1; batches 3 and 5 will be processed with treatment 2, etc.

Solutions 127

Random Assignment

� � � � � � � � ��

Group 1 2 batches

- Treatment 1 50◦C / 60 rpm

B B B B B B B BBN

�

Group 2 2 batches

- Treatment 2 50◦C / 90 rpm

J J J Ĵ��

�*

Group 3 2 batches

- Treatment 3 50◦C / 120 rpm HHHjHHHj Group 4

2 batches - Treatment 4

60◦C / 60 rpm �� *J

J J JĴ

Group 5 2 batches

- Treatment 5 60◦C / 90 rpm

�B

B B B B B B B BN Group 6

2 batches - Treatment 6

60◦C / 120 rpm � � � � � � � ��

Observe yield

3.77 (a) Below. (b) Have each subject do the task twice, once under each temperature condition, randomly choosing which temperature comes Þrst. Compute the difference in each subjectÕs performances at the two temperatures.

Random Assignment

��* Group 1 10 subjects

- Treatment 1 70◦F HHHj

HHHj Group 2 10 subjects

- Treatment 2 90◦F

��* Count number of insertions

3.78 Subjectswho are unwilling to have their therapy chosen for themmaybe systematically different from those who give their consent. In other words, they may have personality (or other) characteristics which might affect the outcome of their therapy. This would defeat the purpose of randomization, which is to have control and experimental groups that are similar (except in the treatment they receive).

3.79 The 1128 letters are a voluntary response sample, which do not necessarily reßect the opinions of her constituents, since persons with strong opinions on the subject are more likely to take the time to write.

3.80 (a) While we would expect some difference in scores between the two samples, the difference we observed was so large that it would rarely occur purely by chance (if both groups had the same mean score). (b) This observational study found an association between running and mood. It was not an experiment and so does not show that running actually changesmood. Perhaps some personality types aremore likely to take up running in the Þrst place.

128 Chapter 3 Producing Data

3.81 Results will vary with the lines chosen in Table B, but probability computations reveal that about 95% of samples will have 3 to 7 defective rats in each sample. [The number of defective rats has a hypergeometric distribution with parameters N = 30, r = 10, n = 15; P(3 ≤ N ≤ 7) = 0.9498.]

3.82 Shown are the true sampling distri- butions (the vertical scale is the same for all three histograms). Changing p affects both the center and spread of the distributions; the spread increases as p grows (although it begins to decrease as p grows past 0.5). The difference in variability between p = 0.3 and p = 0.5 is hard to see. A normal quantile plot for the p = 0.5

sample should look very much like a line. 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

3.83 Each histogram should be centered near 0.6, with the spread decreasing as the sample size increases. Shown are the actual sampling distribution for n = 50 (on the bottom) and normal approximations for n = 200 and n = 800 (middle and top). When n increases by a factor of 4, note that the sampling distribution shrinks to half its former width and at the same time doubles its height. With n = 50, most p̂ values will be between 0.4 and

0.8; with n = 200, most will be between 0.5 and 0.7; and with n = 800, most will be between 0.55 and 0.65.

0.4 0.5 0.6 0.7 0.8

Solutions 129

Chapter 4 Solutions

Section 1: Randomness

4.1 Long trials of this experiment often approach 40% heads. One theory attributes this surprising result to a Òbottle-cap effectÓ due to an unequal rim on the penny. We donÕt know. But a teaching assistant claims to have spent a proÞtable evening at a party betting on spinning coins after learning of the effect.

4.3 (a) We expect probability 1/2 (for the Þrst ßip and for any ßip of the coin). (b) The theoretical probability that the Þrst head appears on an odd-numbered toss of a fair coin

is 12 + ( 1 2

)3 + ( 12)5 + · · · = 23 . 4.4 Obviously, results will vary with the type of thumbtack used. If you try this experiment, note that although it is commonly done when ßipping coins, we do not recommend throwing the tack in the air, catching it, and slapping it down on the back of your other hand . . . .

4.6 In the long run, of a large number of hands of Þve cards, about 2% (one out of 50) will contain a three of a kind. [Note: This probability is actually 884165

.= 0.02113.]

4.7 The theoretical probabilities are (in order) 116 , 4 16 = 14 , 616 = 38 , 416 = 14 , 116 .

4.8 (a) With n = 20, nearly all answers will be 0.40 or greater. With n = 80, nearly all answers will be between 0.58 and 0.88. With n = 320, nearly all answers will be between 0.66 and 0.80.

4.9 (a) Most answers will be between 35% and 65%. (b) Based on 10,000 simulated trialsÑmore than students are expected to doÑthere is about an 80% chance of having a longest run of 4 or more (i.e., either making or missing 4 shots in a row), a 54% chance of getting 5 or more, a 31% chance of getting 6 or more, and a 16% chance of getting 7 or more. The average (ÒexpectedÓ) longest run length is about 6.

4.10 (a) The theoretical probability is about 0.7190. (b) For comparison, the theoretical histogram is the Þrst one on the right. (c) The curve furthest to the right approximates the theoretical histogram. (d) Both are (should be) centered on or near 0.73; the second histogram should be less spread out. 0.5 0.75 1 0.5 0.75 1

130 Chapter 4 Probability: The Study of Randomness

Section 2: Probability Models

4.11 (a) 0. (b) 1. (c) 0.01. (d) 0.6 (or 0.99, but Òmore often than notÓ is a rather weak description of an event with probability 0.99!)

4.12 (a) S = {germinates, does not germinate}. (b) If measured in weeks, for example, S = {0, 1, 2, . . .}. (c) S = {A, B, C, D, F}. (d) S = {misses both, makes one, makes both}, or S = {misses both, makes Þrst/misses second, misses Þrst/makes second, makes both}. (e) S = {1, 2, 3, 4, 5, 6, 7}.

4.13 (a) S = {all numbers between 0 and 24}. (b) S = {0, 1, 2, . . . , 11 000}. (c) S = {0, 1, 2, . . . , 12}. (d) S = {all numbers greater than or equal to 0}, or S = {0, 0.01, 0.02, 0.03, . . .}. (e) S = {all positive and negative numbers}. Note that the rats can lose weight.

4.14 S = {all numbers between and }. The numbers in the blanks may vary. Table 1.8 has values from 86 to 195 cal; the range of values in S should include at least those numbers. Some students may play it safe and say Òall numbers greater than 0.Ó

4.15 (a) The given probabilities have sum 0.96, so P(type AB) = 0.04. (b) P(type O or B) = 0.49+ 0.20 = 0.69.

4.16 (a) The sum of the given probabilities is 0.9, so P(blue) = 0.1. (b) The sum of the given probabilities is 0.7, so P(blue) = 0.3. (c) P(plain M&M is red, yellow, or orange) = 0.2 + 0.2 + 0.1 = 0.5. P(peanut M&M is red, yellow, or orange) = 0.1+ 0.2+ 0.1 = 0.4.

4.17 Model 1: Legitimate. Model 2: Legitimate. Model 3: Probabilities have sum 67 . Model 4: Probabilities cannot be negative.

4.18 (a) Legitimate. (b) Not legitimate, because probabilities sum to more than 1. (c) Not legitimate, because probabilities sum to less than 1.

4.19 No: The probabilities he describes are 0.1, 0.1, 0.3, and 0.6, which add up to 1.1.

4.20 Use the complement rule: 1− 0.46 = 0.54.

4.21 P(either CV disease or cancer) = 0.45+ 0.22 = 0.67; P(other cause) = 1− 0.67 = 0.33.

4.22 (a) P(not forested) = 1−0.35 = 0.65. (b) P(forest or pasture) = 0.35+0.03 = 0.38. (c) P(neither forest nor pasture) = 1− 0.38 = 0.62.

Solutions 131

4.23 (a) The sum is 1, as we expect, since all possible outcomes are listed. (b) 1− 0.41 = 0.59. (c) 0.41+ 0.23 = 0.64. (d) (0.41)(0.41) = 0.1681.

4.24 (a) P(A) = 0.09 + 0.20 = 0.29. P(B) = 0.09 + 0.05 + 0.04 = 0.18. (b) Ac is the event that the farm is 50 or more acres in size; P(Ac) = 1 − 0.29 = 0.71. (c) {A or B} is the event that a farm is either less than 50 or more than 500 acres in size; P(A or B) = 0.29+ 0.18 = 0.47.

4.25 (a)The probabilities sum to 1. (b)Adding up the second row gives P(female) = 0.43. (c) 1− 0.03− 0.01 = 0.96. (d) 0.11+ 0.12+ 0.01+ 0.04 = 0.28. (e) 1− 0.28 = 0.72.

4.26 (a) 1/38. (b) Since 18 slots are red, the probability of a red is P(red) = 1838 .= 0.474.

4.27 (a) There are 10 pairs. Just using initials: {(A,D), (A,J), (A,S), (A,R), (D,J), (D,S), (D,R), (J,S), (J,R), (S,R)} (b) Each has probability 1/10 = 10%. (c) Julie is chosen in 4 of the 10 possible outcomes: 4/10 = 40%. (d) There are 3 pairs with neither Sam nor Roberto, so the probability is 3/10.

4.28 Fight one big battle: His probability of winning is 0.6, compared to 0.83 = 0.512. (Or he could choose to try for a negotiated peace.)

4.29 (1− 0.05)12 = (0.95)12 .= 0.5404.

4.30 No: It is unlikely that these events are independent. In particular, it is reasonable to expect that college graduates are less likely to be laborers or operators.

4.31 (a) P(A) = 38,225166,438 .= 0.230 since there are 38,225 (thousand) people who have

completed 4+ years of college out of 166,438 (thousand). (b) P(B) = 52,022166,438 .= 0.313.

P(A)P(B).

4.32 (1− 0.02)20 = (0.98)20 .= 0.6676.

4.33 Look at the Þrst Þve rolls in each sequence. All have one G and four RÕs, so those probabilities are the same. In the Þrst sequence, you win regardless of the sixth roll; for the second, you win if the sixth roll is G; for the third sequence, you win if it is R. The

respective probabilities are ( 2 6

)4 (4 6

) = 2243

.= 0.00823, ( 2 6

)4 (4 6

)2 = 4729 .= 0.00549, and( 2 6

)5 (4 6

) = 2729

.= 0.00274.

4.34 P(Þrst child is albino) = 12 · 12 = 14 . P(both of two children are albino) = 14 · 14 = 116 . P(neither is albino) =

( 1− 14

)2 = 916 .

132 Chapter 4 Probability: The Study of Randomness

4.35 (a) (0.65)3 .= 0.2746 (under the random walk theory). (b) 0.35 (since performance in separate years is independent). (c) (0.65)2 + (0.35)2 = 0.545.

4.36 (a) P(under 65) = 0.321 + 0.124 = 0.445. P(65 or older) = 1 − 0.445 = 0.555. (b) P(tests done) = 0.321 + 0.365 = 0.686. P(tests not done) = 1 − 0.686 = 0.314. (c) P(A and B) = 0.365; P(A) P(B) = (0.555)(0.686) .= 0.3807. A and B are not independent; tests were done less frequently on older patients than if these events were independent.

Section 3: Random Variables

4.37 P(less than 3) = P(1 or 2) = 26 = 13 .

4.38 (a) BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG. Each has probability 1/8. (b) Three of the eight arrangements have two (and only two) girls, so P(X = 2) = 3/8 = 0.375. (c) See table.

Value of X 0 1 2 3 Probability 1/8 3/8 3/8 1/8

4.39 (a) 1%. (b)All probabilities are between 0 and 1; the probabilities add to 1. (c) P(X ≤ 3) = 0.48+0.38+0.08 = 1−0.01−0.05 = 0.94. (d) P(X < 3) = 0.48+0.38 = 0.86. (e)Write either X ≥ 4 or X > 3. The probability is 0.05+ 0.01 = 0.06.

4.40 (a) All probabilities are between 0 and 1; the probabilities add to 1. Histogram at right. (b) P(X ≥ 5) = 0.07+ 0.03+ 0.01 = 0.11. (c) P(X > 5) = 0.03+ 0.01 = 0.04. (d) P(2 < X ≤ 4) = 0.17+ 0.15 = 0.32. (e) P(X 6= 1) = 1− 0.25 = 0.75. (f) Write either X ≥ 3 or X > 2. The probability is 1− (0.25+ 0.32) = 0.43. 1 2 3 4 5 6 7

4.41 (a) 75.2%. (b) All probabilities are between 0 and 1; the probabilities add to 1. (c) P(X ≥ 6) = 1−0.010−0.007 = 0.983. (d) P(X > 6) = 1−0.010−0.007−0.007 = 0.976. (e) Either X ≥ 9 or X > 8. The probability is 0.068+ 0.070+ 0.041+ 0.752 = 0.931.

4.42 (a) Sample space below. We must assume that we can distinguish between, e.g., Ò(1,2)Ó and Ò(2,1)Ó; otherwise the outcomes are not equally likely. (b) Each pair has probability 1/36. (c) The value of X is given below each pair. Histogram below, right. (d) P(7 or 11) = 636 + 236 = 836 = 29 . (e) P(not 7) = 1− 636 = 56 .

Solutions 133

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) 2 3 4 5 6 7 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 3 4 5 6 7 8 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 4 5 6 7 8 9 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 5 6 7 8 9 10 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 6 7 8 9 10 11 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) 7 8 9 10 11 12

2 3 4 5 6 7 8 9 10 11 12

4.43 (a) (0.6)(0.6)(0.4) = 0.144. (b) The possible combinations are SSS, SSO, SOS, OSS, SOO, OSO, OOS, OOO (S = support, O = oppose). P(SSS) = 0.63 = 0.216, P(SSO) = P(SOS) = P(OSS) = (0.62)(0.4) = 0.144, P(SOO) = P(OSO) = P(OOS) = (0.6)(0.42) = 0.096, and P(OOO) = 0.43 = 0.064. (c) The distribution is given in the table. The probabilities are found by adding the probabilities from (b), noting that (e.g.) P(X = 1) = P(SSO or SOS or OSS). (d) Write either X ≥ 2 or X > 1. The probability is 0.288+ 0.064 = 0.352.

Value of X 0 1 2 3 Probability 0.216 0.432 0.288 0.064

4.44 (a) P(0 ≤ X ≤ 0.4) = 0.4. (b) P(0.4 ≤ X ≤ 1) = 0.6. (c) P(0.3 ≤ X ≤ 0.5) = 0.2. (d) P(0.3 < X < 0.5) = 0.2. (e) P(0.226 ≤ X ≤ 0.713) = 0.713 − 0.226 = 0.487.

4.45 (a) P(X ≤ 0.49) = 0.49. (b) P(X ≥ 0.27) = 0.73. (c) P(0.27 < X < 1.27) = P(0.27 < X < 1) = 0.73. (d) P(0.1 ≤ X ≤ 0.2 or 0.8 ≤ X ≤ 0.9) = 0.1+ 0.1 = 0.2. (e) P(not [0.3 ≤ X ≤ 0.8]) = 1− 0.5 = 0.5. (f) P(X = 0.5) = 0.

4.46 (a) The height should be 12 , since the area under the curve must be 1. The density curve is at the right. (b) P(y ≤ 1) = 12 . (c) P(0.5 < y < 1.3) = 0.4. (d) P(y ≥ 0.8) = 0.6. 0.5 1 1.5 20

134 Chapter 4 Probability: The Study of Randomness

4.47 (a) The area of a triangle is 12bh = 12(2)(1) = 1. (b) P(Y < 1) = 0.5. (c) P(Y < 0.5) = 0.125.

0.5 1 1.5 20 0.5 1 1.5 20

4.48 (a) P( p̂ ≥ 0.5) = P(Z ≥ 0.5−0.30.023 ) .= P(Z ≥ 8.7) .= 0. (b) P( p̂ < 0.25) .= P(Z <

−2.17) = 0.0150. (c) P(0.25 ≤ p̂ ≤ 0.35) .= P(−2.17 ≤ Z ≤ 2.17) = 0.9700.

4.49 (a) P( p̂ ≥ 0.16) = P(Z ≥ 0.16−0.150.0092 ) .= P(Z ≥ 1.09) = 0.1379. (b) P(0.14 ≤ p̂ ≤

0.16) .= P(−1.09 ≤ Z ≤ 1.09) = 0.7242.

Section 4: Means and Variances of Random Variables

4.50 (a) The payoff is either $0 or $3; see table. (b) For each $1 bet, µX = ($0)(0.75) + ($3)(0.25) = $0.75. (c) The casino makes 25 cents for every dollar bet (in the long run).

Value of X 0 3 Probability 0.75 0.25

4.51 µ = (0)(0.10)+ (1)(0.15)+ (2)(0.30)+ (3)(0.30)+ (4)(0.15) = 2.25.

4.52 The missing probability is 0.99058 (so that the sum is 1). This gives mean earnings µX = $303.3525.

4.53 The mean µ of the companyÕs ÒwinningsÓ (premiums) and their ÒlossesÓ (insurance claims) is positive. Even though the company will lose a large amount of money on a small number of policyholders who die, it will gain a small amount on the majority. The law of large numbers says that the average ÒwinningsÓ minus ÒlossesÓ should be close to µ, and overall the company will almost certainly show a proÞt.

4.54 If your number is abc, then of the 1000 three-digit numbers, there are sixÑ abc, acb, bac, bca, cab, cbaÑfor which you will win the box. Therefore, we win nothing with probability 9941000 = 0.994 and win $83.33 with probability 61000 = 0.006. The expected payoff on a $1 bet is µ = ($0)(0.994)+ ($83.33)(0.006) = $0.50.

4.55 (a) Independent: Weather conditions a year apart should be independent. (b) Not independent: Weather patterns tend to persist for several days; todayÕs weather tells us something about tomorrowÕs. (c) Not independent: The two locations are very close together, and would likely have similar weather conditions.

Solutions 135

4.56 (a) Not independent: Knowing the total X of the Þrst two cards tells us something about the total Y for three cards. (b) Independent: Separate rolls of the dice should be independent.

4.57 (a) The wheel is not affected by its past outcomesÑit has no memory; outcomes are independent. So on any one spin, black and red remain equally likely. (b) Removing a card changes the composition of the remaining deck, so successive draws are not independent. If you hold 5 red cards, the deck now contains 5 fewer red cards, so your chance of another red decreases.

4.58 No: Assuming all Òat-batÓs are independent of each other, the 35% Þgure applies only to the Òlong runÓ of the season, not to Òshort runs.Ó

4.59 (a) The total mean is 11+20 = 31 seconds. (b)No: Changing the standard deviations does not affect the means. (c) No: The total mean does not depend on dependence or independence of the two variables.

4.60 The total mean is 40+ 5+ 25 = 70 minutes.

4.61 In 4.51, we had µ = 2.25, so σ 2X = (0 − 2.25)2(0.10) + (1 − 2.25)2(0.15) + (2 − 2.25)2(0.30)+ (3− 2.25)2(0.30)+ (4− 2.25)2(0.15) = 1.3875, and σX =

√ 1.3875 .=

1.178.

4.62 µX = (0)(0.03)+ (1)(0.16)+ (2)(0.30)+ (3)(0.23)+ (4)(0.17)+ (5)(0.11) = 2.68. σ 2X = (0− 2.68)2(0.03)+ (1− 2.68)2(0.16)+ (2− 2.68)2(0.30)+ (3− 2.68)2(0.23)+ (4− 2.68)2(0.17)+ (5− 2.68)2(0.11) = 1.7176, and σX =

√ 1.7176 .= 1.3106.

4.63 The two histograms are superimposed at the right. Means: µH = 2.6 and µF = 3.14 persons. Variances: σ 2 H = 2.02 and σ 2

F = 1.5604. Standard deviations:

σH .= 1.421 and σF .= 1.249 persons. Since families must include at least two people,

it is not too surprising that the average family is slightly larger (about 0.54 persons) than the average household. For large family/household sizes, the differences between the distributions are small.

1 2 3 4 5 6 7

4.64 µX = (µ− σ)(0.5)+ (µ+ σ)(0.5) = µ, and σX = σ since σ 2X = [µ− (µ− σ)]2(0.5)+ [µ− (µ+ σ)]2(0.5) = σ 2(0.5)+ σ 2(0.5) = σ 2.

4.65 Since the two times are independent, the total variance isσ 2total = σ 2pos+σ 2att = 22+42 = 20, so σtotal =

√ 20 .= 4.472 seconds.

136 Chapter 4 Probability: The Study of Randomness

4.66 Since the two times are independent, the total variance is σ 2total = σ 2Þrst + σ 2second = 22 + 12 = 5, so σtotal =

√ 5 .= 2.236 minutes.

4.67 (a) σ 2Y = (300 − 445)2(0.4) + (500 − 445)2(0.5) + (750 − 455)2(0.1) = 19, 225 and σY

.= 138.65 units. (b) σ 2X+Y = σ 2X + σ 2Y = 7, 800, 000 + 19, 225 = 7, 819, 225, so σX+Y

.= 2796.29 units. (c) σ 2Z = σ 22000X + σ 23500Y = (2000)2σ 2X + (3500)2σ 2Y , so σZ

.= $5, 606, 738.

4.68 (a) Randomly selected students would presumably be unrelated. (b) µf−m = µf − µm = 120 − 105 = 15. σ 2f−m = σ 2f + σ 2m = 282 + 352 = 2009, so σf−m .= 44.82. (c)Knowing only themean and standard deviation, we cannot Þnd that probability (unless we assume that the distribution is normal). Many different distributions can have the same mean and standard deviation.

4.69 (a) µX = 550◦Celsius; σ 2X = 32.5, so σX .= 5.701◦C. (b) Mean: 0◦C; standard deviation: 5.701◦C. (c) µY = 95µX + 32 = 1022◦F, and σY = 95σX

.= 10.26◦F.

4.70 (a) µY−X = µY − µX = 2.001 − 2.000 = 0.001 g. σ 2Y−X = σ 2Y + σ 2X = 0.0022 + 0.0012 = 0.000005, so σY−X .= 0.002236 g. (b) µZ = 12µX + 12µY = 2.0005 g. σ 2Z = 14σ 2X + 14σ 2Y = 0.00000125, so σZ

.= 0.001118 g. Z is slightly more variable than Y , since σY < σZ .

4.71 σ 2X = 94, 236, 826.64, so that σX .= $9707.57.

4.72 (a) µT = µX + µY = 2µX = $606.705. σT = √ σ 2X + σ 2Y =

√ 2σ 2X = $13, 728.57.

(b) µZ = 12µT = µX = $303.3525. σZ = √ 1 4σ

2 X + 14σ 2Y =

√ 1 2σ

2 X = $6864.29. (c)With

this new deÞnition of Z : µZ = µX = $303.3525 (unchanged). σZ = √ 1 4σ

2 X = 12σX =

$4853.78 (smaller by a factor of 1/ √ 2).

4.73 (a) For the Þrst program, µA = (600) ( 1 2

) + (0)

( 1 2

) = 300 people. [And for the

second, µB = (400)(1) = 400.] (b) There is no difference (except in the phrasing): saving 400 is the same as losing 200. (c) No: The choice seems to be based on how the options Òsound.Ó

4.74 Below is the probability distribution for L , the length of the longest run of heads or tails. P(You win) = P(run of 1 or 2) = 89512

.= 0.1738, so the expected outcome is µ = ($2)(0.1738) + (−$1)(0.8262) .= −$0.4785. On the average, you will lose about 48 cents each time you play. (Simulated results should be close to this exact result; how close depends on how many trials are used.)

Value of L 1 2 3 4 5 6 7 8 9 10

Probability 1512 88 512

185 512

127 512

63 512

28 512

12 512

5 512

2 512

1 512

Solutions 137

Section 5: General Probability Rules

4.75 P(A or B) = P(A)+ P(B)− P(A and B) = 0.125+ 0.237− 0.077 = 0.285.

4.76 P(A or B) = P(A)+ P(B)− P(A and B) = 0.6+ 0.4− 0.2 = 0.8.

4.77 (a) {A and B}: household is both prosperous and educated; P(A and B) = 0.077 (given). (b) {A and Bc}: household is prosperous but not edu- cated; P(A and Bc) = P(A) − P(A and B) = 0.048. (c) {Ac and B}: household is not prosperous but is edu- cated; P(Ac and B) = P(B) − P(A and B) = 0.160. (d) {Ac and Bc}: household is neither prosperous nor educated; P(Ac and Bc) = 0.715 (so that the probabilities add to 1).

Ac and Bc

0.715

A and Bc

0.048

Ac and B 0.160

A and B 0.077

4.78 (a) This event is {A and B}; P(A and B) = 0.2 (given). (b) This is {A and Bc}; P(A and Bc) = P(A) − P(A and B) = 0.4. (c) This is {Ac and B}; P(Ac and B) = P(B) − P(A and B) = 0.2. (d) This is {Ac and Bc}; P(Ac and Bc) = 0.2 (so that the probabili- ties add to 1).

Ac and Bc

0.2

A and Bc

0.4

Ac and B 0.2

A and B 0.2

4.79 (a) 18,26299,585 .= 0.1834. (b) 7,76718,262

.= 0.4253. (c) 7,76799,585 .= 0.0780.

(d) P(over 65 and married) = P(over 65) P(married | over 65) = (0.1834)(0.4253). (Or look at the fractions and notice the cancellation when we multiply.)

4.80 (a) 11,08099,585 .= 0.1113. (b) 8,63618,262

.= 0.4729. (c) 2,42568,709 .= 0.0353. (d) No: Among other

reasons, if they were independent, the answers to (a) and (b) would be the same. (We would hardly expect them to be independent.)

4.81 (a) 3,04658,929 .= 0.0517. (b) Ò0.241 is the proportion of women who are married among

those women who are age 18 to 24.Ó (c) Ò0.0517 is the proportion of women who are age 18 to 24 among those women who are married.Ó

4.82 (a) 8561626 .= 0.5264. (b) 3074

.= 0.4054. (c) No: If they were independent, the answers to (a) and (b) would be the same.

4.83 (a) 7701626 .= 0.4736. (b) 529770

.= 0.6870. (c) Using the multiplication rule: P(male and bachelorÕs degree) = P(male) P(bachelorÕs degree | male) = (0.4736)(0.6870) = 0.3254. (Answers will vary with how much previous answers had been rounded.) Di- rectly: 5291626

.= 0.3253. [Note that the difference between these answers is inconsequential, since the numbers in the table are rounded to the nearest thousand anyway.]

138 Chapter 4 Probability: The Study of Randomness

4.84 There were 24, 457 + 6, 027 = 30, 484 suicides altogether. (a) 24,45730,484 .= 0.8023.

(b) 15,802+2,36730,484 .= 0.5960. (c) Among men: 15,80224,457

.= 0.6461. Among women: 2,3676,027 .=

0.3927. (d) In choosing a suicide method, men are much more likely than women to use a Þrearm.

4.85 In constructing the Venn diagram, start with the numbers given for Òonly teaÓ and Òall three,Ó then determine other values. For example, P(coffee and cola, but not tea) = P(coffee and cola) − P(all three). (a) 15% drink only cola. (b) 20% drink none of these.

All three 0.05

Tea 0.05

Coffee 0.20

Cola 0.15 None

0.20

Coffee/Tea 0.10

Coffee/Cola 0.20

Tea/Cola 0.05

4.86 P(A and B) = P(A) P(B | A) = 0.1472.

4.87 If F = {dollar falls} and R = {renegotiation demanded}, then P(F and R) = P(F) P(R | F) = (0.4)(0.8) = 0.32.

4.88 (a) P(A) = 0.846, P(B | A) = 0.951, P(B | Ac) = 0.919. (b) At right. (c) P(A and B) = (0.846)(0.951) .= 0.8045. P(Ac and B) = (0.154)(0.919) .= 0.1415. P(B) .= 0.8045+ 0.1415 .= 0.9460. 0.919

0.081

Member of� labor force

0.846

0.154

0.951

0.049

0.804546

0.041454

0.141526

0.012474

Race Employed

4.89 If F = {dollar falls} and R = {renegotiation demanded}, then P(R) = P(F and R)+ P(Fc and R) = 0.32+ P(Fc) P(R | Fc) = 0.32+ (0.6)(0.2) = 0.44.

4.90 P(A | B) = P(A and B) P(B)

.= 0.8045 0.9460

.= 0.8504.

4.91 P(correct) = P(knows answer)+ P(doesnÕt know, but guesses correctly) = 0.75 + (0.25)(0.20) = 0.8.

Know

answer? Guess

correctly?

yes 0.75

no 0.25

yes 0.20

no 0.80

correct 0.75

correct 0.05

incorrect 0.20

4.92 Tree diagram at right. The black candidate expects to get 12% + 36% + 10% = 58% of the vote.

0.5

0.5 Hispanic

Voter

0.4

0.2

White Against

For0.3

0.7

0.12

0.28

Against

For 0.36

0.04

Against

For 0.10

0.10

Race Vote

Black 0.9

0.1

Solutions 139

4.93 P(knows the answer | gives the correct answer) = 0.750.80 = 1516 = 0.9375.

4.94 The event {Y < 1/2} is the bottom half of the square, while {Y > X} is the upper left triangle of the square. They overlap in a triangle with area 1/8, so

P(Y < 12 | Y > X) = P(Y < 12 and Y > X)

P(Y > X) = 1/8 1/2 = 1 4 .

Y 1 2

Y X

4.95 (a) The rat is in state A after trials 1, 2, and 3, and then changes to state B after trial 4. (b) P(X = 4) = (0.8)(0.8)(0.8)(0.2) = 0.1024. (c) P(X = x) = (0.8)x−1(0.2) for any x ≥ 1Ñthe rat fails to learn from the Þrst x − 1 shocks, then learns from the last shock. [This is an example of a geometric distribution.]

4.96 John should choose the surgery, which gives P(A) = 0.646+ 0.073 = 0.719. Medicalmanagement

0.7

0.3

A 0.7

Ac 0.3

Die

Surgery

0.85

0.10

0.05

Survive

Ac A 0.073

0.027

Ac 0.05

Complications 0.73

0.27

Ac A 0.646

0.204

0.76

0.24

4.97 With C = {building a plant is more proÞtable}, we have P(C) = 0.3078+ 0.1728+ 0.01 = 0.4906 and P(Cc) = 1 − P(C) = 0.5094. (It is also a good idea to check oneÕs work by noting that 0.0162 + 0.4032 + 0.09 = 0.5094.) Contracting with a Hong Kong factory has a slight edge.

0.9

0.1

0.30

0.70

0.10

0.90

0.36

0.64

0.95

0.05

0.3078

0.0162

0.1728

0.4032

0.01

0.09

Exercises

4.98 The probability of winning with one ticket is 1+18+120+270100,000 = 0.00409; the mean is µ = ($5000)

( 1

100,000

) + ($200)

( 18

100,000

) + ($25)

( 120

100,000

) + ($20)

( 270

100,000

) = $0.17.

4.99 (a) µX = (1)(0.1)+ (1.5)(0.2)+ (2)(0.4)+ (4)(0.2)+ (10)(0.1) = 3 million dollars. σ 2X = (4)(0.1)+(2.25)(0.2)+(1)(0.4)+(1)(0.2)+(49)(0.1) = 503.375, soσX .= 22.436 million dollars. (b) µY = 0.9µX − 0.2 = 2.5 million dollars, and σY = 0.9σX .= 20.192 million dollars.

140 Chapter 4 Probability: The Study of Randomness

4.100 (a) The probability of winning nothing is 1− (

1 10,000 + 11,000 + 1100 + 120

) = 0.9389.

(b)Themean isµ = ($1000) (

1 10,000

) +($200)

( 1

1,000

) +($50)

( 1 100

) +($10)

( 1 20

) = $1.30.

1 10,000

) + ($198.70)2

( 1

1,000

) + ($48.70)2

( 1 100

) + ($8.70)2

( 1 20

) =

168.31, so σ .= $12.9734.

4.101 (a) Asian stochastic beetle: µ = (0)(0.2) + (1)(0.3) + (2)(0.5) = 1.3 females. Benign boiler beetle: µ = (0)(0.4) + (1)(0.4) + (2)(0.2) = 0.8 females. (b) When a large population of beetles is considered, each generation of Asian stochastic beetles will contain close to 1.3 times as many females as the preceding generation. So the population will grow steadily. Each generation of benign boiler beetles, on the other hand, contains only about 80% as many females as the preceding generation.

4.102 Y = −70+ 120X : We need b = 120 so that σY = bσX = 1. Sinceµa+bX = a+bµX = a + 120(1400) = a + 70, we need a = −70 to make µY = 0.

4.103 (a) S = {3, 4, 5, . . . , 18} (note these are not equally likely). (b) {X = 5} means that the three dice come up (1,1,3), (1,3,1), (3,1,1), (1,2,2), (2,1,2), or (2,2,1). [Here we assume that there is a Þrst, second, and third die, so we distinguish between, e.g., (1,1,3) and (1,3,1). This makes the computation easier.] Each of these possibilities has

probability ( 1 6

) ( 1 6

) = ( 1 6

)3 , so P(X = 5) = 6

( 1 6

)3 = 136 . (c) µX1 = µX2 = µX3 = (1)

( 1 6

) + (2)

( 1 6

) + (3)

( 1 6

) + (4)

( 1 6

) + (5)

( 1 6

) + (6)

( 1 6

) = 3.5, and σ 2Xi = (6.25)

( 1 6

) +

(2.25) ( 1 6

) + (0.25)

( 1 6

) + (0.25)

( 1 6

) + (2.25)

( 1 6

) + (6.25)

( 1 6

) = 2.916, so σXi .= 1.708.

Since the three rolls of the dice are independent, µX = µX1 + µX2 + µX3 = 10.5 and σ 2X = σ 2X1 + σ 2X2 + σ 2X3 = 8.75, so that σX

.= 2.958.

4.104 (a) µZ = 0.5µX + 0.5µY = 0.065. σ 2Z = 0.52σ 2X + 0.52σ 2Y = 0.020225, so σZ

.= 0.1422. (b) For a given choice of α, µZ = αµX + (1− α)µY = 0.02+ 0.09α and σZ =

√ α2σ 2X + (1− α)2σ 2Y =

√ 0.0025− 0.005α + 0.809α2.

4.105 If we imagine throwing the astragali one at a time, there are 24 different ways that we could end with all four sides different (24 = 4 · 3 · 2 · 1: the Þrst astragalus can be any of the four sides, the second must be one of the other three, the third must be one of the remaining two, and the last must be the one missing side.) Any one of these 24 ways has the same probabilityÑ(0.4)(0.4)(0.1)(0.1)Ñso P(roll a Venus) = 24(0.4)(0.4)(0.1)(0.1) = 0.0384.

4.106 (a)Writing (x, y), where x is AnnÕs choice and y is BobÕs choice, the sample space has 16 elements:

(A,A) (A,B) (A,C) (A,D) (B,A) (B,B) (B,C) (B,D) 0 2 −3 0 −2 0 0 3

(C,A) (C,B) (C,C) (C,D) (D,A) (D,B) (D,C) (D,D) 3 0 0 −4 0 −3 4 0

Solutions 141

(b) The value of X is written below each entry in the table. (c) Below. (d) The mean is 0, so the game is fair. The variance is 4.75, so σX

.= 2.1794. Value of X −4 −3 −2 0 2 3 4 Probability 116

2 16

1 16

8 16

1 16

2 16

1 16

4.107 (a) P(X ≥ 50) = 0.14+ 0.05 = 0.19. (b) P(X ≥ 100 | X ≥ 50) = 0.050.19 = 519 .

4.108 If I = {infection} and F = {failure}, then P(I or F) = P(I )+P(F)−P(I and F) = 0.03+0.14−0.01 = 0.16. The requested probability is P(I c and Fc) = 1−P(I or F) = 0.84.

4.109 (a) P(B or O) = 0.13 + 0.44 = 0.57. (b) P(wife has type B and husband has type A) = (0.13)(0.37) = 0.0481. (c) P(one has type A and other has type B) = (0.13)(0.37)+ (0.37)(0.13) = 0.0962. (d) P(at least one has type O) = 1− P(neither has type O) = 1− (1− 0.44)(1− 0.44) = 0.6864.

4.110 (a) P(female | A) = 0.090.14+0.09 = 923 .= 0.3913.

(b) P(female | D or E) = 0.01+0.040.11+0.12+0.01+0.04 = 528 .= 0.1786.

4.111 The response will be ÒnoÓ with probability 0.35 = (0.5)(0.7). If the probability of plagiarism were 0.2, then P(student answers ÒnoÓ) = 0.4 = (0.5)(0.8). If 39% of students surveyed answered Òno,Ó then we estimate that 2 · 39% = 78% have not plagiarized, so about 22% have plagiarized.

Flip coin

Did they plagiarize?

tails 0.5

heads 0.5

yes 0.3

no 0.7

ÒyesÓ 0.5

ÒyesÓ 0.15

ÒnoÓ 0.35

4.112 (a) At right. (b) P(positive) = 0.01485+ 0.00997 = 0.02482. (c) P(has antibody | positive) = 0.009970.02482

.= 0.4017.

0.997

0.003

Has antibody

Member of population

0.99

0.01

No antibody negative

positive

negative

positive

0.015

0.985

0.01485

0.97515

0.00997

0.00003

4.113 (a) The exact distribution is given below; the probability histogram is at the right. Actual simula- tion results will vary, but should have roughly this shape. (b) This probability is about 0.508. Based on 50 simulated trials, most answers will be between 0.30 and 0.72. (c) The true mean is approximately 2.8. Both computed means should be the same.

1 2 3 4 5 6 7 8 9 100

Value of X 0 1 2 3 4 5 6 7 8 9 10

Probability 11024 143 1024

360 1024

269 1024

139 1024

64 1024

28 1024

12 1024

5 1024

2 1024

1 1024

142 Chapter 5 From Probability to Inference

Chapter 5 Solutions

Section 1: Sampling Distributions for Counts and Proportions

5.1 (a) Itmay be binomial ifwe assume that there are no twins or othermultiple births among the next 20 (this would violate requirement 2ÑindependenceÑof the binomial setting), and that for all births, the probability that the baby is female is the same (requirement 4). (b) No: The number of observations is not Þxed. (c) No: It is not reasonable to assume that the opinions of a husband and wife are independent.

5.2 (a) No: There is no Þxed number of observations. (b) A binomial distribution is reasonable here; a Òlarge cityÓ will have a population over 1000 (10 times as big as the sample). (c) In a ÒPick 3Ó game, JoeÕs chance of winning the lottery is the same every week, so assuming that a year consists of 52weeks (observations), this would be binomial.

5.3 (a) Yes: It is reasonable to assume that the results for the 50 students are independent, and each has the same chance of passing. (b) No: Since the student receives instruc- tion after incorrect answers, her probability of success is likely to increase. (c) No: Temperature may affect the outcome of the test.

5.4 (a) The population is three times larger than the sample; it should be at least 10 times larger. (b) np = (500)(0.002) = 1 is too small; it should be at least 10.

5.5 (a) There are 150 independent observations, each with probability of ÒsuccessÓ (response) p = 0.5. (b) µ = np = (150)(0.5) = 75. (c) P(X ≤ 70) = 0.2312, or see table. (d) Use n = 200, since (200)(0.5) = 100.

Normal Table Exact Normal Approx. Table Normal Prob. Approx. with CC Normal with CC 0.2312 0.2071 0.2312 0.2061 0.2327

5.6 (a) There are 200 responses, each independent of the others, and each with equal probability (0.4) of seeking nutritious food. (b) The mean is (200)(0.4) = 80. We could interpret Òbetween 75 and 85Ó as P(75 ≤ X ≤ 85) = 0.5727 (or see line 1 of the table); or we could exclude 75 and 85 and Þnd P(75 < X < 85) = P(76 ≤ X ≤ 84) = 0.4839 (or see line 2). (c) P(X ≥ 100) = 0.0026 (or see line 3).

Normal Table Exact Normal Approx. Table Normal Prob. Approx. with CC Normal with CC 0.5727 0.5295 0.5727 0.5284 0.5704 0.4839 0.4363 0.4840 0.4380 0.4844 0.0026 0.0019 0.0024 0.0019 0.0025

Solutions 143

5.7 (a) p̂ = 0.86 (86%). (b) P(X ≤ 86) = 0.1239 (or see the table). The normal approximation can be used, since Rule of Thumb 2 is just satisÞedÑn(1 − p) = 10. (c) Even when the claim is correct, there will be some variation in sample proportions. In particular, in about 12% of samples we can expect to observe 86 or fewer orders shipped on time.

Normal Table Exact Normal Approx. Table Normal Prob. Approx. with CC Normal with CC 0.1239 0.0912 0.1217 0.0918 0.1210

5.8 (a) This is the probability that 26 to 34 people from the sample jog; P(26 ≤ X ≤ 34) = 0.6273 (or see line 1 of the table). (b) These probabilities (normal approxima- tions only) are given in the last three lines of the table. As sample size increases, the probability that our estimate is accurate increases.

Normal Table Exact Normal Approx. Table Normal Prob. Approx. with CC Normal with CC 0.6273 0.5717 0.6271 0.5704 0.6266 Ñ 0.8869 0.8977 0.8858 0.8968 Ñ 0.9749 0.9771 0.9750 0.9774 Ñ 0.9985 0.9986 0.9984 0.9986

5.9 (a) Find P(0.41 ≤ p̂ ≤ 0.47) = P(123 ≤ X ≤ 141) = 0.7309 (table line 1). (b) For n = 600, P(0.41 ≤ p̂ ≤ 0.47) = P(246 ≤ X ≤ 282) = 0.8719 (table line 2). For n = 1200, P(0.41 ≤ p̂ ≤ 0.47) = P(492 ≤ X ≤ 564) = 0.9663 (table line 3). Larger sample sizes are more likely to produce values of p̂ close to the true value of p.

Normal Table Exact Normal Approx. Table Normal Prob. Approx. with CC Normal with CC 0.7309 0.7048 0.7308 0.7062 0.7286 0.8719 0.8612 0.8719 0.8612 0.8714 0.9663 0.9637 0.9662 0.9634 0.9660

5.10 (a) It would be reduced by a factor of 1/ √ 2 to about 4.9%. (b) The sample would

have to be four times as big: n = 200. A larger sample gives a more accurate estimate of the proportion we seek. [This assumes that the campus is big enough that the binomial approximation is still valid for n = 200; by our rule of thumb, we need at least 2000 students.]

5.11 X , the number of women in our sample who have never been married, has a binomial distribution with n = 10 and p = 0.25. (a) P(X = 2) = 0.2816. (b) P(X ≤ 2) = 0.5256. (c) P(10− X ≥ 8) = P(X ≤ 2) = 0.5256.

5.12 If the universityÕs claim is true, XÑthe number of athletes in our sample who graduatedÑwould have a binomial distribution with n = 20 and p = 0.80 . (a) P(X = 11) = 0.0074. (b) P(X ≤ 11) = 0.0100.

144 Chapter 5 From Probability to Inference

5.13 (a) n = 4 and p = 1/4 = 0.25. (b) The distribution is below; the histogram is at the right. (c) µ = np = 1.

X 0 1 2 3 4 pX .3164 .4219 .2109 .0469 .0039

1 2 3 40

5.14 (a) n = 6 and p = 0.65. (b) The distribution is below; the histogram is at the right. (c) µ = np = 3.9. (d) σ = √np(1− p) .= 1.1683; one standard deviation from µ means P(3 ≤ X ≤ 5) = 0.8072. X 0 1 2 3 4 5 6 pX .0018 .0205 .0951 .2355 .3280 .2437 .0754

1 2 3 4 5 60

5.15 (a) p = 1/4 = 0.25. (b) P(X ≥ 10) = 0.0139. (c) µ = np = 5, σ =√ np(1− p) = √3.75 .= 1.9365. (d) No: The trials would not be independent, since

the subject may alter his/her guessing strategy based on this information.

5.16 (a) Drivers in separate cars should be independent; it is reasonable to believe that all such cars have the same probability of having a male driver. (b) There might be different probabilities that the male is driving in each of these two situations. (c) X has a Bin(10, 0.85) distribution; P(X ≤ 8) .= 0.4557. (d) Y has a Bin(100, 0.85) distribution (assuming that no car is observed by more than one student); P(Y ≤ 80) .= 0.1065.

5.17 (a) The probability that all are assessed as truthful is ( 12 0

) (0.2)0(0.8)12 .= 0.0687; the

probability that at least one is reported to be a liar is 1 − 0.0687 = 0.9313. (b) µ = (12)(0.2) = 2.4, σ = √1.92 .= 1.3856. (c) P(X < µ) = P(X = 0, 1, or 2) = 0.5583.

5.18 (a)µ = (300)(0.21) = 63, σ = √49.77 .= 7.0548. (b) np = 63 and n(1− p) = 237 are both more than 10. The normal approximation gives 0.0080, or 0.0097 with the continuity correction.

5.19 (a) µ = (1500)(0.12) = 180 and σ = √158.4 .= 12.5857. (b) np = 180 and n(1 − p) = 1320 are both more than 10. Normal approximation values for P(X ≤ 170) are in the table.

Normal Table Normal Approx. Table Normal Approx. with CC Normal with CC 0.2134 0.2252 0.2148 0.2266

5.20 (a)µ = (1500)(0.7) = 1050 and σ = √315 .= 17.7482. (b) P(X ≥ 1000) = 0.9976 (0.9978 with continuity correction). (c) P(X > 1200) < 0.00005 (itÕs very small). (d)With n = 1700, P(X > 1200) is about 0.28 or 0.29.

Solutions 145

5.21 (a) P( p̂ ≤ 0.70) = P(X ≤ 70) is on line 1. (b) P( p̂ ≤ 0.70) = P(X ≤ 175) is on line 2. (c) 400 (with n = 100, σ =√ (0.7)(0.3)/100 .= 0.0458; with n =

400, σ = √(0.7)(0.3)/400 .= 0.0229). (d) Yes: Regardless of p, n must be quadrupled to cut the standard deviation in half.

Normal Table Normal Approx. Table Normal Approx. with CC Normal with CC 0.1241 0.1493 0.1251 0.1492 0.0339 0.0398 0.0336 0.0401

5.22 (a) µX = (1000)(0.2) = 200 and σX = √ 160 .= 12.6491. (b) µ p̂ = p = 0.2 and

σ p̂ = √ p(1− p)/1000 = √0.00016 .= 0.0126491. (c) P( p̂ ≥ 0.24) = P(X ≥ 240) =

0.0008 (0.0009 with continuity correction). (d) From a standard normal distribution, P(Z > 2.326) = 0.01, so the subject must score 2.326 standard deviations above the mean: µ p̂ + 2.326σ p̂ = 0.2294. This corresponds to 230 or more successes.

5.23 (a) ( n n

) = n!

n! 0! = 1. The onlyway to distribute n successes among n observations is for all observations to be successes. (b)

( n

n−1 ) = n!

(n − 1)! 1! = n · (n − 1)! (n − 1)! = n. To distribute

n − 1 successes among n observations, the one failure must be either observation 1, 2, 3, . . . , n − 1, or n. (c)

( n k

) = n!

k! (n − k)! = n!

(n − k)! [n − (n − k)]! = (

n n−k

) . Distributing k

successes is equivalent to distributing n − k failures.

Section 2: The Sampling Distribution of a Sample Mean

5.24 (a) σx = σ/ √ 3 .= 5.7735 mg. (b) Solve σ/√n = 5: √n = 2, so n = 4. The average

of several measurements is more likely than a single measurement to be close to the mean.

5.25 (a) P(X < 0) = P(Z < 0−(−3.5)26 ) = P(Z < 0.1346) = 0.5535 (table value: 0.5517). (b)Themean is the populationmean−3.5%. The standard deviation is σ/√n = 26%/

√ 5 = 11.628%. (c) P(average return < 0) = P(Z < 0−(−3.5)

26/ √ 5 ) = P(Z <

0.3010) = 0.6183 (table value: 0.6179). Averages of several observations are more likely to be close to µ than an individual observation.

5.26 (a) P(X ≥ 21) = P(Z ≥ 21−18.65.9 ) = P(Z ≥ 0.4068) = 0.3421 (table value: 0.3409). [Since ACT scores are reported as whole numbers, we might instead compute P(X ≥ 20.5) = P(Z ≥ 0.3220) = 0.3737 (table value: 0.3745).] (b) µx = 18.6 and σx = σ/

√ 50 .= 0.8344. (c) P(x ≥ 21) = P(Z ≥ 21−18.6

5.9/ √ 50 ) = P(Z ≥ 2.8764) = 0.0020.

[In this case, it is not appropriate to Þnd P(x ≥ 20.5), unless x is rounded to the nearest whole number.]

5.27 (a) Normal with mean 123 mg and standard deviation σx = σ/ √ 3 .= 0.0462 mg.

(b) P(X ≥ 124 mg) = P(Z ≥ 124−123 0.08/

√ 3 ) = P(Z ≥ 21.65)Ñessentially 0.

5.28 µx = 40.125 mm and σx = σ/ √ 4 = 0.001 mm.

146 Chapter 5 From Probability to Inference

5.29 (a) P(X < 295 ml) = P(Z < 295−2983 ) = P(Z < −1) = 0.8413. (b) x has a N (298 ml, σ/

√ 6) distribution, so P(x < 295 ml) = P(Z < 295−298

3/ √ 6 ) = P(Z <

−2.4495) = 0.0072 (table value: 0.0071).

5.30 (a) P(X < 3.5) = P(Z < 3.5−3.80.2 ) = P(Z < −1.5) = 0.0668. (b) x has a N (3.8, 0.1) distribution, so P(x < 3.5) = P(Z < 3.5−3.80.1 ) = P(Z < −3) = 0.0013.

5.31 (a) P(X = 1) = 1838 = 919 and P(X = −1) = 1019 . µX = 919 − 1019 = − 119 dollars, and σX =

√ 360/361 .= $0.9986. (b) In the long run, the gamblerÕs average losses will

be close to − 119 .= −$0.0526 per bet. (c) x has a N (− 119 , 0.1412) distribution. 95% of

the time, the mean winnings will fall between−$0.3350 and $0.2298; his total winnings will be between −$16.75 and $11.49. (d) P(x < 0) = P(Z < 0.3727) = 0.6453 (table vale: 0.6443). (e) The total mean winnings have a N (− 119 , 0.003158) distribution, so 95% of the time, the mean winnings are between −$0.05895 and −$0.04632 (using the 68Ð95Ð99.7 rule), or−$0.05882 and−$0.04644 (using z∗ = 1.96). The casino winnings are between $5895 and $4632, or $5882 and $4644.

5.32 (a) Normal with µx = 55, 000 miles and σx = 4500/ √ 8 .= 1591 miles. (b) P(x ≤

51, 800) = P(Z ≤ −2.0113) = 0.0221 (table value: 0.0222).

5.33 x is approximately normal with µx = 1.6 and σx = 1.2/ √ 200 .= 0.0849 ßaws.

P(x > 2) .= P(Z > 4.71) = 0 (essentially).

5.34 (a) x is approximately normal with µx = 2.2 and σx = 1.4/ √ 52 .= 0.1941 accidents.

(b) P(x < 2) .= P(Z < −1.0302) = 0.1515. (c) P(N < 100) = P(x < 10052 ) = P(Z < −1.4264) = 0.0769 (table value: 0.0764). Alternatively, we might use the continuity correction and Þnd P(N < 99.5) = P(x < 99.552 ) = P(Z < −1.4759) = 0.0700 (table value: 0.0694).

5.35 (a) x is approximately normal with µx = 0.9 and σx = 0.15/ √ 125 .= 0.01342 g/mi.

(b) P(Z > 2.326) = 0.01 if Z is N (0, 1), so L = 0.9 + (2.326)(0.01342) = 0.9312 g/mi.

5.36 Over 45 years, x (the mean return) is approximately normal with µx = 9% and σx = 28%/

√ 45 .= 4.1740%. P(x > 15%) = P(Z > 1.4375) = 0.0753 (table value:

0.0749). P(x < 5%) = P(Z < −0.9583) = 0.1690 (table value: 0.1685).

5.37 L = µ− 1.645σ/√n = 12.513.

5.38 (a) R1+R2 is normalwithmean 100+250 = 350Ä and s.d. √ 2.52 + 2.82 .= 3.7537Ä.

(b) P(345 ≤ R1 + R2 ≤ 355) = P(−1.3320 ≤ Z ≤ 1.3320) = 0.8172 (table value: 0.8164).

Solutions 147

5.39 (a) µx = 360 g and µy = 385 g, so µy−x = 385 − 360 = 25 g. σx = 12.298 g and σy = 11.180 g, so σy−x =

√ σ 2y + σ 2x =

√ 276.25 .= 16.62 g. (b) x is

N (360 g, 12.298 g), y is N (385 g, 11.180 g), and y − x is N (25 g, 16.62 g). (c) P(y − x ≥ 25) = P(Z ≥ 0) = 0.5.

5.40 (a) x is normal with µx = 34 and σx = 12/ √ 26 .= 2.3534. (b) y is normal with

µy = 37 and σy = 11/ √ 24 .= 2.2454. (c) y − x is normal with µy−x = 37 − 34 = 3

and σy−x = √ σ 2x + σ 2y .=

√ 10.5801 .= 3.2527. (d) P(y − x ≥ 4) = P(Z ≥ 0.3074) =

0.3793 (table value: 0.3783).

5.41 (a) y is N (µY , σY/ √ m), and x is N (µX , σX/

√ n).

(b) y − x is N ( µY − µX ,

√ σ 2Y m + σ

2 X

) .

5.42 (a) Two standard deviations: d1 = 2(0.002) = 0.004 and d2 = 2(0.001) = 0.002. (b) σX+Y+Z =

√ 0.0022 + 0.0012 + 0.0012 .= 0.002449, so d .= 0.005Ñconsiderably

less than d1 + 2d2 = 0.008.

5.43 If F and L are their respective scores, then F− L has a N (0,√22 + 22) = N (0, 2√2) distribution, so P(|F − L| > 5) = P(|Z | > 1.7678) = 0.0771 (table value: 0.0768).

5.44 (a) X+Y would be normal withµX+Y = 25+25 = 50 and σX+Y = √ 181 .= 13.4536.

(b) P(X + Y ≥ 60) = P(Z ≥ 0.7433) = 0.2287 (table value: 0.2296). (c) The mean is correct, but the standard deviation is not.

5.45 (a) Yes: This is always true; it does not depend on independence. (b) No: It is not reasonable to believe that X and Y are independent.

5.46 (a) Shown is a stemplot for one set of 100 means. This set had mean 139.7 and standard deviation 26.9; of course, these will vary for other samples. (b) For the 72 survival times, µ = 141.847. (c) σ = 108.448. This is found by dividing by n (72) rather than n − 1 (71), since we are viewing the 72 survival times as a population rather than a sample. If we ignore this technical distinc- tion, we can instead use s = 109.209. We expect that the standard deviation of 100 means should be close to σ/ √ 12 = 31.3061 (or s/√12 = 31.5258). (d) According

to the central limit theorem, the sample means will generally look a lot more normal.

8 9 9 7889 10 112346778899 11 01122456899 12 001233345889 13 114578899 14 111223344567788999 15 01222344569 16 0456789 17 25689 18 0023458 19 07 20 2

148 Chapter 5 From Probability to Inference

Section 3: Control Charts

5.47 The center line is at µ = 75◦; the control limits should be at 75◦ ± 3σ/√4, which means 74.25◦ and 75.75◦.

5.48 Center: 0.8750 inch; control limits: µ±3σ/√5 = 0.8750±0.0016, i.e., 0.8734 inch and 0.8766 inch.

5.49 (a) Center: 11.5; control limits: 11.2 and 11.8. (b) Graphs at right and below. Points outside control limits are circled; the ninth point of a run of nine is marked with a square. (c) Set B is from the in-control process. The process mean shifted suddenly for Set A; it appears to have changed on about the 11th or 12th sample. The mean drifted gradually for the process in Set C.

o o

11.1

11.2

11.3

11.4

11.5

11.6

11.7

11.8 UCL

LCL

Set B

o o o o o o

o o o

o o

11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9

12.1 12.0

UCL

LCL

Set C

o o

o o o

o o

o o o

11.1

11.2

11.3

11.4

11.5

11.6

11.7

11.8 UCL

LCL

Set A

5.50 The centerline is µ = 2.2050 cm, with control limits µ ± 3σ/√5 = 2.2037 to 2.2063 cm. The mean of sample number 7 fell below the lower control limit; that would have been the time to correct the process. There is no run of nine.

o o o o

o o

2.202

2.203

2.204

2.205

2.206

B ea

rin g

di am

et er

(c m

)

Solutions 149

5.51 (a) Center: µ = 10 psi; control limits: µ± 3σ/√3 = 7.922 and 12.078. (b) There are no runs that should concern us here. Lot 13 signals that the process is out of control. The two samples that follow the bad one are Þne, so it may be that whatever caused the low average for the 13th sample was an isolated incident (temperature ßuctuations in the oven during the baking of that batch, or a bad batch of ingredients, perhaps). The operator should investigate to see if there is such an explanation, and try to remedy the situation if necessary.

o o

LCLB re

ak in

g st

re ng

th (p

si )

5.52 (a) Center: x = 3.064%. (b) Control limits: x ± 3s/√6 = −8.51% and 14.64%. (c) Three of the Þrst Þve returns are outside the control limits; after that, there are no out-of-control signals. After considerable ßuctuation in the Þrst few years, Wal-Mart stock has had relatively stable returns.

Ð5

Ð10 o

oo o

o o

oo o

o o

oo o

Ð15

M on

th ly

ra te

o f r

et ur

n (%

)

5.53 Control charts focus on ensuring that the process is consistent, not that the product is good. An in-control process may consistently produce some percentage of low-quality products. Keeping a process in control allows one to detect shifts in the distribution of the output (which may have been caused by some correctable error); it does not help in Þxing problems that are inherent to the process.

5.54 Let A = {at least 4 of 5 points fall above µ+ σ/√n } and B = {at least 4 of 5 points fall below µ− σ/√n }. Note that P(A and B) = 0. The probability that any point falls above µ + σ/√n (or below µ − σ/√n) is about

16%Ñhalf of the 32% that fall outside the central 68%Ñso P(A) .= ( 5 4

) (0.16)4(0.84)+(

5 5

) (0.16)5 .= 0.0029. P(B) also equals 0.0029, so P(A or B) = P(A)+P(B) = 0.0058.

5.55 The probability that any point falls within µ ± σ/√n is about 68%, so P(15 points within one sigma level) .= (0.68)15 .= 0.0031.

5.56 c = 3.090 (Looking at Table A, there appear to be three possible answersÑ3.08, 3.09, or 3.10. In fact, the answer is 3.090232....)

150 Chapter 5 From Probability to Inference

5.57 Center: 162 lbs; control limits: 159.4 and 164.6 lbs. The Þrst Þve points, and the eighth, are above the upper control limit; the Þrst 9 points are a Òrun of nineÓ above the centerline. However, the overall impression is that JoeÕs weight returns to being Òin controlÓ; it decreases fairly steadily, and the last eight points are between the control limits.

o o

o o o

158

160

162

164

166

168

W ei

gh t (

lb s)

5.58 (a) x = 8.4005 and s = 0.6233 min. (b) Control limits: x ± 2s = 7.15 to 9.65 min. (c) The times for October 27 and December 5 are both high, for the reasons given in the exercise. There was one day (November 28) with an extraordinarily low time (which is perhaps no cause for concern). The last 10 points are all above the centerline; ice or snow may have slowed him down on some or all of those days. There is no apparent trend.

oo o

o o

o oo

o o

o oo

oo o

6.5

D riv

e tim

e (m

in ut

es )

5.59 (a) Mean: 0.1; standard deviation:√ p(1−p) 400 = 0.015. (b) Approximately

N (0.1, 0.015). (c) Center: 0.1; control limits: 0.055 and 0.145. (d) This process is out of control. Points below the lower control limit would not be a problem here, but beginning with lot number 2, we see many points above the upper control limit, and every value of p̂ is above the center line (with the exception of two points that fall on the center line). A failure rate above 0.1 is strongly indicated.

o o

0.05

0.1

0.15

0.2

Pr op

or tio

n de

fe ct

iv e

5.60 p̂ is approximately normal with mean p and standard deviation √

p(1−p) n , so use cen-

terline p and control limits p ± 3 √

p(1−p) n .

5.61 Center: 0.0225; control limits: 0.0225 ± 3√0.0225)(0.9775)/80 = −0.02724 and 0.07224. Since −0.02724 is a meaningless value for a proportion, the LCL may as well be set to 0, especially since we are concerned with the failure proportion being too high rather than too low.

Solutions 151

Exercises

5.62 X , the number of free throws made, has a binomial distribution with n = 6 and p = 0.7. P(X ≤ 2) = 0.0705; this is fairly small, which gives some reason to doubt that it was just bad luck.

5.63 (a) P(Z > 105−10015 ) = P(Z > 13) = 0.3694 (table value: 0.3707). (b) µx = 100; σx = 15/

√ 60 .= 1.93649. (c) P(Z > 105−1001.93649 ) = P(Z > 2.5820) = 0.0049.

(d) The answer to (a) could be quite different; (b) would be the same (it does not depend on normality at all). The answer we gave for (c) would still be fairly reliable because of the central limit theorem.

5.64 (a) µ = np = 3025 and σ = √np(1− p) .= 51.5652. (b) P(X ≥ 3500) .= P(Z ≥ 9.21), which is basically 0.

5.65 No: There is no Þxed number of trials. (This is called a negative binomial distribution.)

5.66 (a) P(X = 6) = ( 8 6

) ( 3 4

)6 ( 1 4

)2 .= 0.3115. (b) µ = np = 60. (c) P(X ≥ 50) = 0.9954 (normal approximation: 0.9951, or 0.9966 with continuity correction).

5.67 P(75012 < x < 825 12 ) = P(−1.732 < Z < 2.598) = 0.9537 (table value: 0.9535).

5.68 (a) Binomial with n = 500 and p = 0.52. (b) Find P(X ≥ 250) = P( p̂ ≥ 0.5); possible approximations are in the table. Use µX = 260 and σX .= 11.1714, or µ p̂ = 0.52 and σ p̂ .= 0.02234.

Normal Table Normal Approx. Table Normal Approx. with CC Normal with CC 0.8146 0.8264 0.8159 0.8264

5.69 (a) No. Possible reasons: One could never have X = 0. There is no Þxed number of ÒattemptsÓ here. Solving np = 1.5 and √np(1− p) = 0.75 gives p = 0.625 and n = 2.4. (b) No: A count assumes only whole-number values, so it cannot be normally distributed. (c) Approximately normal with µx = 1.5 and σx = 0.75/

√ 700 .= 0.02835.

(d) 700x has (approximately) a N (1050, 19.84) distribution; P(700x > 1075) = P(Z > 1.2599) = 0.1039 (table value: 0.1038). We could also do a continuity correction for this question: P(700x > 1075.5) = P(Z > 1.2851) = 0.0994 (table value: 0.0985).

5.70 Find P( p̂ ≥ 0.5) = P(X ≥ 250); possible approximations are in the table. Use µX = 225 and σX .= 11.1243, or µ p̂ = 0.45 and σ p̂ .= 0.02225.

Normal Table Normal Approx. Table Normal Approx. with CC Normal with CC 0.0123 0.0138 0.0122 0.0139

152 Chapter 5 From Probability to Inference

5.71 (a) The machine that makes the caps and the machine that applies the torque are not the same. (b) T (torque) is N (7, 0.9) and S (cap strength) is N (10, 1.2), so T − S is N (−3,√0.92 + 1.22) = N (−3, 1.5). Then P(T > S) = P(T − S > 0) = P(Z > 2) = 0.0228.

5.72 Center: 10 inch-lb; control limits: 10± 3(1.2)/√6 = 8.53 and 11.47 inch-lb.

5.73 (a) P(W < 2.8 or W > 3.2) = P(Z < −0.1913 or Z > 2.4472) = 1 − P(−0.1913 ≤ Z ≤ 2.4472) = 0.4313 (0.4318 using table). (b) Center: 3.0 µm. Control limits: 3± 3(0.1516)/√5 = 2.797 and 3.203 µm.

5.74 (a) x is N (32, 6/ √ 23) .= N (32, 1.2511), while y is N (29, 5/√23) .= N (29, 1.0426).

(b) Since the two groups are independent, y − x is N (29 − 32, √ (52 + 62)/23) .=

N (−3, 1.6285). (c) P(y > x) = P(y − x > 0) = P(Z > 1.8421) = 0.0327 (ta- ble value: 0.0329).

5.75 X − Y is N (0,√0.32 + 0.32) .= N (0, 0.4243), so P(|X − Y | ≥ 0.8) = P(|Z | ≥ 1.8856) = 1− P(|Z | ≤ 1.8856) = 0.0593 (table value: 0.0588).

Solutions 153

Chapter 6 Solutions

Section 1: Estimating with ConÞdence

6.1 (a) σx = 4.5/ √ 24 .= 0.9186 kg. (b) x = 61.7916, so the 95% conÞdence interval is

x ± 1.96σx .= 59.99 to 63.59 kg. Since 65 kg is well above the upper conÞdence limit, we have good evidence that µ < 65 kg.

6.2 x = 123.8 bu/acre, and σx = 10/ √ 15 .=

2.582 bu/acre. (a)Ð(c) See the table; the intervals are x ± z∗σx , (d) The margin of error increases with the conÞdence level.

Conf. Level z∗ Interval 90% 1.645 119.6 to 128.0 bu/acre 95% 1.960 118.7 to 128.9 bu/acre 99% 2.576 117.1 to 130.5 bu/acre

6.3 (a) 1 kg is 2.2 pounds, so x∗ = (2.2)(61.7916) .= 135.942 lbs. (b) σx∗ .= (2.2)(0.9186) .= 2.021 lbs. (c) Either compute x∗ ± 1.96σx∗ , or convert the conÞdence limits from 6.1: 132.0 to 139.9 lbs.

6.4 99% conÞdence interval: x±2.576σx = 59.43 to 64.16 kg. This is wider than the 95% interval; it must be wider so that we can be more conÞdent that our interval includes µ.

6.5 With n = 60, σx = 10/ √ 60 .= 1.291 bu/acre. (a) 95% conÞdence interval: x ±

1.960σx = 121.3 to 126.3 bu/acre. (b) Smaller: with a larger sample comes more information, which in turns gives less uncertainty (ÒnoiseÓ) about the value ofµ. (c)They will also be smaller.

6.6 (a) 3.4± (1.645)(0.2) = 3.071 to 3.729. (b) 3.4± (1.645)(0.2/√3) = 3.210 to 3.590.

6.7 11.78± (2.576)(3.2/√114) .= 11.78± 0.77, or 11.01 to 12.55 years.

6.8 2.36± (1.960)(0.8/√50) .= 2.36± 0.22, or 2.14 to 2.58.

6.9 35.091± (1.960)(11/√44) .= 35.091± 3.250, or 31.84 to 38.34.

6.10 n = ( (2.576)(3.2)

)2 .= 67.95Ñtake n = 68.

6.11 (a) 1.96σ/ √ 100 = 2.352 points. (b) 1.96σ/√10 .= 7.438 points. (c) n =

(1.96σ 3

)2 .= 61.46Ñtake n = 62, which is under the 100-student maximum.

6.12 n = ( (1.96)(0.2) 0.06

)2 .= 42.68Ñtake n = 43.

154 Chapter 6 Introduction to Inference

6.13 n = ( (1.645)(10)

)2 .= 16.91Ñtake n = 17.

6.14 (a) 10.0023± (2.326)(0.0002/√5) = 10.0021 to 10.0025 g. (b) n =

( (2.326)(0.0002)

0.0001

)2 .= 21.64Ñtake n = 22.

6.15 $23, 453± (2.576)($8721/√2621) .= $23, 453± $439, or $23,014 to $23,892.

6.16 Multiply the interval of 6.15 by 2621: about $60,320,000 to $62,620,000 (60.32 to 62.62 million dollars).

6.17 (a) No: We can only be 95% conÞdent. (b) The interval (27% to 33%) was based on a method that gives correct results (i.e., includes the correct percentage) 95% of the time. (c) For 95% conÞdence, z∗ = 1.960, so σestimate = 3%1.96

.= 1.53%. (d) No, it only accounts for random ßuctuation.

6.18 $34, 076± (1.96)($200) = $33, 684 to $34,468. (Note that $200 is the standard error of the sample median, not the standard deviation of the distribution of incomes. We do not divide by the sample size.)

6.19 (a) (0.95)7 .= 0.698 = 69.8%. (b) ( 7 6

) (0.95)6(0.05)+ (0.95)7 .= 0.956 = 95.6%.

6.20 (a) The interval 52% ± 2% was based on a method that gives correct results (i.e., includes the correct percentage) 95% of the time. (b) Although 52% ± 2% seems to suggest that Ringel has at least 50% of the vote, we are only 95% conÞdent in that interval; it is possible that our sample was an ÒunluckyÓ one that did not give results within 2% of the true proportion.

6.21 Probably not, because the interval is so wide: Such a large margin of error (±$2000) would suggest either a very small sample size or a large standard deviation, but neither of these seems very likelyÑin particular, a large standard deviation would mean a lot of variability in Þrst-year salaries, suggesting that some trainees start out much higher, and some start out much lower. It is more likely that this range was based on looking at the list of Þrst-year salaries and observing that most were between $20,000 and $24,000.

6.22 (a) The proportion of women giving positive responses in our sample will almost certainly not be exactly the same as the proportion in the population; it serves only as an estimate of the population value. (b) The interval was based on a method that gives correct results 95% of the time. (c) The sample size for women was more than twice as large as that for men. Larger sample sizes lead to smaller margins of error (with the same conÞdence level).

Solutions 155

6.23 No: The interval refers to the mean math score, not to individual scores, which will be much more variable (indeed, if more than 95% of students score below 470, they are not doing very well).

6.24 Since the numbers are based on a voluntary response, rather than an SRS, the methods of this section cannot be usedÑthe interval does not apply to the whole population.

6.25 (a) Now x = 63.012 kg and σx = 4.5/5 = 0.9, so the interval is 61.248 to 64.776 kg. (b) The interval from Exercise 6.1 may be better, since 92.3 kg is an obvious outlier and may need to be excluded.

Section 2: Tests of SigniÞcance

6.26 (a) H0: µ = 1250 ft2; Ha: µ < 1250 ft2. (b) H0: µ = 30 mpg; Ha: µ > 30 mpg. (c) H0: µ = 5 mm; Ha: µ 6= 5 mm.

6.27 (a) H0: µ = 18 sec; Ha: µ < 18 sec. (b) H0: µ = 50; Ha: µ > 50. (c) H0: µ = 24; Ha: µ 6= 24

6.28 (a) H0: pm = pf ; Ha: pm > pf , where pm is the proportion of males who enjoy math, and pf is that proportion for females. (b) H0: µA = µB; Ha: µA > µB , where µA is the mean score for group A and µB is the group B mean. (c) H0: ρ = 0; Ha: ρ > 0, where ρ is the (population) correlation between income and percent of disposable income saved.

6.29 (a) H0: µ = $52, 500; Ha: µ > $52, 500. (b) H0: µ = 2.6 hr; Ha: µ 6= 2.6 hr.

6.30 Even if calcium were not effective in lowering blood pressure, there might be some difference in blood pressure between the two groups. However, in this case the difference was so great that it is unlikely to have occurred by chance (if we assume that calcium is not effective). Therefore we reject the assumption that calcium has no effect on blood pressure.

6.31 Whilewemight expect somedifference in the amount of ethnocentrismbetween church attenders and nonattenders, the observed difference was so large that it is unlikely to be due to chance (i.e., it would happen less than 5% of the time if there were no difference between the groups).

6.32 (a) Letµ1 be the mean for the exercise group andµ2 be the mean for the control group. We might then test H0: µ1 = µ2 vs. Ha: µ1 6= µ2. (The alternative might be one-sided if we have reason to believe the effect will go in one particular direction.) (b)No: P = 0.87 gives no reason to reject H0. (c) There is no (or Òvery littleÓ) difference between the two groupsÕ means. (d) E.g., sample size(s), how the study was designed, how exercise was

156 Chapter 6 Introduction to Inference

incorporated (were students in an exercise program for the whole term, or did they just jog around the block before going to take the Þnal?).

6.33 There almost certainly was some difference between the sexes and between blacks and whites; the observed difference between men and women was so large that it is unlikely to be due to chance. For black and white students, however, the difference was small enough that it could be attributed to random variation.

6.34 z = 11.2−6.9 2.7/ √ 5

.= 3.56, which has P = 0.0002; we conclude that the means (and the authors) are different.

6.35 (a) z = 135.2−115 30/ √ 20

.= 3.01, which gives P = 0.0013. We reject H0 and conclude that the older students do have a higher mean score. (b)We assume the 20 students were an SRS, and that the population is (nearly) normalÑnear enough that the distribution of x is close to normal. The assumption that we have an SRS is more important.

6.36 z = 123.8−120 10/ √ 40

.= 2.40, which gives P = 0.0164. This is strong evidence that this yearÕs mean is different. Slight nonnormality will not be a problem since we have a reasonably large sample size.

6.37 (a) H0: µ = 20; Ha: µ > 20. z = 22.1−206/√53 .= 2.548, so P = P(Z > 2.548) .= 0.0054.

This is strong evidence thatµ > 20Ñthe students have a higher average than past students have. (b) Randomly assign some (25Ð30) students to take the course, and compare their ACT mean score with those who did not take the course.

6.38 (a) H0: µ = 9.5 mg/dl; Ha: µ 6= 9.5 mg/dl. (b) z = 9.58−9.50.4/√180 .= 2.68 and P .= 0.0074.

This is strong evidence against H0; the pregnant womenÕs calcium level is different from 9.5 mg/dl. (c) 9.58± (1.96)(0.4/√180) .= 9.52 to 9.64 mg/dl.

6.39 (a) H0: µ = 32; Ha: µ > 32. (b) x = 35.091, so z = 35.091−3211/√44 .= 1.86 and

P .= 0.0314. This is fairly good evidence that children in this district have a mean score higher than the national averageÑobservations this extreme would occur in only about 3 out of 100 samples if H0 were true.

6.40 (a) z = 0.4365−0.5 0.2887/

√ 100

.= −2.20. (b) SigniÞcant at 5% (z < −1.960). (c) Not signiÞcant at 1% (z ≥ −2.576).

6.41 (a) SigniÞcant at 5% (z > 1.645). (b) SigniÞcant at 1% (z > 2.326).

6.42 (a) Not signiÞcant at 5% (|z| ≤ 1.960). (b) Not signiÞcant at 1% (|z| ≤ 2.576).

6.43 When a test is signiÞcant at the 1% level, it means that if the null hypothesis is true, outcomes similar to those seen are expected to occur less than once in 100 repetitions of the

Solutions 157

experiment or sampling. ÒSigniÞcant at the 5% levelÓmeanswe have observed something which occurs in less than 5 out of 100 repetitions (when H0 is true). Something that occurs Òless than once in 100 repetitionsÓ also occurs Òless than 5 times in 100 repetitions,Ó so signiÞcance at the 1% level implies signiÞcance at the 5% level (or any higher level).

6.44 Since 3.291 is close to 3.3, the P-value is close to (and slightly less than) 2(0.0005) = 0.001.

6.45 Since 0.215 < 0.674, P > 0.25. (In fact, P = P(Z > 0.215) = 0.4149). [This assumes that the test gave some (weak) evidence in favor of the alternative, e.g., we had H0: µ = 10 vs. Ha: µ > 10. If the alternative had been, e.g., Ha: µ < 10, then P = P(Z < 0.215), which is even biggerÑthat is, it gives even less reason to reject H0.]

6.46 (a) Reject H0 if z > 1.645. (b) Reject H0 if |z| > 1.96. (c) For tests at a Þxed signiÞcance level (α), we reject H0 when we observe values of our statistic that are so extreme (far from the mean, or other ÒcenterÓ of the sampling distribution) that they would rarely occur when H0 is true. (SpeciÞcally, they occur with probability no greater than α.) For a two-sided alternative, we split the rejection regionÑthis set of extreme valuesÑinto two pieces, while with a one-sided alternative, all the extreme values are in one piece, which is twice as large (in area) as either of the two pieces used for the two-sided test.

6.47 Since 1.282 < 1.37 < 1.645, the P-value is between 2(0.05) = 0.10 and 2(0.10) = 0.20. From Table A, P = 2(0.0853) = 0.1706.

6.48 (a) The interval is 104.13 ± (1.96)(9/√12) = 99.04 to 109.23 pci/L. (b) Test H0: µ = 105 pci/L vs. Ha: µ 6= 105 pci/L; since 105 is in the interval from (a), we do not have enough evidence to reject H0.

6.49 (a) x ± 1.96σx .= 61.79± 1.80, or 59.99 to 63.59 kg. (b) No, since 61.3 is inside the conÞdence interval. (c) No, since 63 is inside the conÞdence interval.

6.50 (a) Test H0: µ = 7 mg vs. Ha: µ 6= 7 mg; since 7 is not in the interval (1.9 to 6.5 mg), we have evidence against H0. (b) No, since 5 is in the interval.

6.51 P = 0.1292. Although this sample showed some difference in market share between pioneers with patents or trade secrets and those without, the difference was small enough that it could have arisen merely by chance. The observed difference would occur in about 13% of all samples even if there is no difference between the two types of pioneer companies.

6.52 (a) H0: p = 0.5 vs. Ha: p > 0.5. (b) Binomial with parameters n = 5 and p = 0.5. (c) P = P(X ≥ 4) = 0.1875.

158 Chapter 6 Introduction to Inference

Section 3: Use and Abuse of Tests

6.53 A test of signiÞcance answers question (b).

6.54 There is evidence that vitamin C is effective, but not necessarily that the effect is Òstrong.Ó The large sample sizes could make even a small effect signiÞcant.

6.55 (a) z = 478−475 100/ √ 100 = 0.3, so P = P(Z > 0.3) = 0.3821. (b) z = 478−475

100/ √ 1000

.= 0.95, so P = P(Z > 0.95) = 0.1711. (c) z = 478−475

100/ √ 10000 = 3, so P = P(Z > 3) = 0.0013.

6.56 The interval is 478 ± (2.576)(100/√n). n = 100: 452.24 to 503.76. n = 1000: 469.85 to 486.15. n = 10, 000: 475.42 to 480.58.

6.57 (a) z = 1.64 < 1.645Ñnot signiÞcant at 5% level (P = 0.0505). (b) z = 1.65 > 1.645ÑsigniÞcant at 5% level (P = 0.0495).

6.58 Since the numbers are based on a voluntary response, rather than an SRS, the methods of this section cannot be usedÑthe interval does not apply to the whole population.

6.59 (a)No: In a sample of size 500, we expect to see about 5 people who have a ÒP-valueÓ of 0.01 or less. These four might have ESP, or they may simply be among the ÒluckyÓ ones we expect to see. (b) The researcher should repeat the procedure on these four to see if they again perform well.

6.60 Using α/6 = 0.0083 as the cutoff, the fourth (P = 0.008) and sixth (P = 0.001) are signiÞcant.

6.61 Using α/12 = 0.00416 as the cutoff, the Þfth (P = 0.001), sixth (P = 0.004), and eleventh (P = 0.002) are signiÞcant.

6.62 (a) X has a binomial distribution with n = 77 and p = 0.05. (b) P(X ≥ 2) = 1− P(X ≤ 1) = 1− (0.95)77 −

( 77 1

) (0.95)76(0.05) .= 0.9027.

Section 4: Power and Inference as a Decision

6.63 z ≥ 2.326 is equivalent to x ≥ 450+ 2.326(100/√500) .= 460.4, so the power is

P(reject H0 when µ = 460) = P(x ≥ 460.4 when µ = 460) = P

( Z ≥ 460.4−460

100/ √ 500

) = P(Z ≥ 0.0894) = 0.4644.

This is quite a bit less than the Ò80% powerÓ standard; this test is not very sensitive to a 10-point increase in the mean score.

Solutions 159

6.64 z ≤ −1.645 is equivalent to x ≤ 300− 1.645(3/√6) .= 297.99. (a) P(x ≤ 297.99whenµ = 299) = P

( Z ≤ 297.99−299

3/ √ 6

) = P(Z ≤ −0.8287) = 0.2036.

(b) P(x ≤ 297.99 when µ = 295) = P ( Z ≤ 297.99−295

3/ √ 6

) = P(Z ≤ 2.437) = 0.9926.

(c) The power against µ = 290 would be greaterÑit is further from µ0 (300), so it easier to distinguish from the null hypothesis.

6.65 We reject H0 when x ≤ 300− 1.645(3/√n). (a) P

( Z ≤ 299.013−299

3/ √ 25

) = P(Z ≤ 0.0216) = 0.5086.

(b) P ( Z ≤ 299.5065−299

3/ √ 100

) = P(Z ≤ 1.6883) = 0.9543.

6.66 (a) We reject H0 if x ≤ 124.54 or x ≥ 131.46; these numbers are 128 ± (1.96)(15/

√ 72). The power against µ = 134 is 1 − P

( 124.54−134 15/ √ 72 ≤ Z ≤ 131.46−134

15/ √ 72

) .= 1− P(−5.35 ≤ Z ≤ −1.43) .= 0.9236. (b) Power: 0.9236 (same as (a)). Over 90% of the time, this test will detect a difference of 6 (in either the positive or negative direction). (c) The power would be higherÑit is easier to detect greater differences than smaller ones.

6.67 (a) P(x > 0 when µ = 0) = P(Z > 0) = 0.50. (b) P(x ≤ 0 when µ = 0.3) = P

( Z ≤ 0−0.3

1/ √ 9

) = P(Z ≤ −0.9) = 0.1841. (c) P(x ≤ 0 when µ = 1) =

P ( Z ≤ 0−1

1/ √ 9

) = P(Z ≤ −3) = 0.0013.

6.68 P(Type I error) = 0.05 = α. P(Type II error) = 1− 0.9926 = 0.0074.

6.69 P(Type I error) = 0.01 = α. P(Type II error) = 1− 0.4644 = 0.5356.

6.70 (a) P(Type I error) = P(X 6= 4 and X 6= 6 when the distribution is p0) = 0.5. (b) P(Type II error) = P(X = 4 or X = 6 when the distribution is p1) = 0.3.

6.71 (a) H0: the patient is ill (or Òthe patient should see a doctorÓ); Ha: the patient is healthy (or Òthe patient should not see a doctorÓ). A Type I error means a false negativeÑclearing a patient who should be referred to a doctor. A Type II error is a false positiveÑsending a healthy patient to the doctor. (b) One might wish to lower the probability of a false negative so that most ill patients are treated. On the other hand, if money is an issue, or there is concern about sending toomany patients to see the doctor, lowering the probability of false positives might be desirable.

160 Chapter 6 Introduction to Inference

6.72 (b) P(lot is accepted) = (0.1)(0.95) + (0.9)(0.08) = 0.167. (c) P(lot is bad, given it was accepted) = P(bad and accepted)/P(accepted) = (0.9)(0.08)/0.167 .= 0.4311.

good

bad

reject

0.05

0.95

0.08

0.92

0.90

0.10

Exercises

6.73 x = 5.36 mg/dl, so x ± 1.645σ/√6 is 4.76 to 5.97 mg/dl.

6.74 There is some evidence, but not strong evidence, since the conÞdence interval (just) includes 4.8. (An interval with a higher conÞdence level would overlap the 2.6Ð4.8 mg/dl range even more.)

6.75 (a) The plot is reasonably symmetric for such a small sample. (b) x = 30.4 µg/l; 30.4 ± (1.96)(7/√10) gives 26.06 to 34.74 µg/l. (c) H0: µ = 25; Ha: µ > 25. z = 2.44; so P = 0.007. (We knew from (b) that it had to be smaller than 0.025). This is fairly strong evidence against H0; the beginnersÕ mean threshold is higher than 25 µg/l.

2 034 2 3 01124 3 6 4 3

6.76 (a) Wider; raising the conÞdence level increases the interval size. (b) Yes: $35,000 falls outside the 90% conÞdence interval, indicating that P < 0.10.

6.77 Divide everything by 52.14: $653.55± $6.21, or $647.34 to $659.76.

6.78 (a) 145± (1.645)(8/√15), or 141.6 to 148.4 mg/g. (b) H0: µ = 140; Ha: µ > 140. z = 145−140

8/ √ 15

.= 2.42; the P-value is about 0.0078. This strongly supports Ha over H0. (c) We must assume that the 15 cuttings in our sample are an SRS. Since our sample is not too large, the population should be normally distributed, or at least not extremely nonnormal.

6.79 12.9 ± (1.96)(1.6/√26), or 12.3 to 13.5 g/100 ml. This assumes that the babies are an SRS from the population. The population should not be too nonnormal (although a sample of size 26 will overcome quite a bit of skewness).

6.80 (a) The intended population is probably Òthe American publicÓ; the population which was actually sampledwas Òcitizens of Indianapolis (with listed phone numbers).Ó (b)Take x ± 1.96s/√201. Food stores: 15.22 to 22.12. Mass merchandisers: 27.77 to 36.99. Pharmacies: 43.68 to 53.52. (c) The conÞdence intervals do not overlap at all; in par- ticular, the lower conÞdence limit of the rating for pharmacies is higher than the upper conÞdence limit for the other stores. This indicates that the pharmacies are really higher.

Solutions 161

6.81 (a) x has a N (0, 55%/ √ 104) = N (0, 5.3932%)

distribution. (b) z = 6.9−0 55/ √ 104

.= 1.28, so P = 0.1003 (or 0.1004, using software). (c) Not signiÞcant at α = 0.05. The study gives some evidence of increased compensation, but it is not very strongÑit would happen 10% of the time just by chance.

5.39% 10.79%

16.18%

-5.39%

-10.79%

-16.18% 0

6.9%

6.82 (a) The width of the interval (which equals twice the margin of error) decreases. (b) The P-value decreases (the evidence against H0 becomes stronger). (c) The power increases (the test becomes better at distinguishing between H0 and Ha).

6.83 H0: p = 1838; Ha: p 6= 1838 .

6.84 No: ÒSigniÞcant at α = 0.05Ó doesmean that the null hypothesis is unlikely, but only in the sense that the evidence (from the sample) would not occur very often if H0 were true. There is no probability associated with H0 [unless one is a Bayesian statistician]; it is either true or it is not.

6.85 Yes: SigniÞcance tests allow us to discriminate between random differences (Òchance variationÓ) that might occur when the null hypothesis is true, and differences that are unlikely to occur when H0 is true.

6.86 (a) The difference observed in the study would occur in less than 1% of all samples if the two groups actually have the same proportion. (b) The interval is constructed using a method that is correct (i.e., contains the actual proportion) 95% of the time. (c) NoÑ treatments were not randomly assigned, but instead were chosen by the mothers. Mothers who choose to attend a job training program may be more inclined to get themselves out of welfare.

6.87 For each sample, Þnd x , then take x ± 1.96(5/√5) = x ± 4.383. We ÒexpectÓ to see that 95 of the 100 intervals will include 20 (the true value of µ);

binomial computations show that about 99% of the time, 90 or more of the 100 intervals will include 20.

6.88 For each sample, Þnd x , then compute z = x−20 5/ √ 5 . Choose a signiÞcance level α and the

appropriate cutoff pointÑe.g., with α = 0.10, reject H0 if |z| > 1.645; with α = 0.05, reject H0 if |z| > 1.96. If, for example, α = 0.05, we ÒexpectÓ to reject H0 (i.e., make the wrong decision)

only 5 of the 100 times.

6.89 For each sample, Þnd x , then compute z = x−22.5 5/ √ 5 . Choose a signiÞcance level α and

the appropriate cutoff point (z∗)Ñe.g., with α = 0.10, reject H0 if |z| > 1.645; with α = 0.05, reject H0 if |z| > 1.96.

162 Chapter 6 Introduction to Inference

Since Z = x−20 5/ √ 5 has a N (0, 1) distribution, the probability that we will accept H0

is P ( −z∗ < x−22.5

5/ √ 5 < z∗

) = P(−z∗ < Z − 1.118 < z∗) = P(1.118 − z∗ < Z <

1.118+ z∗). If α = 0.10 (z∗ = 1.645), this probability is 0.698; if α = 0.05 (z∗ = 1.96), this probability is 0.799. For smaller α, the probability will be larger. Thus we ÒexpectÓ to (wrongly) accept H0 a majority of the time, and correctly reject H0 about 30% of the time or less.

6.90 Note to instructors: Before assigning this problem to students, it might be good to check that the software or calculator they will use makes this process relatively easyÑ and tell them how to do it (if they are not using Minitab). (b) m = 1.96σ/√n = 1.96(100)/

√ 100 = 19.6. (d) The number of intervals containing 460 has a binomial

distribution with n = 25 and p = 0.95; about 99.3% of the time, 21 or more of the intervals should include the true mean. In separate simulations, the number of intervals containing 460 could vary; in the long run, about 95% of intervals would contain µ.

6.91 Note to instructors: Before assigning this problem to students, it might be good to check that the software or calculator they will use makes this process relatively easyÑ and tell them how to do it (if they are not using Minitab). (b) Since σx = 10, compute z = x−46010 . Reject H0 if |z| > 1.96. (c) The number of rejections has a binomial distribution with parameters n = 25 and p = 0.05. Rarely (only about 0.7% of the time) would more than 4 of the 25 samples lead you to reject H0. In the long run, about 5% of samples would wrongly reject H0.

6.92 (b) Since σx = 10, compute z = x−46010 . Again use α = 0.05, so we reject H0 if |z| > 1.96. Based on the power computed in the next part, the number of rejections has a binomial distribution with parameters n = 25 and p .= 0.516. Most (98.4%) of the time, between 7 and 18 of the 25 samples will result in rejection. (c) The power is

P(reject H0 when µ = 480) = 1− P ( −1.96+ 460−48010 < Z < 1.96+ 460−48010

) = 1− P(−3.96 < Z < −0.04) .= 0.5160

In the long run, about 51.6% of samples with µ = 480 would reject H0: µ = 460.

Solutions 163

Chapter 7 Solutions

Section 1: Inference for the Mean of a Population

7.1 (a) df = 11, t∗ = 1.796. (b) df = 29, t∗ = 2.045. (c) df = 17, t∗ = 1.333.

7.2 (a) df = 54, t∗ = 2.6700. (b) df = 34, t∗ = 1.6909. (c) df = 89, t∗ = 1.9870. If software is unavailable, the answers are (a) df = 50, t∗ = 2.678. (b) df = 30, t∗ = 1.697. (c) df = 80, t∗ = 1.990.

7.3 (a) df = 14. (b) 1.761 and 2.145. (c) 0.05 and 0.025 (respectively). (d) 0.025 < P < 0.05. (e) SigniÞcant at 5%, but not at 1%. (f) P = 0.0345.

7.4 (a) df = 29. (b) 1.055 < 1.12 < 1.311; these have right-tail probabilities 0.15 and 0.10 (respectively). (d) 0.20 < P < 0.30. (e) It is not signiÞcant at either level. (f) P = 0.272.

7.5 (a) df = 11. (b) 0.01 < P < 0.02. (c) P = 0.0161.

7.6 x = 544.75, s .= 79.7, SEx .= 39.85. A conÞdence interval is not really appropriate; what would it represent? If it is intended to capture the mean LSAT score for all students, then these four are certainly not a random sample.

7.7 (a)The stemplot shownhas stems in 1000s, split 5ways. The data are right-skewed, with a high outlier of 2433 (and possibly 1933). The quantile plot shows these two outliers, but otherwise it is not strikingly different from a line. (b) x = 926, s = 427.2, SEx = 69.3 (all in mg). (c) Using 30 degrees of freedom, we have 926± (2.042)(69.3), or 784.5 to 1067.5 mg; Minitab reports 785.6 to 1066.5 mg.

0 3 0 4444455 0 666667777 0 88899999 1 0011 1 22223 1 44 1 1 9 2 2 2 4

oo oooo

o o

o o o

o oo

o o

500

1000

1500

2000

2500

-3 -2 -1 0 1 2 3

C al

ci um

in ta

z score

164 Chapter 7 Inference for Distributions

7.8 (a) Without the outliers, the stemplot shows some details not pre- viously apparent. The normal quantile plot is essentially the same as before (except that the two points that deviated greatly from the line are gone). (b) x = 856.2, s = 306.7, SEx = 51.1 (all in mg). (c) Using 30 degrees of freedom, we have 856.2 ± (2.042)(51.1), or 751.9 to 960.5 mg; Minitab reports 752.4 to 960.0 mg.

3 7 4 01346 5 47 6 25789 7 1478 8 008 9 04779 10 56 11 05 12 0556 13 2 14 22

7.9 (a) The transformed data have x = 77.17%, s = 35.6%, and SEx = 5.78%; the conÞdence interval is 65.37% to 88.97% (using t∗ = 2.042 with df = 30), or 65.46% to 88.87% (from Minitab). (b) After dividing the intervals from Exercise 7.7 by 12, the intervals are the same (up to rounding error).

7.10 (a) H0 is µ = 1200; the alternative might be either Ha: µ 6= 1200 or µ < 1200Ñ the latter since we are likely more concerned with low calcium intake than with high intake. (b) t = (x − µ)/SEx = (926 − 1200)/69.3 .= −3.95. With df = 37, we have P = 0.0003Ñor half of that, for the one-sided alternative. (c)Whichever alternative we use, we conclude that the daily intake is signiÞcantly different from (less than) the RDA.

7.11 (a) The stemplot (with split stems) is right-skewed with high outliers of 63 and 79. [In fact, according to the 1.5IQR outlier test, 48 is an outlier, too.] The quantile plot suggests that the distribution is not normal. (b) x = 23.56, s = 12.52, SEx = 1.77 can openers. Using df = 40, we have t∗ = 2.021 and the interval is 19.98 to 27.14 can openers; Minitab reports 20.00 to 27.12 can openers. (c)With such a large sample size (the text says we need n ≥ 40), the t distribution is fairly good in spite of the skewness and outliers.

1 01233344 1 5566667778999999 2 00124444 2 5555566667 3 244 3 5 4 1 4 8 5 5 6 3 6 7 7 9

ooo o o oo

o oo

o o

o ooo

o o

ooo o

o o

ooo

oo o

o o o

o o

-3 -2 -1 0 1 2 3

C an

o pe

ne r s

al es

z score

7.12 (a) Each store sells an average of 23.56 can openers, so the average proÞt is ($2.15)(23.56) .= $50.65. (b) Multiply the 95% conÞdence interval from Exercise 7.11 by $2.15. Using df = 40, this gives $42.96 to $58.35. Using df = 49, this gives $43.00 to $58.31.

Solutions 165

7.13 (a) Each store averaged $50.65, giving a total of ($50.65)(3275) = $165, 879. (b)Multiply the interval from Exercise 7.12 by 3275. Using df = 40, this gives $140,684 to $191,100. Using df = 49, this gives $140,825 to $190,959.

7.14 t∗ = 2.080 for df = 21, so the interval is $2.08±(2.080)($0.176), or $1.714 to $2.446 per bushel.

7.15 For large df, use normal distribution critical values: x ± 1.645 SEx , or 87.6 to 104.4 days.

7.16 Use t∗ = 2.581 (df = 1000, from the table), or t∗ = 2.5793 (df = 1405, from software). Either choiceÑor even using the normal distribution critical valueÑgives the same interval: x ± t∗ s/√n = 3.83 to 3.97.

7.17 (a) Methods of displaying will vary. Below is a stemplot where the digits are the stems, and all leaves are Ò0ÓÑthis is essentially the same as a histogram. The scores are slightly left-skewed. The normal quantile plot looks reasonably straight, except for the granularity of the data. (b) x = 3.618, s = 3.055, SEx = 0.524. (c) Using df = 30, we have t∗ = 2.042 and the interval is 2.548 to 4.688. Minitab reports 2.551 to 4.684.

−3 0 −2 00 −1 0 −0 0 0 0 1 0 2 000 3 00000 4 0000000 5 00 6 000 7 00000 8 9 00

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7.18 Test H0: µ = 0 vs. Ha: µ > 0, where µ is the mean improvement in scores. t = (x − µ)/SEx = 3.618/0.524 .= 6.90, which has P < 0.0005; we conclude that scores are higher. The conÞdence interval from Exercise 7.17 tells us that the mean improvement is about 2.5 to 4.7 points.

Output from Minitab: Test of mu = 0.000 vs mu > 0.000

Variable N Mean StDev SE Mean T P-Value Scores 34 3.618 3.055 0.524 6.90 0.0000

7.19 H0: µ = 0; Ha: µ < 0, where µ is the mean change in vitamin C content. Subtract ÒAfterÓ from ÒBeforeÓ to give −53, −52, −57, −52, and −61 mg/100g. Then x = −55, s = 3.94, and SEx = 1.76 mg/100g, so t .= −31.24, which is signiÞcant for any reasonable α (using df = 4). Cooking does decrease the vitamin C content.

166 Chapter 7 Inference for Distributions

7.20 (a) The mean change is −55 ± (2.776)(1.76) = −59.9 to −50.1 mg/100g. (b) The ÒAfterÓ measurements are 20.4%, 27.6%, 29.6%, 36.7%, and 17.3% of the speciÞcation. For these percents, x = 26.33%, s = 7.68%, and SEx = 3.44%, so the interval is 16.8% to 35.9% of the speciÞcation.

7.21 (a) H0: µ = 0; Ha: µ > 0. t = 342108/√250 = 50.07; since P .= 0, we reject H0 and

conclude that the new policy would increase credit card usage. (b) Using t∗ = 1.984 (df = 100): $328 to $356. Using t∗ = 1.9695 (df = 249, from software): $329 to $355. (The lower [upper] limits of these two intervals actually differ by only about $0.10.) (c) The sample size is very large, and we are told that we have an SRS. This means that outliers are the only potential snag, and there are none. (d)Make the offer to an SRS of 250 customers, and choose another SRS of 250 as a control group. Compare the mean increase for the two groups.

7.22 x = 44.44, s = 20.74, and SEx = 9.28 (all in µg), and t∗2.776, so the interval is 18.68 to 70.20 µg.

7.23 (a) x = 5.36 mg/dl, while s .= 0.6653 so SEx .= 0.2716 mg/dl. (b) df = 5, t∗ = 2.015, and the interval is 4.819 to 5.914 mg/dl.

7.24 (a) x = 1.75mg/dl, while s .= 0.1291 so SEx .= 0.0645msec. (b) df = 3, t∗ = 2.353, and the interval is 1.6 to 1.9 msec.

7.25 H0: µ = 4.8; Ha: µ > 4.8 mg/dl. t = 5.36−4.80.2716 .= 2.086. For df = 5, we have

0.025 < P < 0.05 (Minitab gives 0.046). This is fairly strong, though not overwhelming, evidence that the patientÕs phosphate level is above normal.

7.26 H0: µ = 1.3; Ha: µ > 1.3 msec. t = 1.75−1.30.0645 .= 6.98. For df = 3, we have

0.0025 < P < 0.005 (Minitab gives 0.003). This is strong evidence that the mean refractory period has increased.

7.27 (a) 114.9±(2.056)(9.3/√27), or 111.2 to 118.6mmHg. (b)The essential assumption is that the 27 men tested can be regarded as an SRS from a population, such as all healthy white males in a stated age group. The assumption that blood pressure in this population is normally distributed is not essential, because x from a sample of size 27 will be roughly normal in any event, as long as the population is not too greatly skewed and has no outliers.

7.28 (a) 1.67 ± (2.120)(0.25/√17), or 1.54 to 1.80. (b) The essential assumption is that the 17 Mexicans tested can be regarded as an SRS from the population of all Mexicans. The assumption that ARSMA scores are normally distributed is clearly not satisÞed but is not essential since scores range from 1 to 5, so there are no outliers and skewness is limited.

Solutions 167

7.29 (a) At right. (b) H0: µ = 105; Ha: µ 6= 105. x = 104.13 and s = 9.40 pCi/l, so t = 104.13−105

9.40/ √ 12

.= −0.32. With df = 11, we have P > 2(0.25) = 0.50 (Minitab reports P = 0.76), which gives us little reason to doubt that µ = 105 pCi/l.

9 1 9 5679 10 134 10 5 11 1 11 9 12 2

7.30 x = 22.125, s .= 2.09, and SEx .= 1.045. The margin of error, 1.045t∗, varies with the choice of conÞdence level; note that df = 3. For 90% conÞdence, m.e. .= ±2.46. For 95% conÞdence, m.e. .= ±2.33. For 99% conÞdence, m.e. .= ±6.11. Explanation: The procedure we used gives results that lie within± of the correct mean % of the time.

7.31 (a) ÒSEMÓ = Òstandard error of the meanÓ (SEx ). (b) s = √ 3 · SEx .= 0.0173.

7.32 (a) H0: µ = 0 vs. Ha: µ < 0 mg/100g, where µ is the change (Haiti minus Factory). (b) t = −4.96 with df = 26, which has P < 0.0005. The mean is signiÞcantly less than 0. (c) See the Minitab output. Note that there is no simple relationship between the Factory and Haiti conÞdence intervals, and the Change interval; the latter cannot be determined by looking at the Þrst two.

Output from Minitab: Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð SigniÞcance Test Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð

Test of mu = 0.00 vs mu < 0.00

Variable N Mean StDev SE Mean T P-Value Change 27 -5.33 5.59 1.08 -4.96 0.0000

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð ConÞdence Intervals Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Variable N Mean StDev SE Mean 95.0 % C.I. Factory 27 42.852 4.793 0.923 ( 40.955, 44.749) Haiti 27 37.519 2.440 0.469 ( 36.553, 38.484) Change 27 -5.33 5.59 1.08 ( -7.54, -3.12)

7.33 (a) For each subject, randomly select (e.g., by ßipping a coin) which knob (right or left) that subject should use Þrst. (b) H0: µ = 0 vs. Ha: µ < 0, where µ is the mean of (right-thread time − left-thread time). (c) x = −13.32 sec; SEx = 22.94/

√ 25 .= 4.59

sec, so t = −2.90. With df = 24, we see that 0.0025 < P < 0.005; Minitab reports P = 0.0039. We have good evidence that the mean difference really is negative, i.e., that the mean time for right-threaded knobs is less than the mean time for left-threaded knobs.

7.34 t∗ = 1.711, so the interval for the mean difference is −13.32 ± (1.711)(4.59), or about −21.2 to −5.5 sec. We have xRH = 104.12 and xLH = 117.44; xRH/xLH = 88.7%. Right-handersworking

with right-handed knobs can accomplish the task in about 90% of the time needed by those working with left-handed knobs. [Note: Another way we could answer the second question is to Þnd the mean of (right-hand time)/(left-hand time), which is 91.7%.]

168 Chapter 7 Inference for Distributions

7.35 (a) H0: µ = 0 vs. Ha: µ > 0, where µ is the mean improvement in score (posttest − pretest). (b) The stemplot of the differences, with stems split 5 ways, shows that the data are slightly left-skewed, with no outliers; the t test should be reliable. (c) x = 1.450; SEx = 3.203/

√ 20 .= 0.716, so t .= 2.02. With df = 19, we see that

0.025 < P < 0.05; Minitab reports P = 0.029. This is signiÞcant at 5%, but not at 1%Ñwe have some evidence that scores improve, but it is not overwhelming. (d) Minitab gives 0.211 to 2.689; using t∗ = 1.729 and the values of x and SEx above, we obtain 1.45± 1.238, or 0.212 to 2.688.

−0 54 −0 32 −0 11 0 11 0 2223333 0 4455 0 7

7.36 (a) For each subject, randomly select (e.g., by ßipping a coin) which test should be administered Þrst. (b) H0: µ = 0 vs. Ha: µ 6= 0, where µ is the mean difference in scores (ARSMA− BI). SEx = 0.2767/

√ 22 .= 0.05899, so t .= 4.27. With df = 21, we

see that P < 0.0005. We have good evidence that the scores differ (i.e., that the mean difference is not 0). (c) 0.2519± (2.080)(0.05899) = 0.1292 to 0.3746 points.

7.37 H0: µ = 0; Ha: µ > 0, whereµ is the mean of (variety A−variety B). t = 0.340.83/√10 .=

1.295; with df = 9, we see that 0.10 < P < 0.15 (Minitab gives P .= 0.11). We do not have enough evidence to conclude that Variety A has a higher yield.

7.38 (a) Two independent samples (3). (b) Matched pairs (2). (c) Single sample (1). (d) Two independent samples (3).

7.39 With all 50 states listed in the table, we have information about the entire population in question; no statistical procedures are needed (or meaningful).

7.40 (a) The critical value is t∗ = 2.423 (using df = 40 from the table), or 2.4049 (using df = 49, from software). (b) Reject H0 if t ≥ t∗, so x > 0+ t∗(108/

√ 50); this is either

x ≥ 37.01 (table) or x ≥ 36.73 (software). (c) The power is P(x ≥ 37.01 when µ = 100) = P

( x−100 108/ √ 50 ≥ 37.01−100

108/ √ 50

) .= P(Z ≥ −4.12) > 0.9999. Using the software t∗, the power is P(Z ≥ −4.14)Ñslightly greater. A sample size of 50 will almost always detect µ = 100.

7.41 (a) This is a one-sided test; we reject H0: µ = 0 if t ≥ 1.833. This translates to x ≥ 0+ (1.833)(0.83/√10) = 0.4811. The power against µ = 0.5 lb/plant is P(x ≥ 0.4811 when µ = 0.5) = P

( x−0.5 0.83/ √ 10 ≥ 0.4811−0.5

0.83/ √ 10

) = P(Z ≥ −0.072) .= 0.5279.

(b)We reject H0 if t ≥ 1.711, which translates to x ≥ 0+ (1.711)(0.83/ √ 25) = 0.2840.

The power against µ = 0.5 lb/plant is P(x ≥ 0.2840 when µ = 0.5) = P

( x−0.5 0.83/ √ 25 ≥ 0.2840−0.5

0.83/ √ 25

) = P(Z ≥ −1.301) .= 0.9032.

Solutions 169

7.42 (a) t∗ = 2.080. (b) We reject H0 if |t | ≥ 2.080, which translates to |x | ≥ 2.080s/

√ 22 .= 0.133. (c) The power against µ = 0.2 is

P(|x | ≥ 0.133 when µ = 0.2) = 1− P(−0.133 ≤ x ≤ 0.133) = 1− P

(−0.133−0.2 0.3/ √ 22 ≤ x−0.2

0.3/ √ 22 ≤ 0.133−0.2

0.3/ √ 22

) = 1− P(−5.21 ≤ Z ≤ −1.05) .= 0.8531.

7.43 (a) H0: population median = 0 vs. Ha: population median < 0, or H0: p = 1/2 vs. Ha: p < 1/2, where p is the proportion of (right − left) differences that are positive. (Equivalently one could take Ha: p > 1/2, where p is the proportion of negative differences.) (b) One pair of the 25 had no difference; of the remaining 24, only 5 differences were positive. If X (the number of positive differences) has a Bin(24, 1/2) distribution, the P-value is P(X ≤ 5), for which the normal approximation gives P(Z < −2.86) = 0.0021 (without the continuity correction) or P(Z < −2.65) = 0.0040 (with the continuity correction). [In fact, P = 0.0033.] In any case, this is strong evidence against H0, indicating that the median right-threaded knob time is shorter.

Output from Minitab: Sign test of median = 0.00000 versus L.T. 0.00000

N BELOW EQUAL ABOVE P-VALUE MEDIAN RH-LH 25 19 1 5 0.0033 -12.00

7.44 Test H0: population median = 0 vs. Ha: population median > 0. Six of the 20 differences are negative. If X (the number of negative differences) has a Bin(20, 1/2) distribution, the P-value is P(X ≤ 6) = 0.0577Ñwhich is not quite signiÞcant (if we have α = 0.05).

Output from Minitab: Sign test of median = 0.00000 versus G.T. 0.00000

N BELOW EQUAL ABOVE P-VALUE MEDIAN Post-Pre 20 6 0 14 0.0577 2.000

7.45 We cannot use the sign test, since we cannot determine the number of positive and negative differences in the original data.

7.46 (a) x .= 141.85, s .= 109.2, and SEx .= 12.87 days. Use t∗ = 2.000 (df = 60, from the table) or t∗ = 1.994 (df = 71, from software). The former gives 116.1 to 167.6; the latter 116.2 to 167.5. (b) A stemplot and quantile plot are shown. These were based on common (base 10) logarithms; for natural logs, the quantile plot differs only in vertical scale, but the stemplot has a slightly different appearance. (c) We now have x .= 2.07205, s .= 0.243015, and SEx .= 0.028640. For t∗ = 2.000, the interval is 2.0148 to 2.1293; for t∗ = 1.994, it is 2.0149 to 2.1292. (If using natural logs, these intervals are 4.6392 to 4.9030, or 4.6395 to 4.9026.)

170 Chapter 7 Inference for Distributions

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7.47 Using common (base 10) logarithms: x = 2.5552, s = 0.0653, and SEx = 0.0292, giving the interval 2.4929 to 2.6175. Using natural (base e) logarithms: x = 5.8836, s = 0.1504, and SEx = 0.0672, giving the interval 5.7402 to 6.0270. [Note that these intervals are equivalent; if we exponentiate to undo the logarithms, we obtain the interval 311.1 to 414.5 hours.]

Section 2: Comparing Two Means

7.48 (a) H0: µ1 = µ2; Ha: µ1 > µ2. x1 = 48.705 and s1 = 1.534 mg/100g, while x2 = 21.795 and s2 = 0.7707 mg/100g; thus t = 22.16. Using df = 1, we have 0.01 < P < 0.02 (Minitab gives 0.014). Software approximation: df = 1.47, and P = 0.0039. This is fairly strong evidence that vitamin C is lost in storage. (b) 90% conÞdence interval: 19.2 to 34.6 mg/100g vitamin C lost (using df = 1), or 22.3 to 31.5 mg/100g (using df = 1.47).

7.49 (a) H0: µ1 = µ2; Ha: µ1 > µ2. x1 = 95.3 and s1 = 0.990 mg/100g, while x2 = 95.85 and s2 = 2.19 mg/100g. Since x2 > x1, we have no evidence against H0; further analysis is not necessary. (However, just for reference, t = −0.323.) (b) 90% conÞdence interval: −11.3 to 10.2 mg/100g vitamin E lost (using df = 1), or −7.36 to 6.26 mg/100g (using df = 1.39).

7.50 Small samples may lead to rejection of H0, if (as in Exercise 7.48) the evidence is very strong. (The weakness of small samples is that they are not very powerful; the rejection in 7.48 occurred because the evidence suggests that the true means are quite different.)

Solutions 171

7.51 (a) Control scores are fairly symmetrical, while piano scores are slightly left-skewed. Scores in the piano group are generally higher than scores in the control group. (b) Below. (c) H0: µ1 = µ2; Ha: µ1 > µ2. t = 5.06. Whether df = 33 or df = 61.7, P < 0.0001, so we reject H0 and conclude that piano lessons improved the test scores.

n x s SEx Piano 34 3.618 3.055 0.524 Control 44 0.386 2.423 0.365

Control Piano 0 −6 −5

0 −4 000 −3 0 00 −2 00

0000000 −1 0 00000 −0 0 000000 0 0 000000 1 0 0000000 2 000

0 3 00000 000 4 0000000 0 5 00

6 000 7 00000 8 9 00

7.52 The standard error of the difference is

SED = √ s21/n1 + s22/n2 .= 0.6387

and the interval is (x1 − x2) ± t∗SED. Answers will vary with the degrees of freedom used; see the table.

df t∗ Interval 30 2.042 1.928 to 4.536 33 2.0345 1.933 to 4.531 61.7 1.9992 1.955 to 4.509

7.53 Having the control group in 7.51 and 7.52 makes our conclusions more reliable, since it accounts for increases in scores that may come about simply from the passage of time. Between signiÞcance tests and conÞdence intervals, preferences might vary somewhat. Arguably, there is an advantage to the test since we have a one-sided alternative; the conÞdence interval by its nature is two-sided.

7.54 (a) The back-to-back stemplot shows a roughly normal shape for the healthy Þrms, while failed- Þrm ratios are generally lower and have a slightly less normal (skewed right) distribution. (b) H0: µ1 = µ2 vs. Ha: µ1 > µ2. Summary statistics for the two groups are below; the t value is 7.90, with either df = 32 or df = 81.7. Either way, P < 0.0005, so we conclude that failed ÞrmsÕ ratios are lower. (c) One cannot impose the ÒtreatmentsÓ of failure or success on a Þrm.

n x s SEx Healthy 68 1.726 0.639 0.078 Failed 33 0.824 0.481 0.084

Failed Healthy 11100 0 1

22 0 2 5544 0

6 0 66 9999988888 0 899999

111111 1 00011 33 1 2223 4 1 4445555 6 1 66666777

1 88888889999 0 2 0000111

2 222223 2 455 2 6677 2 8 3 01

7.55 (a) H0: µ1 = µ2; Ha: µ1 6= µ2. For the low-Þtness group, x1 = 4.64 and s1 = 0.69. For the high-Þtness group, x2 = 6.43 and s2 = 0.43. t = −8.23, so P < 0.0001 (using either df = 13 or df = 21.8); this difference is signiÞcant at 5% and at 1% (and much

172 Chapter 7 Inference for Distributions

lower). (b) All the subjects were college faculty members. Additionally, all the subjects volunteered for a Þtness program, which could add some further confounding.

7.56 For H0: µ1 = µ2 vs. Ha: µ1 > µ2, we have t = 5.99 (with df 11 or 17.6). This is signiÞcant (P < 0.0005), so we conclude that the treatment was effective.

n x s SEx Control 13 3.38 1.19 0.33 Treatment 12 1.167 0.577 0.17

7.57 (a) We test H0: µ1 = µ2 vs. Ha: µ1 < µ2. t = −7.34, which gives P < 0.0001 whether df = 133 or df = 140.6. Cocaine use is associated with lower birth weights. (b) The standard error of the difference is SED

.= 52.47, and the interval is (x1 − x2) ± t∗SED. Answers will vary with the degrees of freedom used; see the table. (c) The ÒOtherÓ group may include drug users, since some in it were not tested. Among drug users, there may have been other (ÒconfoundingÓ) factors that affected birthweight. Note that in this situation, an experiment is out of the question.

df t∗ Interval 100 1.984 −489.1 to −280.9 g 133 1.9780 −488.8 to −281.2 g 140.6 1.9770 −488.7 to −281.3 g

7.58 (a) s1 .= 0.2182 and s2 .= 0.1945. (b) We test H0: µ1 = µ2 vs. Ha: µ1 6= µ2.

t = −17.83, which gives P < 0.0001 whether df = 44 or df = 97.7. There is strong evidence that mean wheat prices differ between July and September.

7.59 (x2 − x1) ± t∗ √ s21/n1 + s22/n2 = $0.66 ± t∗($0.037014), where t∗ is chosen with

either df = 44 or df = 97.7. Whatever choice of df is made, t∗ .= 2, so the interval is about $0.59 to $0.73.

7.60 (a) SED .= 2.1299. Answers will vary

with the df used; see the table. (b) Because of random ßuctuations between stores, we might (just by chance) have seen a rise in the average number of units sold even if actual mean sales had remained unchangedÑor even if they dropped slightly.

df t∗ Interval 50 2.009 −1.28 to 7.28 units 52 2.0067 −1.27 to 7.27 units 121.9 1.9796 −1.22 to 7.22 units

7.61 (a) H0: µA = µB; Ha: µA 6= µB ; t = −1.484. Using t (149) and t (297.2) distri- butions, P equals 0.1399 and 0.1388, respectively; not signiÞcant in either case. The bank might choose to implement Proposal A even though the difference is not signiÞcant, since it may have a slight advantage over Proposal B. Otherwise, the bank should choose whichever option costs them less. (b) Because the sample sizes are equal and large, the t procedure is reliable in spite of the skewness.

Solutions 173

7.62 (a) We test H0: µ1 = µ2 vs. Ha: µ1 > µ2. t .= 1.654; using t (18) and t (37.6) distributions, P equals 0.0578 and 0.0532, respectively. We have some evidence of a higher mean hemoglobin level for breast-fed infants, but not quite enough to be signiÞcant at the 5% level. (b) SED

.= 0.5442, and the interval is (x1 − x2) ± t∗SED. The two possible answers are given in the table. (c) We are assuming that we have two SRSs from each pop- ulation, and that underlying distributions are normal. Since the sample sizes add to 42, normality is not a crucial assumption.

df t∗ Interval 18 2.101 −0.243 to 2.043 37.6 2.0251 −0.202 to 2.002

7.63 (a) H0: µ1 = µ2 vs. Ha: µ1 < µ2, where µ1 is the beta-blocker population mean pulse rate and µ2 is the placebo mean pulse rate. We Þnd t .= −2.4525; with a t (29) distribution, we have 0.01 < P < 0.02 (in fact, P = 0.01022), which is signiÞcant at 5% but not at 1%. With a t (57.8) distribution, we have P = 0.0086, which is signiÞcant at 5% and at 1%. (b) See table at right.

df t∗ Interval 29 2.756 −10.83 to 0.63 bpm 57.8 2.6636 −10.64 to 0.44 bpm

7.64 (a) H0 : µskilled = µnovice vs. Ha : µs > µn. (b) The t statistic we want is the ÒUnequalÓ value: t = 3.1583 with df = 9.8. Its P-value is 0.0052 (half of that given). This is strong evidence against H0. (c) See table at right.

df t∗ Interval 9 1.833 0.4922 to 1.8535 9.8 1.8162 0.4984 to 1.8473

7.65 H0: µskilled = µnovice vs. Ha: µs 6= µn (use a two-sided alternative since we have no preconceived idea of the direction of the difference). Use the ÒUnequalÓ values: t = 0.5143 with df = 11.8; its P-value is 0.6165. There is no reason to reject H0; skilled and novice rowers seem to have (practically) the same mean weight.

7.66 With such large samples, the t distribution is practically indistinguishable from the normal distribution, and in fact, for df = 19, 882 or 38,786, t∗ = 2.576. Thus the interval is 27.915 to 32.085 points. (If one takes the conservative approach, with df = 1000, the interval is 27.911 to 32.089.)

174 Chapter 7 Inference for Distributions

7.67 (a) Using back-to-back stemplots, we see that both distri- butions are slightly skewed to the right, and have one or two moderately high outliers. Normal quantile plots (not shown) are fairly linear. A t procedure may be (cautiously) used in spite of the skewness, since the sum of the sample sizes is almost 40. (b) H0: µw = µm; Ha: µw > µm. Summary statistics (below) lead to t = 2.0561, so P = 0.0277 (with df = 17) or P = 0.0235 (with df = 35.6).

n x s Women 18 141.056 26.4363 Men 20 121.250 32.8519

This gives fairly strong evidenceÑsigniÞcant at 5% but not 1%Ñthat the womenÕs mean is higher. (c) For µm − µw: −36.56 to −3.05 (df = 17) or −36.07 to −3.54 (df = 35.6).

Women Men 7 05 8 8 9 12

931 10 489 5 11 3455

966 12 6 77 13 2 80 14 06 442 15 1 55 16 9 8 17

18 07 19

0 20

7.68 (a) A back-to-back stemplot shows no particular skewness or outliers. (b) H0: µ1 = µ2 vs. Ha: µ1 < µ2. Summary statistics give t .= −2.47, so 0.01 < P < 0.02 (with df = 19) or P = 0.0092 (with df = 36.9).

n x s Control 20 366.30 50.8052 Experimental 20 402.95 42.7286

This gives fairly strong evidence that the high-lysine diet leads to increased weight gain. It is signiÞcant at the 10% and 5% levels either way, and at the 1% level using the higher df. (c) The interval (for µ2 − µ1) is 5.58 to 67.72 g (df = 19) or 6.57 to 66.73 g (df = 36.9).

Control Exper. 7 2 8 2 1 3 1 22 3 23

55544 3 66 3 67 988 3 99 10 4 00001 3 4 22233 5 4 4 6 4 67

7.69 (a) t = 1.604 with df = 9 or df = 15.6; the P-value is either 0.0716 or 0.0644, respectively. Both are similar to the P-value in Example 7.20, and the conclusion is essentially the same. (b) With df = 9: −0.76 to 11.30 (margin of error: 6.03). With df = 15.6: −0.48 to 11.02 (margin of error: 5.75). Both margins of errors are similar to (but slightly larger than) the margin of error in Example 7.21.

7.70 (a) SED .= 7.9895; see table. (b) We know

that we can reject H0, since 0 is well outside our conÞdence interval. (We assume here that the alternative is two-sided, but since the interval is so far from 0, we would still reject H0 in favor of µ1 > µ2.) (c) We assume that the hot dogs are SRSs of each population, and that the distributions are not extremely skewed (or otherwise nonnormal). Both assumptions seem reasonable in this case.

n x s Beef 20 156.850 22.6420 Poultry 17 122.471 25.4831

df t∗ Interval 16 2.120 17.4 to 51.3 cal 32.4 2.0360 18.1 to 50.6 cal

Solutions 175

7.71 (a) SEx2 = 50.74/ √ 20 .= 11.35. SEx1−x2 =

√ 33.892/10+ 50.742/20 .= 15.61.

(b) H0: µ1 = µ2 vs. Ha: µ1 6= µ2; t = 1.249. Using t (9) and t (25.4) distributions, P equals 0.2431 and 0.2229, respectively; the difference is not signiÞcant. (c) −15.8 to 54.8 msec (df = 9) or −12.6 to 51.6 msec (df = 25.4). These intervals had to contain 0 because according to (b), the observed difference would occur in more than 22% of samples when the means are the same; thus 0 would appear in any conÞdence interval with a conÞdence level greater than about 78%.

7.72 If they did this for many separate tests, there would be a fair chance that they would wrongly reject H0 for one or more of their tests. If they are using α = 0.05, and do (e.g.) 20 comparisons, then even if all 20 null hypotheses are true, we ÒexpectÓ to reject one of them (since 0.05 · 20 = 1).

7.73 (a) Using t∗ = 1.660 (df = 100), the interval is $412.68 to $635.58. Using t∗ = 1.6473 (df = 620), the interval is $413.54 to $634.72. Using t∗ = 1.6461 (df = 1249.2), the interval is $413.62 to $634.64. (b) Because the sample sizes are so large (and the sample sizes are almost the same), deviations from the assumptions have little effect. (c) The sample is not really random, but there is no reason to expect that the method used should introduce any bias into the sample. (d) Students without employment were excluded, so the survey results can only (possibly) extend to employed undergraduates. Knowing the number of unreturned questionnaires would also be useful.

7.74 t .= 17.6− 9.5√ 6.342

6 + 1.952

.= 2.99 and df .= ( 6.342

6 + 1.952

)2 1 5

( 6.342

)2 +15 ( 1.952

)2 .= 5.9. 7.75 s2p = 27.75, sp .= 5.2679, and t = 0.6489 with df = 293, so P = 0.5169Ñnot signiÞcant. The conclusion is similar to that in Example 7.16, where we found P > 0.5.

7.76 (a) We test H0: µ1 = µ2 vs. Ha: µ1 > µ2. s2p .= 3.0475, sp .= 1.7457, and t .= 1.663 with df = 40, so P = 0.0520 (similar to the values from Exercise 7.62). We have some evidence of a higher mean hemoglobin level for breast-fed infants, but not quite enough to be signiÞcant at the 5% level. (b) Using t∗ = 2.021, the interval is (x1 − x2)± t∗sp

√ 1 23 + 119 = −0.194 to 1.994.

7.77 With equal variances, t = 0.5376 (df = 16), which gives P = 0.5982. As before, there is no reason to reject H0; skilled and novice rowers seem to have (practically) the same mean weight.

7.78 (a)We test H0: µ1 = µ2 vs. Ha: µ1 6= µ2. s2p .= 2116.18, sp .= 46.002, and t .= 1.094 with df = 28, so P = 0.2831. The difference is not signiÞcant. (b) Using t∗ = 2.048, the interval is (x1 − x2)± t∗sp

√ 1 10 + 120 = −17.0 to 56.0 msec. (c) The t- and P-values

are similar to those in Exercise 7.71, where we had t = 1.249 and P equals either 0.2431 (df = 9) or 0.2229 (df = 25.4). The t-value is smaller here because sp

√ 1 10 + 120

.= 17.82

176 Chapter 7 Inference for Distributions

is slightly bigger than SED .= 15.61; this correspondingly makes P larger. The larger

standard error also makes the conÞdence interval widerÑin 7.71, we had −15.8 to 54.8 msec (df = 9) or −12.6 to 51.6 msec (df = 25.4)

Section 3: Optional Topics in Comparing Distributions

7.79 (a) From an F(9, 20) distribution, F∗ = 2.39. (b) P is between 2(0.025) = 0.05 and 2(0.05) = 0.10; F = 2.45 is signiÞcant at the 10% level but not at the 5% level.

7.80 (a) Comparing to an F(20, 25) distribution, we Þnd that F∗ = 2.30 for p = 0.025 (the critical value for a 5% two-sided test). Since 2.88 > F∗, this is signiÞcant. (b) P is between 2(0.001) = 0.002 and 2(0.01) = 0.02. With Minitab or other software, we Þnd P = 2(0.0067) = 0.0134.

7.81 H0: σ1 = σ2; Ha: σ1 6= σ2. F = 3.0552/2.4232 .= 1.59; referring to an F(33, 43) distribution, we Þnd P = 0.1529 (from the table, use an F(30, 40) distribution and observe that P > 0.1). We do not have enough evidence to conclude that the standard deviations are different.

7.82 Test H0: σ1 = σ2 vs. Ha: σ1 6= σ2. F = (90 · 0.0232)/(45 · 0.0292) .= 1.258; comparing to an F(89, 44) distribution, we Þnd P = 0.4033 (from the table, use an F(60, 40) distribution and observe that P > 0.1). We cannot conclude that the standard deviations are different.

7.83 (a) An F(1, 1) distribution; with a two-sided alternative, we need the critical value for p = 0.025: F∗ = 647.79. This is a very low-power test, since large differences between σ1 and σ2 would rarely be detected. (b) H0: σ1 = σ2 vs. Ha: σ1 6= σ2. F = (s21/s22) = (1.5342/0.77072) .= 3.963. Not surprisingly, we do not reject H0.

7.84 (a) H0: σ1 = σ2; Ha: σ1 6= σ2. (b) Put the larger standard deviation on top: F = (s22/s21) = (0.958952/0.479062) .= 4.007. Comparing to an F(7, 9) distribution, we Þnd 0.05 < P < 0.10; Minitab gives 0.0574. There is some evidence of inequality, but not quite enough to reject H0 (at the 5% level).

7.85 (a) H0: σ1 = σ2; Ha: σ1 6= σ2. (b)Put the larger standard deviation on top: F .= 2.196 from an F(7, 9) distribution, so P > 0.20 (in fact, P = 0.2697).

7.86 F = (87/74)2 .= 1.382; this comes from an F(19882, 19936) distribution, so we compare to F(1000, 1000) and Þnd P < 0.002. (In fact, P is a lot smaller than that.) With such large samples, the estimated standard deviations are very accurate, so ifσ1 = σ2, then s1 and s2 should be nearly equal (and F should be very close to 1).

Solutions 177

7.87 (a) H0: σm = σw; Ha: σm > σw. (b) F = (32.8519/26.4363)2 .= 1.544 from an F(19, 17) distribution. (c) Using the F(15, 17) entry in the table, we Þnd P > 0.10 (in fact, P = 0.1862). We do not have enough evidence to conclude that menÕs SSHA scores are more variable.

7.88 For testing H0: σ1 = σ2 vs. Ha: σ1 6= σ2, we have F = (50.74/33.89)2 .= 2.242 from an F(19, 9) distribution. Using the F(15, 9) entry in the table, we Þnd P > 0.20 (in fact, P = 0.2152). The difference is standard deviations is not signiÞcant.

7.89 df = 198; we reject H0 if t > 1.660 (from the table, with df = 100), or t > 1.6526 (using df = 198). The noncentrality parameter is δ .= 3.2636; the power is about 95% (actually, 0.946), regardless of which t∗ value is used. (The normal approximation agrees nicely with the ÒtrueÓ answer in this case.)

Output from G•Power: Post-hoc analysis for "t-Test (means)", one-tailed: Alpha: 0.0500 Power (1-beta): 0.9460 Effect size "d": 0.4615 Total sample size: 200 (n 1:100, n 2: 100) Critical value: t(198) = 1.6526 Delta: 3.2636

7.90 δ = 300/(650√2/n .= 0.32636√n. The table shows the values of δ, the t∗

values (for df = 48, 98, 148, 198, and 248), and the power computed using the normal approximation (ÒPower1Ó) and the G•Power software (ÒPower2Ó). To reliably detect a difference of 300 g, we should choose at least n = 75. (This number will vary based on what we consider to be Òreliable.Ó)

n δ t∗ Power1 Power2

25 1.6318 1.6772 0.4819 0.4855 50 2.3077 1.6606 0.7412 0.7411 75 2.8263 1.6546 0.8794 0.8787 100 3.2636 1.6526 0.9464 0.9460 125 3.6488 1.6510 0.9771 0.9769

7.91 The standard error is 650 √ 2/n, and df = 2n − 2. The

critical values and margins of error are given in the table. Graph not shown; plot margin of error vs. sample size.

n t∗ m.e. 25 2.0106 369.6 50 1.9845 258.0 75 1.9761 209.8 100 1.9720 181.3 125 1.9696 161.9

7.92 Note: One might reasonably do this computation with a two-sided Ha (since the original alternative of Exercise 7.55 was two-sided), or a one-sided Ha (since the data in that exercise suggested that µ2 > µ1). Both answers are shown. Two-sided Ha: (a) df = 38; we reject H0 if |t | > 2.750 (from the table, with df = 30),

or |t | > 2.7116 (using df = 38). The noncentrality parameter is δ .= 2.2588. Note that since Ha is two-sided, the power is P(|T | > t∗) = P(T < −t∗ or T > t∗); the normal approximation would therefore be P(Z < −t∗ − δ or Z > t∗ − δ). G•Power reports that power .= 0.3391 (see output below). The normal approximation

178 Chapter 7 Inference for Distributions

gives P(Z < −5.01 or Z > 0.4912) .= 0.3116 (using t∗ = 2.750), or P(Z < −4.97 or Z > 0.4528) .= 0.3253 (t∗ = 2.7116). Regardless of the method used, we conclude that we will detect a difference of 0.5 only about one-third of the time. (b) df = 58; we reject H0 if |t | > 2.009 (df = 50), or |t | > 2.0017 (df = 58). The noncentrality parameter is δ .= 2.7664. G•Power reports power .= 0.7765 (see output below). The normal approximation gives P(Z < −4.78 or Z > −0.7574) .= 0.7756 (t∗ = 2.009), or P(Z < −4.77 or Z > −0.7647) .= 0.7778 (t∗ = 2.0017). We will detect a difference of 0.5 about three-fourths of the time. One-sided Ha: (a)With Ha: µ1 < µ2, t∗ = 2.457 (df = 30) or t∗ = 2.4286 (df = 38),

and the power is 0.4412 (G•Power), with normal approximations 0.4214 (df = 30) or 0.4326 (df = 38). (b) t∗ = 1.676 (df = 50) or t∗ = 1.6716 (df = 58), and the power is 0.8619 (G•Power), with normal approximations 0.8622 (df = 50) or 0.8632 (df = 58).

Output from G•Power: Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð 20 players, 1% signiÞcance Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Post-hoc analysis for "t-Test (means)", two-tailed: Alpha: 0.0100 Power (1-beta): 0.3391 Effect size "d": 0.7143 Total sample size: 40 (n 1:20, n 2: 20) Critical value: t(38) = 2.7116 Delta: 2.2588 Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð 30 players, 5% signiÞcance Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Post-hoc analysis for "t-Test (means)", two-tailed: Alpha: 0.0500 Power (1-beta): 0.7765 Effect size "d": 0.7143 Total sample size: 60 (n 1:30, n 2: 30) Critical value: t(58) = 2.0017 Delta: 2.7664

Exercises

7.93 Back-to-back stemplots below. The distributions appear similar; the most striking difference is the relatively large number of boys with low GPAs. Testing the difference in GPAs, we obtain SED

.= 0.4582 and t = −0.91, which is not signiÞcant, regardless of whether we use df = 30 (0.15 < P < 0.20) or 74.9 (P = 0.1811). For the difference in IQs, we Þnd SED .= 3.1138 and t = 1.64, which is fairly strong evidence, although it is not quite signiÞcant at the 5% level: 0.05 < P < 0.10 (df = 30), or P = 0.0503 (df = 56.9).

GPA IQ n x s x s

Boys 47 7.2816 2.3190 110.96 12.121 Girls 31 7.6966 1.7208 105.84 14.271

Solutions 179

GPA: Girls Boys 0 5 1 7 2 4

4 3 689 7 4 068

952 5 0 4200 6 019

988855432 7 1124556666899 998731 8 001112238 95530 9 1113445567

17 10 57

IQ: Girls Boys 42 7

7 79 8

96 8 31 9 03 86 9 77

433320 10 0234 875 10 556667779

44422211 11 00001123334 98 11 556899 0 12 03344 8 12 67788 20 13

13 6

7.94 The table below gives means and standard deviations for the two groups, as well as 95% conÞdence intervals. For H0: µOL = µDL vs. Ha: µOL 6= µDL, we have SED .= 7.1446, and t .= 5.19, which is signiÞcant (P < 0.0005 whether we use df = 14 or df = 28.1). We conclude that offensive linemen are heavier (on the average). Based on the conÞdence intervals, we believe that the

mean weight of offensive linemen is about 20 to 50 lb more than that of defensive linemen.

n x s OL 15 288.2 21.627 DL 18 251.1 18.908

df t∗ Interval 14 2.145 21.8 to 52.4 lb 28.1 2.0480 22.5 to 51.7 lb

OL DL 22 0

5 23 055 24 000555 25 00 26 05

500 27 5 44 28 005

555550 29 50 30

31 32

5 33

7.95 Both distributions have two high outliers. When we include those houses, SED

.= $12, 160.0 and t .= −0.5268. For testing H0: µ3 = µ4 vs. Ha: µ3 < µ4, P

.= 0.3 whether we take df = 21 or df = 35.8, so we have little reason to reject H0. The 95% conÞdence interval for the difference µ3 − µ4 is about −$31,694 to $18,882 (df = 21) or −$31,073 to $18,261 (df = 35.8). Without the outliers, SED

.= $7, 448.37 and t .= −0.1131, so we have little reason to believe that µ3 < µ4 (P

.= 0.455 with either df = 19 or df = 34.5). The 95% conÞdence interval for the difference µ3 − µ4 is about −$16,432 to $14,747 (df = 19) or −$15,971 to $14,286 (df = 34.5).

With outliers Without outliers n x s n x s

3BR 34 $171,717 $36,382.3 32 $164,668 $22,881.5 4BR 22 $178,123 $48,954.4 20 $165,510 $27,970.3

3BR 4BR 1 1 1

3322 1 222 55554444 1 445 777777666 1 667777 9999888888 1 88899

2 00 2 2

7 2 9 2 8

3 1

180 Chapter 7 Inference for Distributions

7.96 Let µ1 and µ2 be the Òtrue mean number of deathsÓ in 1989 and 1990, respectively. (Understanding what this means might make for good class discussion.) The standard error of the difference n1 − n2 (the counts in 1989 and 1990, respectively) would be√ 50+ 47 = √97, and for testing H0: µ1 = µ2 vs. Ha: µ1 > µ2, we Þnd z =

(50− 47)/√97 .= 0.30, which is not signiÞcant. The conÞdence interval for µ1 − µ2 is (50− 47)± 2√97 = −16.7 to 22.7 deaths. Note for instructors: In case you are interested, the assumption underlying this exercise

is that manatee deaths in a given year are a Poisson process (see a probability text for a description). A Poisson distribution with parameterµ has meanµ and standard deviation√ µ; since we have a single observation n from this distribution, our best estimate for the

mean is n, and our best estimate for the standard deviation is √ n.

7.97 It is reasonable to have a prior belief that people who evacuated their pets would score higher, so we test H0: µ1 = µ2 vs. Ha: µ1 > µ2. We Þnd SED .= 0.4630 and t = 3.65, which gives P < 0.0005 no matter how we choose degrees of freedom (115 or 237.0). As one might suspect, people who evacuated their pets have a higher mean score. One might also compute a 95% conÞdence interval for the difference: 0.77 to 2.61

points (df = 115) or 0.78 to 2.60 (df = 237.0).

7.98 (a)We are interested in weight change; the pairs are the ÒbeforeÓ and ÒafterÓ measure- ments. (b) The mean weight change was a loss. The exact amount lost is not speciÞed, but it was big enough that it would rarely happen by chance for an ineffective weight-loss program. (c) Comparing to a t (40) distribution, we Þnd P < 0.0005.

7.99 We test H0: µ1 = µ2 vs. Ha: µ1 < µ2. SED .= 305.817 and t .= −0.7586, which is not signiÞcant regardless of df (in fact P = 0.2256 with df = 57.3). There is not enough evidence to conclude that nitrites decrease amino acid uptake.

7.100 (a) We test H0: µ1 = µ2 vs. Ha: µ1 < µ2. SED

.= 0.2457 and t .= −8.95, which is signiÞcant (P < 0.0005) for either df = 411 or df = 933.8. The mean for the experienced workers is greater. (b) With large sample sizes the t procedure can be used. (c) For a normal distribution, about 95% of all observations fall with 2 standard deviations of the mean; we can (cautiously) use this in spite of the skewness: 37.32 ± 2(3.83) = 29.66 to 44.98Ñabout 30 to 45. (d) The side-by-side boxplots show that the 15th-minute distribution is more symmetric, more spread out, and generally higher than the Þrst minute.

1st minute 15th minute 20

Solutions 181

7.101 (a) The stemplot (after truncating the decimal) shows that the data are left-skewed; there are some low observations, but no particular out- liers. (b) x = 59.58 percent, SEx .= 6.255/

√ 9 .= 2.085, and for df = 8,

t∗ = 2.306, so the interval is 54.8% to 64.4%.

4 9 5 1 5 5 4 5 5 6 0 6 33 6 445

7.102 (a) Òs. e.Ó is standard error (of the mean). To Þnd s, multiply by √ n. (b) No: SED

.= 65.1153 and t .= −0.3532, so P = 0.3624 (df = 82) or 0.3622 (df = 173.9)Ñin either case, there is little evidence against H0. (c) Not very signiÞcantÑ SED

.= 0.1253, t .= −1.1971, and P = 0.2346 (df = 82) or 0.2334 (df = 128.4). (d) 0.39 ± t∗(0.11) = 0.207 to 0.573Ñwhether we use t∗ = 1.664 (df = 80) or t∗ = 1.6636 (df = 82). (e) −0.3119 to 0.0119 (using t (82)) or −0.3114 to 0.0114 (using t (128.4)).

Calories Alcohol n x s x s

Drivers 98 2821 435.58 0.24 0.59397 Conductors 83 2844 437.30 0.39 1.00215

7.103 The similarity of the sample standard deviations suggests that the population standard deviations are likely to be similar. The pooled standard deviation is sp

.= 436.368, and t .= −0.3533, so P = 0.3621 (df = 179)Ñstill not signiÞcant.

7.104 (a) The large sample sizes make the t procedure usable. (b) The F test is not robust against nonnormality, so it should not be used with this distribution.

7.105 No: Counties in California could scarcely be considered an SRS of counties in Indiana.

7.106 (a) Testing H0: µ = 86 vs. Ha: µ < 86, we Þnd t = 83−8610/√40 .= −1.897. With

df = 39, we estimate 0.025 < P < 0.05 (software gives 0.0326). This is fairly strong evidence that the mean is lower. (b) E.g., take several soil samples; use the standard method on half, and the new method on the other half; do a matched pairs analysis on the differences.

7.107 (a) Test H0: µ1 = µ2 vs. Ha: µ1 > µ2; SED .= 16.1870 and t .= 1.1738, so P = 0.1265 (using df = 22) or 0.1235 (df = 43.3). Not enough evidence to reject H0. (b) −14.57 to 52.57 mg/dl (df = 22), or −13.64 to 51.64 mg/dl (df = 43.3). (c) 193 ± (2.060)(68/√26) = 165.53 to 220.47 mg/dl. (d) We are assuming that we have two SRSs from each population, and that underlying distributions are normal. It is unlikely that we have random samples from either population, especially among pets.

182 Chapter 7 Inference for Distributions

7.108 (a) H0: µr = µc; Ha: µr < µc. (b) Use a matched pairs procedure on the (city − rural) differences. (c) There were 26 days when readings were available from both locations; the stemplot of these differences shows two high outliers. (d) We drop the outliers and Þnd x = 1 and s .= 2.106, so t .= 2.33 (df = 23, P = 0.015). This is good evidence that the rural mean is lower (especially given that we have removed the two strongest individual pieces of evidence against H0). If we use the t procedures in spite of the outliers, we get x = 2.192, s = 4.691, and t .= 2.38 (df = 25, P = 0.013). (e) Without the outliers, the 90% conÞdence interval is 0.263 to 1.737; with them, it is 0.621 to 3.764.

−0 32 −0 11110 0 01111111 0 2222222 0 5 0 7 0 1 1 1 5 1 1 8

7.109 We test H0: population median = 0 vs. Ha: population median > 0, or H0: p = 1/2 vs. Ha: p > 1/2, where p is the proportion of (city− rural) differences that are positive. Ignoring missing values and the two ÒzeroÓ differences, there are 6 negative differences and 18 positive differences. If X (the number of positive differences) has a Bin(24, 1/2) distribution, the P-value is P(X ≤ 6) = 0.0113; the normal approximation gives P(Z < −2.45) = 0.0072 (without the continuity correction) or P(Z < −2.25) = 0.0124 (with the continuity correction). In any case, this is strong evidence against H0, indicating that the median city level is higher.

Output from Minitab: Sign test of median = 0.00000 versus G.T. 0.00000

N N* BELOW EQUAL ABOVE P-VALUE MEDIAN City-Rur 26 10 6 2 18 0.0113 1.000

7.110 The stemplot shows the distribution to be fairly symmetric, with a slightly low outlier of 4.88 (it is not an ÒofÞcialÓ outlier). There is nothing to keep us from using the t procedure. x .= 5.4479 and s .= 0.2209; 5.4479 serves as our best estimate of the earthÕs density, with margin of error t∗s/

√ 29 (this is 0.084 for 95% conÞdence, for

example).

48 8 49 50 7 51 0 52 6799 53 04469 54 2467 55 03578 56 12358 57 59 58 5

7.111 Note that SED .= 0.9501 for abdomen skinfolds, while SED .= 0.7877 for thigh

measurements. With 95% conÞdence intervals, for example, the mean abdomen skinfold difference is between 11.62 and 15.38 mm (using df = 103.6). With the same df, the mean thigh skinfold difference is between 9.738 and 12.86 mm.

Solutions 183

7.112 (a) There is a high outlier (2.94 g/mi), but the distribu- tion looks reasonably normal. (b) See the table. Intervals marked with an asterisk (*) were computed using the table value t∗ = 2.704 for df = 40. (c) We test H0: µ = 1; Ha: µ > 1. If we include the outlier, t .= 4.60; without it, t .= 4.64. Either way the P value is very small. To the supervisor, we explain that if the mean NOX emissions were only 1 g/mi, we would almost never see average emissions as high as these. Therefore, we must conclude that mean emissions are higher than 1 g/mi; based on the evidence, we believe that the mean is between about 1.1 and 1.5 g/mi.

n x s Interval All points 46 1.3287 0.4844 1.1366 to 1.5208

or 1.1356 to 1.5218* No outlier 45 1.2929 0.4239 1.1227 to 1.4630

or 1.1220 to 1.4638*

0 455 0 6777 0 899 1 0011111 1 222223333333 1 4444445 1 777 1 888 2 0 2 22 2 2 2 9

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7.113 (a) Back-to-back stemplots and summary statistics below. With a pooled variance, sp .= 83.6388, t .= 3.9969 with df = 222, so P < 0.0001. With unpooled variances,

SED .= 11.6508, t .= 4.0124 with df = 162.2, and again P < 0.0001 (or, with df =

78, we conclude that P < 0.0005). The test for equality of standard deviations gives F .= 1.03 with df 144 and 78; the P-value is 0.9114, so the pooled procedure should be appropriate. In either case, we conclude that male mean SATM scores are higher than female mean SATM scores. A 99% conÞdence interval for the male− female difference (using sp) is 16.36 to 77.13. (b) Back-to-back stemplots and summary statistics below. With a pooled variance,

sp .= 92.6348, t .= 0.9395 with df = 222, so P = 0.3485. With unpooled variances,

SED .= 12.7485, t .= 0.9547 with df = 162.2, so P = 0.3411 (or, with df = 78,

P = 0.3426). The test for equality of standard deviations gives F .= 1.11 with df 144 and 78; the P-value is 0.6033, so the pooled procedure should be appropriate. In either case, we cannot see a difference between male and female mean SATV scores. A 99% conÞdence interval for the male− female difference (using sp) is −21.49 to 45.83. (c) The results may generalize fairly well to students in different years, less well to

students at other schools, and probably not very well to college students in general.

184 Chapter 7 Inference for Distributions

SATM SATV n x s x s

Men 145 611.772 84.0206 508.841 94.3485 Women 79 565.025 82.9294 496.671 89.3849

MenÕs SATM WomenÕs SATM 3 0 3 5

400 4 1334 99999888776 4 56777888999

44444333322222111000 5 0111123334 99999988888877766555555555 5 55555556667777788899999

444444444433333332222222211100000000 6 00011222233334444 999999987777776666655555 6 55555789

3222211100000 7 1124 77766655555 7

0 8

MenÕs SATV WomenÕs SATV 98 2 9

4322 3 33 9999988766 3 55669

444444444332111100000 4 0122223333444 999998888888888877777666666555 4 5666666777777888888899999 44443333322221110000000000000 5 01111122334

998888777777666666555 5 566777777889 43333332111100000 6 0000

9987775 6 668 420 7 00 6 7 5

7.114 (a) A stemplot of the differences (right) shows no outliers (although it also does not look very normal). D .= 0.0046 and s .= 0.01487. (b) With df = 49, t∗ = 2.0096, giving the interval 0.00037 to 0.00883. With df = 40 and t∗ = 2.021, we get 0.00035 to 0.00885. (c) Testing H0: µ = 0 vs. Ha: µ 6= 0, we Þnd t .= 2.19, which has 0.02 < P < 0.04 (using df = 40) or P .= 0.034 (using df = 49). There is fairly strong evidence that the mean difference is not 0.

−2 0000000 −1 0000000 −0 0000 0 00000 1 0000000000000 2 00000000000 3 000

7.115 (a) We test H0: µB = µD vs. Ha: µB < µD. Pooling is appropriate; sp

.= 6.5707. [If we do not pool, SED

.= 1.9811.] Whether or not we pool, t .= 2.87 with df = 42 [or 21, or 39.3], so P = 0.0032 [or 0.0046, or 0.0033]. We conclude that the mean score using DRTA is higher than the mean score with the Basal method. The difference in the average scores is 5.68; a 95% conÞdence interval for the difference in means is about 1.7 to 9.7 points.

n x s Basal 22 41.0455 5.63558 DRTA 22 46.7273 7.38842 Strat 22 44.2727 5.76675

(b)We test H0: µB = µS vs. Ha: µB < µS. Pooling is appropriate; sp .= 5.7015. [If we do not pool, SED

.= 1.7191.] Whether or not we pool, t .= 1.88 with df = 42 [or 21, or 42.0], so P = 0.0337 [or 0.0372, or 0.0337]. We conclude that the mean score using

Solutions 185

Strat is higher than the Basal mean score. The difference in the average scores is 3.23; a 95% conÞdence interval for the difference in means is about −0.24 to 6.7 points.

7.116 Answers will vary with choice of α, and with whether Ha is one- or two-sided. See the table for some combinations. We would reject H0: µb = µg in favor

of Ha: µb 6= µg if |t | ≥ t∗, or |xg − xb| ≥ t∗σ √ 2/n, where t∗ varies with our

choice of α, and with df = 2n − 2. The power against the (two-sided) alternative d = |µg − µb| = 0.4 is P(|T | > t∗) = P(T < −t∗ or T > t∗); the normal approximation would therefore be P(Z < −t∗ − δ or Z > t∗ − δ). The noncentrality parameter is δ = 0.4/(σ√1/n + 1/n) = 0.16√n/2. From this point, one must either use special software (like G•PowerÑoutput

below) or trial and error to Þnd the appropriate n. Since sample sizes end up being fairly large, the normal approximation is quite goodÑboth for estimating the power, and also for approximating t∗ using z∗ from a normal distribution. For example, with α = 0.05, t∗ .= z∗ = 1.96, and we Þnd for n = 613,

Power .= P(Z < −4.7611 or Z > −0.8411) .= 0.7999 while for n = 614,

Power .= P(Z < −4.7634 or Z > −0.8434) .= 0.8005 If we use the one-sided alternative Ha, the power is P(T > t∗) and the normal approximation is P(Z > t∗ − δ). For example, with α = 0.05 and n = 483, t∗ .= z∗ = 1.645, and Power .= P(Z > −4.7382) .= 0.79995, and with n = 484, Power .= P(Z > −4.7405) .= 0.80067. Note that G•Power reports the total sample size; divide this by 2 to get n.

Ha: µb < µg Ha: µb 6= µg α = 0.05 484 615 α = 0.01 786 915

Output from G•Power: A priori analysis for "t-Test (means)", two-tailed: Alpha: 0.0500 Power (1-beta): 0.8000 Effect size "d": 0.1600 Total sample size: 1230 Actual power: 0.8005 Critical value: t(1228) = 1.9619 Delta: 2.8057

A priori analysis for "t-Test (means)", one-tailed: Alpha: 0.0500 Power (1-beta): 0.8000 Effect size "d": 0.1600 Total sample size: 968 Actual power: 0.8002 Critical value: t(966) = 1.6464 Delta: 2.4890

7.117 The table and plot (below) show the power computed by G•Power for |µ1 − µ2| varying between 0.01 and 0.10. These values can also be approximated using the normal distribution; e.g., for a dif-

ference of 0.05, we will reject H0 if |t | > 2.0244 (with a two-sided alternative). The

186 Chapter 7 Inference for Distributions

noncentrality parameter is δ .= 2.2588; the power is approximately

P(Z < −t∗ − δ or Z > t∗ + δ) = 0.5927

or about 60%. One-sided Two-sided

Diff δ Power Power 0.01 0.4518 0.1149 0.0725 0.02 0.9035 0.2244 0.1425 0.03 1.3553 0.3769 0.2620 0.04 1.8070 0.5517 0.4214 0.05 2.2588 0.7168 0.5954 0.06 2.7105 0.8454 0.7522 0.07 3.1623 0.9279 0.8690 0.08 3.6140 0.9715 0.9408 0.09 4.0658 0.9905 0.9773 0.10 4.5175 0.9974 0.9927

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two-ta iled

7.118 Some software handles this task more easily than others. Shown (below, left) is one possible graph; note that the critical values for df ≤ 5 are missing from the graph, in order to show the detail. We see that the critical values get closer to 1.96 as df grows.

For 7.118.

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7.119 Plot above, right. The margin of error is t∗ s/ √ n = t∗/√n, taking t∗ from t distri-

butions with 4, 9, 14, . . . , 99 degrees of freedom. For df ≤ 29, these values are in the table; in order to get all the t∗ values, software is needed. For reference, t∗ = 2.7764 with df = 4 and t∗ = 1.9842 with df = 99; the margin of error gradually decreases from 1.2416 when n = 5 to 0.19842 when n = 100.

Solutions 187

Chapter 8 Solutions

Section 1: Inference for a Single Proportion

8.1 (a) No: n(1− p̂) = 30(0.1) = 3 is less than 10. (b) Yes: n p̂ = n(1− p̂) = 25(0.5) = 12.5. (c) No: n p̂ = 100(0.04) = 4 is less than 10. (d) Yes: n p̂ = (600)(0.6) = 360 and n(1− p̂) = 600(0.4) = 240.

8.2 (a) p̂ = 1584 .= 0.1786, and SE p̂ =

√ p̂(1− p̂)/84 .= 0.0418. (b) p̂ ± 1.645 SE p̂ =

0.1098 to 0.2473.

8.3 No: Some of those who lied about having a degree may also have lied about their major. At most 24 applicants lied about having a degree or about their major.

8.4 (a) p̂ = 5421711 .= 0.3168; about 31.7% of 15+ year-old bicyclists killed between

1987 and 1991 had alcohol in their systems at the time of the accident. (b) SE p̂ =√ p̂(1− p̂)/1711 .= 0.01125; the interval is p̂± 1.960 SE p̂ = 0.2947 to 0.3388. (c) No:

We do not know, for example, what percentage of cyclists who were not involved in fatal accidents had alcohol in their systems.

8.5 p̂ = 3861711 .= 0.2256, and SE p̂ =

√ p̂(1− p̂)/1711 .= 0.0101, so the 95% conÞdence

interval is 0.2256± (1.96)(0.0101), or 0.2058 to 0.2454.

8.6 (a) p̂ = 421500 = 0.842, and SE p̂ = √ p̂(1− p̂)/500 .= 0.0163. (b) 0.842 ±

(1.96)(0.0163), or 0.8100 to 0.8740.

8.7 p̂ = 86100 = 0.86, and SE p̂ = √ p̂(1− p̂)/100 .= 0.0347, so the 95% conÞdence interval

is 0.86± (1.96)(0.0347), or 0.7920 to 0.9280.

8.8 p̂ = 41216 .= 0.1898, and SE p̂ =

√ p̂(1− p̂)/216 .= 0.0267, so the 99% conÞdence

interval is 0.1898± (2.576)(0.0267), or 0.1211 to 0.2585.

8.9 p̂ = 132200 = 0.66, and SE p̂ = √ p̂(1− p̂)/200 .= 0.0335, so the 95% conÞdence interval

is 0.66± (1.96)(0.0335), or 0.5943 to 0.7257.

8.10 (a) No: np0 = 4 is less than 10. (b) Yes: np0 = 60 and n(1 − p0) = 40. (c) No: n(1− p0) = 4 is less than 10. (d) Yes: np0 = 150 and n(1− p0) = 350.

8.11 We want to know if p is signiÞcantly different from 36%, so we test H0: p = 0.36 vs. Ha: p 6= 0.36. We have p̂ = 0.38; under H0, σ p̂ =

√ (0.36)(0.64)/500 .= 0.0215,

so z = 0.38−0.360.0215 = 0.9317. This is clearly not signiÞcant (in fact, P .= 0.35).

188 Chapter 8 Inference for Proportions

8.12 (a)Wewant to know if p (the proportion of urban respondents) is signiÞcantly different from 64%, so we test H0: p = 0.64 vs. Ha: p 6= 0.64. (b) We have p̂ = 0.62; under H0, σ p̂ =

√ (0.64)(0.36)/500 .= 0.0215, so z = 0.62−0.640.0215 = −0.9317. This is clearly not

signiÞcant (in fact, P .= 0.35). (c) The results are the same as the previous exercise; in general, performing a test on a proportion p will give the same results as the equivalent test on p′ = 1− p.

8.13 (a) p̂ = 7501785 .= 0.4202, and SE p̂ =

√ p̂(1− p̂)/200 .= 0.0117, so the 99% conÞdence

interval is 0.4202± (2.576)(0.0117), or 0.3901 to 0.4503. (b)YesÑthe interval does not include 0.50 or more. (c) n =

( 2.576 0.01

)2 (0.4202)(0.5798) .= 16166.9Ñuse n = 16, 167.

8.14 n = ( 1.96 0.03

)2 (0.44)(0.56) .= 1051.7Ñuse n = 1052.

8.15 p̂ = 1375 = 0.173, and SE p̂ = √ p̂(1− p̂)/75 .= 0.0437, so the 95%conÞdence interval

is 0.173± (1.96)(0.0437), or 0.0877 to 0.2590.

8.16 We want to know if p (the proportion of respondents with no children) is signiÞcantly different from 48%, so we test H0: p = 0.48 vs. Ha: p 6= 0.48. We have p̂ = 0.44; under H0, σ p̂ =

√ (0.48)(0.52)/500 .= 0.0223, so z = 0.44−0.480.0223

.= −1.79. This has P .= 2(0.0367) = 0.0734; we donÕt have quite enough evidence to conclude that the telephone survey reached households without children in a different proportion than such households are found in the population.

8.17 (a) Testing H0: p = 0.5 vs. Ha: p =6= 0.5, we have p̂ = 506710000 = 0.5067, and σ p̂ =

√ (0.5)(0.5)/10000 = 0.005, so z = 0.00670.005 = 1.34. This is not signiÞcant at

α = 0.05 (or even α = 0.10). (b) SE p̂ = √ p̂(1− p̂)/10000 .= 0.005, so the 95%

conÞdence interval is 0.5067± (1.96)(0.005), or 0.4969 to 0.5165.

8.18 (a)WetestH0: p = 0.5vs.Ha: p > 0.5; p̂ = 3150 = 0.62, andσ p̂ = √ (0.5)(0.5)/50 .=

0.0707, so z = 0.120.0707 .= 1.70, and P = 0.0446. This is signiÞcant at the 5% levelÑbut

just barely. If one more person had preferred instant, the results would not have been

signiÞcant. (b) SE p̂ = √ p̂(1− p̂)/50 .= 0.0686, so the 90% conÞdence interval is

0.62± (1.645)(0.0686), or 0.5071 to 0.7329.

8.19 (a) H0: p = 0.384 vs. Ha: p > 0.384. (b) p̂ = 2540 = 0.625, and σ p̂ =√ (0.384)(0.616)/40 .= 0.0769, so z = 0.625−0.3840.0769

.= 3.13. (c)Reject H0 since z > 1.645; P = 0.0009. (d) SE p̂ =

√ p̂(1− p̂)/40 .= 0.0765, so the 90% conÞdence interval is

0.625 ± (1.645)(0.0765), or 0.4991 to 0.7509. There is strong evidence that Leroy has improved. (e) We assume that the 40 free throws are an SRS; more speciÞcally, each shot represents an independent trial with the same probability of success, so the number

Solutions 189

of free throws made has a binomial distribution. To use the normal approximation, we also need (for the test) np0 = 15.36 > 10 and n(1 − p0) = 24.64 > 10, and (for the conÞdence interval) n p̂ = 25 > 10 and n(1− p̂) = 15 > 10.

8.20 n = ( 1.96 0.05

)2 (0.35)(0.65) .= 349.6Ñuse n = 350.

8.21 n = ( 1.96 0.05

)2 (0.2)(0.8) .= 245.9Ñuse n = 246.

8.22 (a) Higher: For more conÞdence, we need more information. (b) Higher: For more precision, we need more information. (c) Lower: Standard errors are smaller for more extreme p∗ values (close to 0 or 1). (d) Same: This has no effect on margin of error.

8.23 n = ( 1.645 0.04

)2 (0.7)(0.3) .= 355.2Ñuse n = 356. With p̂ = 0.5, SE p̂ .= 0.0265, so the

true margin of error is (1.645)(0.0265) = 0.0436.

8.24 n = ( 2.576 0.015

)2 (0.2)(0.8) .= 4718.8Ñuse n = 4719. With p̂ = 0.1, SE p̂ .= 0.00437, so

the true margin of error is (2.576)(0.00437) = 0.0112.

8.25 (a) The margins of error are 1.96 √ p̂(1− p̂)/100 = 0.196

√ p̂(1− p̂).

p̂ 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 m.e. .0588 .0784 .0898 .0960 .0980 .0960 .0898 .0784 .0588

(b) No: n p̂ = 100(0.04) = 4 is less than 10.

8.26 The margins of error are 1.96 √ p̂(1− p̂)/500 = 0.196

√ p̂(1− p̂)/5.

p̂ 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 m.e. .0263 .0351 .0402 .0429 .0438 .0429 .0402 .0351 .0263

With n = 500, we could use a normal approximation with p̂ = 0.04, since n p̂ = 500(0.04) = 20. The letter to the benefactor should mention the greatly reduced margins of error.

Section 2: Comparing Two Proportions

8.27 (a) p̂f = 4860 = 0.8, so SE p̂ .= 0.05164 for females. p̂m = 52132 = 0.39, so SE p̂

.= 0.04253 for males. (b) SED =

√ 0.051642 + 0.042532 .= 0.0669, so the interval is

( p̂f − p̂m) ± (1.96)(0.0669), or 0.2749 to 0.5372. There is (with high conÞdence) a considerably higher percentage of juvenile references to females than to males.

8.28 (a) We have p̂f = 27191 .= 0.1414 and p̂m = 5151520

.= 0.3388, which gives SED =

√ p̂f(1− p̂f)/191+ p̂m(1− p̂m)/1520 .= 0.02798. The interval is ( p̂f − p̂m) ±

190 Chapter 8 Inference for Proportions

(1.645)(0.02798), or −0.2435 to −0.1514 (i.e., the female proportion is substantially lower). (b) The female SE p̂ contributes the greater amount, because there were consid- erably fewer women in the sampleÑdividing by 1520 makes the male SE p̂ very small by comparison.

8.29 (a) We have p̂1 = 1584 .= 0.1786 and p̂2 = 21106

.= 0.1981, which gives SED =

√ p̂1(1− p̂1)/84+ p̂2(1− p̂2)/106 .= 0.0570, so the interval is ( p̂1 − p̂2) ±

(1.645)(0.0570), or −0.1132 to 0.0742. Since this interval includes 0, we have little evidence here to suggest that the two proportions are different.

8.30 Testing H0: pf = pm vs. Ha: pf 6= pm, we have p̂f = 0.8, p̂m = 0.39, and p̂ = 48+5260+132

.= 0.5208. This gives sp = √ p̂(1− p̂)( 160 + 1132)

.= 0.0778, so z = ( p̂f − p̂m)/sp

.= 5.22. With P < 0.0001, we have strong evidence that the two proportions are different.

8.31 Test H0: pf = pm vs. Ha: pf 6= pm (assuming we have no belief, before seeing the data, that the difference will lie in a particular directionÑe.g., that pf < pm). The pooled estimate of p is p̂ = 27+515191+1520

.= 0.3168, which gives sp .= 0.0357, so z = 0.1414−0.33880.0357 .=

−5.53. This gives P < 0.0001; it is signiÞcant at any α, so we conclude (with near certainty) that there is a difference between the proportions.

8.32 Testing H0: p1 = p2 vs. Ha: p1 6= p2, we have p̂1 .= 0.1786, p̂2 .= 0.1981, and p̂ = 15+2184+106

.= 0.1895. This gives sp = √ p̂(1− p̂)( 184 + 1106)

.= 0.0572, so z = ( p̂1 − p̂2)/sp .= −0.34. With P .= 2(0.3669) = 0.7338, we have no reason to believe that the two proportions are different.

8.33 (a) H0: p1 = p2; Ha: p1 6= p2. (b) p̂1 = 64160 = 0.4, p̂2 = 89261 .= 0.3410, and

p̂ = 64+89160+261 .= 0.3634, which gives sp .= 0.0483, so z = ( p̂1 − p̂2)/sp .= 1.22. This

gives P .= 2(0.1112) = 0.2224; there is little evidence to suggest a difference between rural and urban households. (c) SED

.= 0.04859, so the interval is 0.0590 ± 0.0799, or −0.0209 to 0.1389.

8.34 (a) p̂h = 4980 = 0.6125 and p̂a = 4382 .= 0.5244. (b) SED .= 0.0775. (c) The interval

is ( p̂h − p̂a)± (1.645)(0.0775) = −0.0394 to 0.2156. Since this interval contains 0, we are not convinced that the true proportions are different.

8.35 (a) p̂1 = 263263+252 .= 0.5107 and p̂2 = 260260+377

.= 0.4082. (b) SED .= 0.0294. (c) 0.1025 ± (2.576)(0.0294), or 0.0268 to 0.1783. Since 0 is not in this interval, there appears to be a real difference in the proportions (though it might be fairly small).

Solutions 191

8.36 (a) p̂ = 49+4380+82 .= 0.5679. (b) sp .= 0.0778. (c) H0: ph = pa vs. Ha: ph > pa.

(d) z = ( p̂h− pa)/sp .= 1.13, so P = 0.1292. There is not enough evidence to conclude that the Yankees were more likely to win at home.

8.37 (a) p̂ = 263+260263+252+260+377 .= 0.4540. (b) sp .= 0.0295. (c) H0: p1 = p2 vs. Ha:

p1 6= p2 (assuming we have no prior information about which might be higher). (d) z = ( p̂1− p2)/sp .= 3.47, which gives P .= 2(0.0003) = 0.0006. We reject H0 and conclude that there is a real difference in the proportions.

8.38 Note that the rules of thumb for the normal approximation are not satisÞed here (the number of birth defects is less than 10). Additionally, one might call into question the assumption of independence, since there may have been multiple births to the same set of parents included in these counts (either twins/triplets/etc., or ÒordinaryÓ siblings). If we carry out the analysis in spite of these issues, we Þnd p̂1 = 16414

.= 0.03865 and p̂2 = 3228

.= 0.01316. We might then Þnd a 95% conÞdence interval: SED .= 0.01211, so the interval is p̂1 − p̂2 ± (1.96)(0.01211) = 0.00175 to 0.04923. (Note that this does not take into account the presumed direction of the difference.) We could also perform a signiÞcance test of H0: p1 = p2 vs. Ha: p1 > p2: p̂ = 19642

.= 0.02960, sp .= 0.01398, z .= 1.82, P = 0.0344. Both the interval and the signiÞcance test suggest that the two proportions are different,

but we must recognize that the issues noted above make this conclusion questionable.

8.39 We have p̂1 = 3814096 .= 0.0930 and p̂2 = 828

.= 0.2857; test H0: p1 = p2 vs. Ha: p1 < p2. p̂ = 381+84096+28

.= 0.0943 and s .= 0.0554; so z .= −3.48, which gives P .= 0.0002. We reject H0 and conclude that there is a real difference in the proportions; abnormal chromosomes are associated with increased criminality. (One could also construct, e.g., a 95% conÞdence interval, but this does not take into account the presumed direction of the difference.) Note that here, as in the previous exercise, one of our counts is less than 10, meaning

that the normality assumption might not be valid for the abnormal-chromosome group.

8.40 We test H0: p1 = p2 vs. Ha: p1 6= p2. p̂1 .= 0.6030, p̂2 .= 0.5913, p̂ .= 0.5976; therefore, sp

.= 0.0441 and z .= 0.27, so P .= 2(0.3936) = 0.7872ÑH0 is quite plausible given this sample.

8.41 p̂1 = 104267 .= 0.3895 and p̂2 = 75230

.= 0.3261; SED .= 0.04297, so the conÞdence interval is 0.0634± (1.96)(0.04297), or −0.0208 to 0.1476.

8.42 (a) H0: pm = pf vs. Ha: pm 6= pf. p̂m .= 0.9009, p̂f .= 0.8101, and p̂ .= 0.8574. Then sp

.= 0.01790 and z .= 5.07, so P < 0.0001. There is strong evidence that the two proportions differ. (b) SED

.= 0.01795, so the interval is pm− pf± (2.576)(0.01785) = 0.0445 to 0.1370. Whether this difference is ÒimportantÓ or not is a matter of opinion.

192 Chapter 8 Inference for Proportions

8.43 (a)For H0: p1 = p2 vs. Ha: p1 6= p2, we have p̂1 = 3583 .= 0.4217, p̂2 = 15136

.= 0.1103, p̂ = 35+1583+136

.= 0.2283, and sp .= 0.0585. Then z .= 5.33, so P < 0.0001. We reject H0 and conclude that there is a real difference in the proportions for the two shield types. (b) SED

.= 0.0605, so the interval is 0.3114± (1.645)(0.0605), or 0.2119 to 0.4109. The ßip-up shields are much more likely to remain on the tractor.

8.44 (a) p̂1 .= 0.8077, p̂2 .= 0.5584. (b) SED .= 0.0721; the interval is 0.1080 to 0.3905.

(c) H0: p1 = p2; Ha: p1 > p2. p̂ .= 0.6839, and sp .= 0.0747, so z .= 3.34, so P .= 0.0004. There is strong evidence that aspirin was effective.

8.45 (a) p̂1 = 918 = 0.5 and p̂2 = 1318 = 0.72. (b) −0.2222± (1.645)(0.1582), or −0.4825 to 0.0380. (c) H0: p1 = p2; Ha: p1 < p2. p̂ = 9+1318+18 = 0.61, sp

.= 0.1625, and z .= −1.37; the P-value is 0.0853. There is some evidence that the proportions are different, but it is not signiÞcant at the 5% level; if the two proportions were equal, we would observe such a difference between p̂1 and p̂2 about 8.5% of the time.

8.46 (a)Again testing H0: pm = pf vs. Ha: pm 6= pf, we have p̂m = 0.9, p̂f .= 0.8082, and p̂ .= 0.8562. Then sp .= 0.0568 and z .= 1.62, so P .= 0.1052. We cannot conclude that the two proportions differ. (b)With the larger sample size, the difference was signiÞcant; a smaller sample size means more variability, so large differences are more likely to happen by chance.

8.47 (a) p̂1 = 0.5, p̂2 = 0.72, p̂ = 0.61, sp .= 0.1149, and z .= −1.93; the P-value is 0.0268 (recall the alternative hypothesis is one-sided). This is signiÞcant evidence of a differenceÑspeciÞcally, that p1 < p2. (b) With the larger sample size, the difference p̂1 − p̂2 = −0.2 is less likely to have happened by chance.

Exercises

8.48 We test H0: pf = pm vs. Ha: pf 6= pm for each text, where, e.g., pf is the proportion of juvenile female references. We can reject H0 for texts 2, 3, 6, and 10. The last three texts do not stand out as different from the Þrst seven. Texts 7 and 9 are notable as the only two with a majority of juvenile male references, while 6 of the 10 texts had juvenile female references a majority of the time.

Text p̂f p̂m p̂ z P 1 .4000 .2059 .2308 0.96 .3370 2 .7143 .2857 .3286 2.29 .0220 3 .4464 .2154 .3223 2.71 .0068 4 .1447 .1210 .1288 0.51 .6100 5 .6667 .2791 .3043 1.41 .1586 6 .8000 .3939 .5208 5.22 .0000 7 .9500 .9722 .9643 −0.61 .5418 8 .2778 .1818 .2157 0.80 .4238 9 .6667 .7273 .7097 −0.95 .3422 10 .7222 .2520 .3103 4.04 .0000

Solutions 193

8.49 The proportions, z-values, and P-values are

Text 1 2 3 4 5 6 7 8 9 10 p̂ .872 .900 .537 .674 .935 .688 .643 .647 .710 .876 z 4.64 6.69 0.82 5.31 5.90 5.20 3.02 2.10 6.60 9.05 P ≈ 0 ≈ 0 .413 ≈ 0 ≈ 0 ≈ 0 .002 .036 ≈ 0 ≈ 0

We reject H0: p = 0.5 for all texts but Text 3 and (perhaps) Text 8. (And maybe also for Text 7, if we are using, e.g., BonferroniÕs procedureÑsee Chapter 6). The last three texts do not seem to be any different from the Þrst seven; the gender of

the author does not seem to affect the proportion.

8.50 The null hypothesis is H0: p1 = p2; the alternative might reasonably be p1 6= p2 or p1 < p2Ñthe latter since we might suspect that older children are more likely to sort correctly. p̂1 = 0.2, p̂2 .= 0.5283, p̂ .= 0.3689; therefore, sp .= 0.0951 and z .= −3.45. Whichever alternative we use, the P-value is small (0.0003 or 0.0006), so we conclude that the older children are better at sorting. The standard error for the conÞdence interval is SED

.= 0.0889, so the interval is p̂1 − p̂2 ± (1.645)(0.0889) = −0.4745 to −0.1821.

8.51 No: The percentage was based on a voluntary response sample, and so it cannot be assumed to be a fair representation of the population. Such a poll is likely to draw a higher-than-actual proportion of people with a strong opinion, especially a strong neg- ative opinion. A conÞdence statement like the one given is not reliable under these circumstances.

8.52 Test H0: p = 0.11 vs. Ha: p < 0.11. p̂ = 0.053 and σ p̂ .= 0.01806, so z .= −3.14. This gives P .= 0.0008; we have strong evidence that the nonconformity rate is lower (i.e., that the modiÞcation is effective). Here we assume that each item in our sample is independent of the others.

8.53 p̂ = 16300 = 0.053 and SE p̂ .= 0.0130; the conÞdence interval for p is 0.053 ±

(1.96)(0.0130), or 0.0279 to 0.0788. Note that the conÞdence interval for p − p0 is not constructed using the procedure

for a difference of two proportions, since p0 is not based on a sample, but is taken as a constant. This conÞdence interval is found by subtracting 0.11 from the previous interval: −0.0821 to −0.0312. In other words, we are 95% conÞdent that the new process has a nonconformity rate that is 3.12% to 8.21% lower than the old process.

8.54 (a) p̂ = 444950 .= 0.4674 and SE p̂ .= 0.0162; the conÞdence interval for p is 0.4674 ±

(2.576)(0.0162) = 0.4257 to 0.5091. (b) Between 42.6% and 50.9% of students change their majors. (c) We expect that between 14,900 and 17,800 students will change their majors.

194 Chapter 8 Inference for Proportions

8.55 (a) p0 = 214851 .= 0.2515. (b) p̂ = 1530 = 0.5. (c) H0: p = p0; Ha: p > p0.

σ p̂ = √ p0(1− p0)/30 .= 0.0792 and z = (0.5 − p0)/σ p̂ .= 3.14, so P = 0.0008; we

reject H0 and conclude that women are more likely to be among the top students than their proportion in the class.

8.56 We test H0: p1 = p2 vs. Ha: p1 6= p2. p̂1 .= 0.4719, p̂2 .= 0.6054, p̂ .= 0.5673; therefore, sp

.= 0.0621, z .= −2.15, and P = 0.0316. Wehave fairly strong evidence that the proportions of vegetarians differ between black and white Seventh-Day Adventists. We should not assume that this extends to blacks and whites in general.

8.57 (a) p̂1 = 553338 .= 0.0165 and p̂2 = 212676

.= 0.0078; SED .= 0.0028, so the conÞdence interval is 0.0086±(1.96)(0.0028), or 0.0032 to 0.0141. (b) H0: p1 = p2; Ha: p1 > p2. p̂ = 55+213338+2676

.= 0.0126, and sp .= 0.0029. Then z .= 2.98, so P = 0.0014. We reject H0; this difference is unlikely to occur by chance, so we conclude that high blood pressure is associated with a higher death rate.

8.58 For the British study, p̂1 = 1483429 .= 0.0432 and p̂2 = 791710

.= 0.0462. To test H0: p1 = p2 vs. Ha: p1 6= p2, we compute p̂ .= 0.0442, sp .= 0.0061, and z .= −0.50, so P = 2(0.3085) = 0.617Ñthere is very little evidence of a difference. For the American study, p̂1 = 10411037

.= 0.0094 and p̂2 = 18911034 .= 0.0171. Testing the

same hypotheses as above, we compute p̂ .= 0.0133, s .= 0.0015, and z .= −5.00, so P is essentially 0. This is strong evidence of a difference: aspirin reduced the risk of a fatal heart attack. The difference in the conclusions can be attributed to the larger sample size for the

American study (important for something as rare as a heart attack), as well as the shorter duration of the study and the lower dosage (taking the aspirin every other day rather than every day).

8.59 (a) H0: p1 = p2; Ha: p1 6= p2. p̂1 = 2882 .= 0.3415, p̂2 = 3078

.= 0.3846, and p̂ = 0.3625, so sp .= 0.0760. Then z .= −0.57 and P .= 0.5686. (b) Gastric freezing is not signiÞcantly more (or less) effective than a placebo treatment.

8.60 The pooled estimate of p is p̂ = (n p̂1 + n p̂2)/(n + n) = ( p̂1 + p̂2)/2 = 0.5, so sp =

√ p̂(1− p̂)(1/n + 1/n) =√

0.5/n, and z = (0.6 − 0.4)/sp = 0.2 √ 2n. The P-value is

2P(Z > z). The difference p̂1 − p̂2 is not signiÞcant for small n, but

it grows more and more signiÞcant as n increases.

n z P 15 1.095 0.2733 25 1.414 0.1573 50 2.000 0.0455 75 2.449 0.0143 100 2.828 0.0047 500 6.325 0.0000

Solutions 195

8.61 SED = √ p̂1(1− p̂1)/n1 + p̂2(1− p̂2)/n2 =

√ 0.24/n + 0.24/n = √0.48/n. With z∗ = 1.96, the

95% conÞdence interval is 0.2± 1.96√0.48/n, and the margin of error is 1.96

√ 0.48/n.

The interval narrows as n increases.

n CI m.e. 15 −0.151 to 0.551 0.351 25 −0.072 to 0.472 0.272 50 0.008 to 0.392 0.192 75 0.043 to 0.357 0.157 100 0.064 to 0.336 0.136 500 0.139 to 0.261 0.061

8.62 SED = √ p̂1(1− p̂1)/n1 + p̂2(1− p̂2)/n2 =

√ 0.25/n + 0.25/n = √0.5/n, and the margin of error is

2.576 SED. [Note that when n = 10, the normal approximation should not really be used: n p̂ = n(1− p̂) = 5.] The margin of error decreases as n increases (speciÞcally, it is

inversely proportional to √ n).

n m.e. 10 0.5760 30 0.3326 50 0.2576 100 0.1822 200 0.1288 500 0.0815

8.63 (a) The margin of error is z∗ √ 0.5(1− 0.5)/n + 0.5(1− 0.5)/n = z∗√0.5/n. With

z∗ = 1.96, this means we need to choose n so that 1.96√0.5/n ≤ 0.05. The smallest such n is 769. (b) Solving z∗

√ 0.5/n ≤ m gives n ≥ 0.5(z∗/m)2.

8.64 The margin of error is 1.645 √ 0.5(1− 0.5)

20 + 0.5(1− 0.5)

n2 = 1.645√0.0125+ 0.25/n2.

We therefore need to solve 1.645 √ 0.0125+ 0.25/n2 = 0.1Ñbut there is no such value

of n2 (except n2 .= −28.4, which makes no sense here). No matter how big n2 is, the

margin of error will always be greater than 1.645 √ 0.0125 .= 0.1840.

8.65 It is likely that little or no useful information would come out of such an experiment; the proportion of people dying of cardiovascular disease is so small that out of a group of 200, we would expect very few to die in a Þve- or six-year period. This experiment would detect differences between treatment and control only if the treatment was very effective (or dangerous)Ñi.e., if it almost completely eliminated (or drastically increased) the risk of CV disease.

8.66 (a) p0 = 143,611181,535 .= 0.7911. (b) p̂ = 339870

.= 0.3897, σ p̂ .= 0.0138, and z = ( p̂ − p0)/σ p̂

.= −29.1, so P .= 0 (regardless of whether Ha is p > p0 or p 6= p0). This is very strong evidence against H0; we conclude that Mexican Americans are underrepresented on juries. (c) p̂1 = 339870

.= 0.3897, while p̂2 = 143,611−339181,535−870 .= 0.7930. Then p̂ .= 0.7911

(the value of p0 from part (a)), sp = 0.0138, and z .= −29.2Ñand again, we have a tiny P-value and reject H0.

196 Chapter 9 Inference for Two-Way Tables

Chapter 9 Solutions

9.1 (a) At right.

(b) The expected counts are (48)(511) 1317 = 18.624,

(48)(806) 1317 = 29.376,

(1269)(511) 1317 = 492.376, and (1269)(806)1317 = 776.624. Then

X 2 = (7− 18.624) 2

18.624 + (41− 29.376)2

29.376 + (504− 492.376)2

492.376 + (765− 776.624)2

776.624 = 12.303. Comparing to a χ2(1) distribution, we Þnd P < 0.0005; we conclude that there is an association between age and whether or not the employee was terminatedÑ speciÞcally, older employees were more likely to be terminated.

Number of Proportion Standard Over 40? Employees Terminated Error No 511 0.0137 0.005142 Yes 806 0.0509 0.007740

9.2 The analysis might include, for ex- ample, expected counts and column percents (shown in the table). We note that older employees are almost twice as likely as under-40 employees to fall into the two lowest performance appraisal categories (partially/fully meets expectations), and are only about one-third as likely to have the highest appraisal. The differences in the percentages are signiÞcant: X 2 = 13.893 + 9.091 + 0.880 + 0.576 + 15.941 + 10.431 = 50.812 (df = 2) has P < 0.0005.

Under 40 Over 40 Partially/ fully meets expectations

82 230 312 123.41 188.59 16.5% 30.3% 24.9%

Usually exceeds expectations

353 496 849 335.81 513.19 71.2% 65.4% 67.7%

Continually exceeds expectations

61 32 93 36.78 56.22 12.3% 4.2% 7.4% 496 758 1254

9.3 (a) Use column percents, because we suspect that ÒsourceÓ is explana- tory. See the table. (b) The expected counts are in the table. The test statistic is X 2 = 1.305 + 0.666 + 2.483+0.632+0.323+1.202 = 6.611; comparing to a χ 2(2) distribution, we Þnd 0.025 < P < 0.05 (software gives 0.037). The conclusion depends on the chosen value of α. With α = 0.05, e.g., so that P < α, we conclude that there is an association between the source of a cat, and whether or not the pet ends up in the animal shelter.

Private Pet Store Other Cases 124 16 76 216

111.92 13.05 91.03 36.2% 40% 27.2% 32.6%

Control 219 24 203 446 231.08 26.95 187.97 63.8% 60% 72.8% 67.4% 343 40 279 662

Solutions 197

9.4 Expected counts and column percents are given in the table. X 2 = 0.569+9.423+9.369+0.223+ 3.689 + 3.668 = 26.939 (df = 2); this has P < 0.0005. We conclude that there is an association between the source of a dog and whether or not the dog ends up in the animal shelter.

Private Pet Store Other Cases 188 7 90 285

198.63 21.10 65.27 26.6% 9.3% 38.8% 28.1%

Control 518 68 142 728 507.37 53.90 166.73 73.4% 90.7% 61.2% 71.9% 706 75 232 1013

9.5 This is a 2 × 3 table, with each household classiÞed by pet (cat or dog) and by source. If we view ÒsourceÓ as explanatory for pet type, then we should look at the conditional distribution of pet type, given the source (i.e., column percents), as given in the table. It appears that cats are more likely to come from an ÒotherÓ source. The test statistic bears this out: X 2 = 13.283+8.138+ 3.431+2.102+39.482+24.188 = 90.624 (df = 2), so that P < 0.0005. We conclude that there is a relationship between source and pet type.

Private Pet Store Other Cats 219 24 203 446

279.98 34.95 131.06 29.7% 26.1% 58.8% 38%

Dogs 518 68 142 728 457.02 57.05 213.94 70.3% 73.9% 41.2% 62% 737 92 345 1174

9.6 (a) These are the percentages in the top row of the table. (b) H0: There is no relationship between intervention and response rate; Ha: There is a relationship. (c) X 2 = 4.906+56.765+41.398+2.872+ 33.234+ 24.237 = 163.413, df = 2, P < 0.0005. The differences between the response rates are signiÞcant; speciÞcally, letters and phone calls both increase the response rate, with the latter being more effective.

Phone Letter Call None

Yes 171 146 118 435 144.38 79.02 211.59 43.7% 68.2% 20.6% 36.9%

No 220 68 455 743 246.62 134.98 361.41 56.3% 31.8% 79.4% 63.1% 391 214 573 1178

9.7 (a) With a letter, 51.2% responded; without, the response rate was 52.6%. (b) H0: there is no relationship between whether or not a letter is sent and whether or not the subject responds; Ha: There is a relationship. The test statistic is X 2 = 0.461+0.460+0.497+0.496 = 1.914; comparing to a χ2(1) distribution, we Þnd 0.15 < P < 0.20 (software gives 0.167). There is little reason to reject the null hypothesis.

Letter No Letter Yes 2570 2645 5215

2604.65 2610.35 51.2% 52.6% 51.9%

No 2448 2384 4832 2413.35 2418.65 48.8% 47.4% 48.1% 5018 5029 10047

198 Chapter 9 Inference for Two-Way Tables

9.8 Responses may vary. Both surveysÑespecially the Þrst oneÑmay be somewhat dated. The questions asked of the college students was one that might have general interest to them,whereas the survey sent to the physicianswasmore important to themprofessionally (this might account for the higher response rate among physicians). Viewed from this perspective, we might expect our survey response to be more like the college student results, since Internet accessibility will likely (for most of our population) be of general, not professional, interest.

9.9 (a) No: No treatment was imposed. (b) See the column percents in the table. Pet owners seem to have better survival rates. (c) H0 says that there is no relationship between patient status and pet ownership (i.e., that survival is independent of pet ownership). Ha says that there is a relationship between survival and pet ownership. (d) X 2 = 0.776+ 0.571+ 4.323+ 3.181 = 8.851 (df = 1), so 0.0025 < P < 0.005 (in fact, P = 0.003). (e) Provided we believe that there are no confounding or lurking variables, we reject H0 and conclude that owning a pet improves survival.

No Pet Pet Alive 28 50 78

33.07 44.93 71.8% 94.3% 84.8%

Dead 11 3 14 5.93 8.07

28.2% 5.7% 15.2% 39 53 92

9.10 (a) In table. These percents show how January performance can predict rest-of-year performance: Among those years in which the S&P index was up in January, the index rose in the rest of the year 72.9% of the time, etc. (b) Since the table is symmetric, each pair of row percents is the same as the corresponding column pair (e.g., the Þrst row is 72.9% and 27.1%Ñthe same as the Þrst column). These show, e.g., that if the index was up for the rest of the year, then there is a 72.9% chance that it was up in January, as well. (c) H0 says that there is no relationship between January performance and rest-of-year performance. Ha says that there is a relationship. (d) The expected counts (in the table) are higher than observed in the Down/Up and Up/Down cellsÑ suggesting that we are less likely than we might expect to see these combinationsÑ and lower than observed in the Up/Up and Down/Down cellsÑsuggesting that these are more likely than we expect. This is in line with the January indicator. (e) X 2 = 0.596+ 1.060+ 1.060+ 1.885 = 4.601, df = 1, P = 0.032. This is fairly strong evidence of a relationship. (f) The data support the January indicator, but mostly for the Up/Up case. That is, in years when the market was down in January, we have little indication of performance for the rest of the year; historically, it has been about 50% up, 50% down. If the market is up in January, however, history suggests it is more likely to be up for the whole year.

January Up Down

Up this year

35 13 48 30.72 17.28 72.9% 48.1% 64%

Down this year

13 14 27 17.28 9.72 27.1% 51.9% 36%

48 27 75

Solutions 199

9.11 (a) At right. (b) Use column percents (here reported as proportions rather than percents): His batting average was .262 during the regular season, and .357Ñmuch higherÑduring the World Series. (c) H0 says that the regular season and World Series distributions (batting averages) are the same; the alternative is that the two distributions are different. X 2 = 0.033 + 3.311 + 0.012 + 1.181 = 4.536, df = 1, P = 0.033. We have fairly strong (though not overwhelming) evidence that Jackson did better in the World Series.

Regular World Hit? Season Series Yes 2584 35 2619 No 7280 63 7343

9864 98 9962

9.12 (a) & (b) See table. Percentage of children receiving tetracycline seems to rise as we move from urban to rural counties. (c) H0: There is no relationship between county type and prescription practice; Ha: There is a relationship. (d) X 2 = 7.370+0.372+ 7.242 + 5.440 + 0.275 + 5.345 = 26.044, df = 2, P < 0.0005. The differences between the tetracycline prescription practices are signiÞcant; doctors in rural counties were most likely to prescribe tetracycline to young children, while urban doctors were least likely to do so.

Urban Intermed. Rural Tetra. 65 90 172 327

90.88 95.98 140.14 30.4% 39.8% 52.1% 42.5%

No tetra.

149 136 158 443 123.12 130.02 189.86 69.6% 60.2% 47.9% 57.5% 214 226 330 770

9.13 Expected counts and column percents in ta- ble. 69.7% of the second-year ÒwinnersÓ also had been winners in the Þrst year, while only 29.7% of second-year losers had been winners in the Þrst year. This suggests some persistence in performance. The test statistic supports this: X 2 = 9.443 + 9.763 + 9.443 + 9.763 = 38.411, df = 1, P < 0.0005. We have strong evidence to support persistence of fund perfor- mance.

Next year Winner Loser

Winner this year

85 35 120 61.00 59.00 69.7% 29.7% 50%

Loser this year

37 83 120 61.00 59.00 30.3% 70.3% 50% 122 118 240

9.14 p̂1 = 85122 .= 0.6967, p̂2 = 35118

.= 0.2966, and the ÒpooledÓ estimate of p is p̂ = 120240 = 0.5. The test statistic for H0: p1 = p2 vs. Ha: p1 6= p2 is z = ( p̂1 − p̂2)/

√ (0.5)(0.5)

( 1 122 + 1118

) .= 6.197. This agrees with the previous result: z2 = 38.411 and the P-value is 2P(Z > 6.197) .= 0.000374.

9.15 With the retrospective approach, we have p̂1 = 85120 = 0.7083, p̂2 = 37120 = 0.3083, and ÒpooledÓ estimate p̂ = 122240 = 0.5083. The test statistic is z = ( p̂1 − p̂2)/

√ p̂(1− p̂)

( 1 120 + 1120

) .= 6.197. This agrees with the previous result: z2 = 38.411.

200 Chapter 9 Inference for Two-Way Tables

9.16 Expected counts and column percents in table. 39.8% of the second-year ÒwinnersÓ also had been winners in the Þrst year, while 59.9% of second-year losers had been winners in the Þrst year. This is evidence against persistence; note also that the expected counts are higher than observed in the Win/Win and Lose/Lose cells. The test statistic is X 2 = 4.981+4.860+4.981+ 4.860 = 19.683, df = 1, P < 0.0005. There is signiÞcant evidence against the null hypothesis (no relationship), but in this case, it is evidence of Òantipersistence.Ó

Next year Winner Loser

Winner this year

96 148 244 120.50 123.50 39.8% 59.9% 50%

Loser this year

145 99 244 120.50 123.50 60.2% 40.1% 50% 241 247 488

9.17 There is no reason to consider one of these as explanatory, but a conditional distribution is useful to determine the nature of the associa- tion. Each cell in the table contains a pair of percentages; the Þrst is the column percent, and the second is the row percent. For example, among nonsmokers, 34.5% were nondrinkers; among nondrinkers, 85.4% were nonsmokers. The percentages in the right margin gives the distribution of alcohol consumption (the overall column percent), while the percentages in the bottom margin are the distribu- tion of smoking behavior. X 2 = 42.252 (df = 6) so P < 0.0005; we conclude that alcohol and nicotine

consumption are not independent. The chief deviation from independence (based on comparison of expected and actual counts) is that nondrinkers are more likely to be nonsmokers than we might expect, while those drinking 0.11 to 0.99 oz/day are less likely to be nonsmokers than we might expect. One possible graph is below.

0 mg 1Ð15 mg 16+ mg 0 oz 105 7 11 123

82.73 17.69 22.59 34.5% 10.8% 13.3% 27.2% 85.4% 5.7% 8.9%

0.01Ð 0.10 oz

58 5 13 76 51.12 10.93 13.96 19.1% 7.7% 15.7% 16.8% 76.3% 6.6% 17.1%

0.11Ð 0.99 oz

84 37 42 163 109.63 23.44 29.93 27.6% 56.9% 50.6% 36.1% 51.5% 22.7% 25.8%

1.00+ oz 57 16 17 90 60.53 12.94 16.53 18.8% 24.6% 20.5% 19.9% 63.3% 17.8% 18.9% 304 65 83 452

67.3% 14.4% 18.4% 100%

None 1Ð15 mg/day 16+ mg/day 0

Pe rc

en t o

f s ub

je ct

Nicotine consumption

No alcohol

0.01Ð0.10 oz/day

0.11Ð0.99 oz/day

1.00+ oz/day

Solutions 201

9.18 Based on the background information given in the problem, there is no reason to consider one of these as explanatory; each is linked to the other. Thus it may be useful to look at both conditional distributions (rows and columns). Comparing the column percents (the Þrst percentage in each cell, along with those in the right margin), we note that children classiÞed as ÒNormalÓ are more likely to be healthy than those whose nutrition is inadequate, but there is not much difference between the three inadequate nutrition groups. Looking at row percents (the second percentage in the cells, and those in the bottom

margin), we observe that thosewith no illness are considerablymore likely to have normal nutrition. Comparing the three illness combinations, there is little variation for nutritional status I (31.6% to 31.8%) or for status III/IV (14.2% to 16.2%), and only small differences in the percentages classiÞed as Normal (14.2% to 21.0%) or as status II (31.9% to 40.0%). The differences in these percentages are statistically signiÞcant: X 2 = 101.291, df =

9, P < 0.0005.

Normal I II III & IV URI 95 143 144 70 452

111.74 133.85 141.61 64.79 33.0% 41.4% 39.5% 41.9% 38.8% 21.0% 31.6% 31.9% 15.5%

Diarrhea 53 94 101 48 296 73.17 87.66 92.74 42.43 18.4% 27.2% 27.7% 28.7% 25.4% 17.9% 31.8% 34.1% 16.2%

Both 27 60 76 27 190 46.97 56.27 59.53 27.24 9.4% 17.4% 20.8% 16.2% 16.3% 14.2% 31.6% 40.0% 14.2%

None 113 48 44 22 227 56.12 67.22 71.12 32.54 39.2% 13.9% 12.1% 13.2% 19.5% 49.8% 21.1% 19.4% 9.7% 288 345 365 167 1165

24.7% 29.6% 31.3% 14.3% 100%

9.19 (a) Blood pressure is explanatory. Of the low blood pressure group, 0.785% died from cardiovascular disease, compared to 1.648% of the high BP group, suggesting that high BP increases the risk of cardiovascular disease. (b) H0: p1 = p2; Ha: p1 < p2. p̂1 = 21/2676, p̂2 = 55/3338, and the ÒpooledÓ estimate of p is p̂ = 76/6014. The test statistic is z = ( p̂1 − p̂2)/

√ p̂(1− p̂)

( 1

2676 + 13338 ) .= −2.98, so that

P = 0.0014; we conclude that the high blood pressure group has a greater risk. (c) Shown at right. The χ2 test is not appropriate since the alternative is one-sided. (d) SE p̂1− p̂2 =

√ p̂1(1− p̂1)/2676+ p̂2(1− p̂2)/3338 .= 0.002786. The 95%

conÞdence interval is p̂1 − p̂2 ± 1.960 SE p̂1− p̂2 = −0.0141 to −0.0032.

Died? Low BP High BP Yes 21 55 76 No 2655 3283 5938

2676 3338 6014

202 Chapter 9 Inference for Two-Way Tables

9.20 (a) H0: p1 = p2; Ha: p1 < p2. p̂1 = 457/1003, p̂2 = 437/620, and the ÒpooledÓ estimate of p is p̂ = 894/1623. The test statistic is z = ( p̂1 − p̂2)/

√ p̂(1− p̂)

( 1

1003 + 1620 ) .= −9.81,

so that P < 0.0001; we conclude that the second proportion is lower (the program is effective). (c) Shown at right. The χ 2 test is not appropriate since the alternative is one-sided. (d) SE p̂1− p̂2 =

√ p̂1(1− p̂1)/1003+ p̂2(1− p̂2)/620 .= 0.002414. The

95% conÞdence interval is p̂1 − p̂2 ± 1.960 SE p̂1− p̂2 = −0.2965 to −0.2019.

First Second Fed? Survey Survey Yes 457 437 894 No 546 183 729

1003 620 1623

9.21 25% of those with low antacid use, 62.5% of the medium-use group, and 80% of the high-use group had AlzheimerÕs, suggesting a con- nection. X 2 = 7.118 (df = 3), so P = 0.069Ñthere is some evidence for the connection, but it is not statistically signiÞcant.

None Low Med High AlzheimerÕs patient

112 3 5 8 128 113.00 6.00 4.00 5.00 49.6% 25% 62.5% 80% 50%

Control group

114 9 3 2 128 113.00 6.00 4.00 5.00 50.4% 75% 37.5% 20% 50% 226 12 8 10 256

9.22 Use column percents, since we view gender as explanatory. Women appear to be more likely to have dropped out. H0: There is no relationship between

gender and student status; Ha: There is a relationship. X 2 = 13.398, df = 2, P = 0.001Ñthere is strong evidence of a relationship. Other factors to consider would be

anything that might account for someone leaving a degree programÑe.g., age of students entering program.

Men Women Completed 423 98 521

404.49 116.51 53.2% 42.8% 50.9%

Still enrolled

134 33 167 129.65 37.35 16.9% 14.4% 16.3%

Dropped out

238 98 336 260.86 75.14 29.9% 42.8% 32.8% 795 229 1024

9.23 71.3% of Irish, 76.0% of Portuguese, 69.5% of Norwegians, and 75.0% of Italians can taste PTC; there seems to be some variation in the percentages among the countries. X 2 = 5.957 (df = 3), so P = 0.114Ñthe observed differences between the percentages are not signiÞcant (for typical choices of α).

Ireland Portugal Norway Italy Tasters 558 345 185 402 1490

572.18 331.76 194.38 391.68 71.3% 76% 69.5% 75% 73.1%

Non- tasters

225 109 81 134 549 210.82 122.24 71.62 144.32 28.7% 24% 30.5% 25% 26.9% 783 454 266 536 2039

Solutions 203

9.24 The differences between ethnic groups (as described by column percents, and rep- resented in the graph) are sig- niÞcant: X 2 = 1078.6, df = 9, P < 0.0005. (With such large samples, even small differences would almost cer- tainly be found signiÞcant.)

Hawaiian Hawaiian- white

Hawaiian- Chinese

White 0

Pe rc

en t o

f s ub

je ct

Hawaiian- Hawaiian- Hawaiian White Chinese White

Type O 1903 4469 2206 53759 62337 2006.89 4289.68 2314.16 53726.27 40.7% 44.8% 41.0% 43% 43.0%

Type A 2490 4671 2368 50008 59537 1916.75 4097.00 2210.21 51313.04 53.3% 46.8% 44.0% 40% 41.0%

Type B 178 606 568 16252 17604 566.75 1211.41 653.52 15172.33 3.8% 6.1% 10.5% 13% 12.1%

Type AB 99 236 243 5001 5579 179.61 383.92 207.11 4808.36 2.1% 2.4% 4.5% 4% 3.8% 4670 9982 5385 125020 145057

9.25 For the British study, X 2 = 0.249 (df = 1), which gives P = 0.618Ñthere is very little evidence of an association. For the American study, X2 = 25.014 (df = 1), which gives P < 0.0005. This is strong evidence of an association: aspirin reduced the risk of a fatal heart attack. The difference in the conclusions can

be attributed to the larger sample size for the American study (important for something as rare as a heart attack), as well as the shorter duration of the study and the lower dosage (taking the aspirin every other day rather than every day).

British study Aspirin No aspirin

Heart attack

148 79 227 151.47 75.53 4.3% 4.6% 4.4%

No heart attack

3281 1631 4912 3277.53 1634.47 95.7% 95.4% 95.6% 3429 1710 5139

PhysicianÕs Health Study Aspirin No aspirin

Heart attack

104 189 293 146.52 146.48 0.9% 1.7% 1.3%

No heart attack

10933 10845 21778 10890.48 10887.52 99.1% 98.3% 98.7% 11037 11034 22071

204 Chapter 9 Inference for Two-Way Tables

9.26 (a) H0: p1 = p2; Ha: p1 6= p2. p̂1 = 28/82, p̂2 = 30/78, and the ÒpooledÓ esti- mate of p is p̂ = 58/160. The test statistic is z = ( p̂1 − p̂2)/

√ p̂(1− p̂)

( 1 82 + 178

) .= −0.5675, so that P > 0.5686 = 2(0.2843) (software gives P = 0.57); there is no reason to believe that the proportions are different. (b) Table at right (with expected counts and column percents). H0: There is no relationship between treatment and relief; Ha: There is a relationship. X 2 = 0.322 (which does equal z2, up to rounding error), df = 1, P = 0.570. (c) Gastric freezing is not effective (or Òis no more effective than a placeboÓ).

Gastric Freezing Control

Relief 28 30 58 29.73 28.27 34.1% 38.5% 36.3%

No relief 54 48 102 52.28 49.72 65.9% 61.5% 63.8%

82 78 160

9.27 (a) X 2 = 2.186 (df = 1), which gives 0.10 < P < 0.15 (in fact, P = 0.140); we do not have enough evidence to conclude that the observed difference in death rates is due to something other than chance. (b) Good condition: X 2 = 0.289 (df = 1), which gives P > 0.25 (in fact, P = 0.591). Poor condition: X 2 = 0.019 (df = 1), which gives P > 0.25 (in fact, P = 0.890). In both cases, we cannot reject the hypothesis that there is no difference between the hospitals. (c) No.

9.28 The study needs samples of thousands, not hundreds. Since cardiovascular disease is relatively rare, sample sizes must be quite largeÑotherwise, it is quite possible that we would observe no heart attacks in one or both of our groups, even if we track them for several years. See also the answer to Exercise 8.65.

9.29 For the sex/SC table (top): X 2 = 23.450, df = 1, P < 0.0005. This is strong evidence of a link between gender and social comparison. For the sex/mastery table (bottom): X 2 =

0.030, df = 1, P > 0.25 (in fact, P = 0.863). There is no evidence of a link between gender and mastery. It appears that the difference between male and

female athletes observed in Example 9.4 is in social comparison, not in mastery.

Female Male HSC 21 49 70

35.00 35.00 31.3% 73.1% 52.2%

LSC 46 18 64 32.00 32.00 68.7% 26.9% 47.8%

67 67 134

HM 35 36 71 35.50 35.50 52.2% 53.7% 53%

LM 32 31 63 31.50 31.50 47.8% 46.3% 47%

67 67 134

Solutions 205

9.30 Since we suspect that student loans may explain career choice, we examine column percents (in the table below, left). We observe that thosewith loans are slightlymore likely to be in Agriculture, Science, and Technology Þelds, and less likely to be inManagement. However, the differences in the table are not signiÞcant: X 2 = 6.525, df = 6, P = 0.368. For 9.30.

Loan No Loan Agric. 32 35 67

8.7% 7.0% 7.7% CDFS 37 50 87

10.1% 10.1% 10.1% Eng. 98 137 235

26.6% 27.6% 27.2% LA/Educ. 89 124 213

24.2% 24.9% 24.6% Mgmt. 24 51 75

6.5% 10.3% 8.7% Science 31 29 60

8.4% 5.8% 6.9% Tech. 57 71 128

15.5% 14.3% 14.8% 368 497 865

For 9.31. Low Medium High

Agric. 5 27 35 67 13.5% 6.8% 8.2% 7.7%

CDFS 1 32 54 87 2.7% 8.0% 12.6% 10.1%

Eng. 12 129 94 235 32.4% 32.3% 22.0% 27.2%

LA/Educ. 7 77 129 213 18.9% 19.3% 30.1% 24.6%

Mgmt. 3 44 28 75 8.1% 11.0% 6.5% 8.7%

Science 7 29 24 60 18.9% 7.3% 5.6% 6.9%

Tech. 2 62 64 128 5.4% 15.5% 15.0% 14.8% 37 400 428 865

9.31 For the table (above, right), X 2 = 43.487 (df = 12), so P < 0.0005, indicating that there is a relationship between PEOPLE score and Þeld of study. Among other observations we could make: Science has a large proportion of low-

scoring students, while liberal arts/education has a large percentage of high-scoring stu- dents. (These two table entries make the largest contributions to the value of X 2.)

9.32 Death rates (deaths/1000 cases) are given in the table and illustrated in the graph. The statistic for testing the association is X 2 = 19.715 (df = 7, P = 0.007). The differences in death rates are signiÞcant; speciÞcally, the risk of complications is greatest for children under 5, and adults over 25. We cannot study the association between catching measles and age because we do not

know the total number of people who were alive in each age group.

Age Death Rate < 1 year 4.44677 1Ð4 5.17483 5Ð9 1.35685 10Ð14 1.58646 15Ð19 2.93794 20Ð24 2.70880 25Ð29 5.99600 30+ 8.48485 <1 1Ð4 5Ð9 10Ð14 15Ð19 20Ð24 25Ð29 30+

D ea

th s

pe r 1

00 0

ca se

Age (years)

206 Chapter 9 Inference for Two-Way Tables

9.33 For example, there were 966 .= (0.883)(1094) hyper- tensive hypokalemic patients, and therefore there were 128 non- hypertensive hypokalemic patients. The respective X 2 values are

83.147, 48.761, 12.042, and 13.639, all with df = 2, which are all signiÞcant (the largest P-value is 0.003). If we drop the hyperkalemic group, the X 2

values are 57.764, 33.125, 11.678, and 8.288, all with df = 2, which are also all signiÞcant (the largest P-value is 0.004). Thus it appears that there is an

association between potassium level and each of the four risk factors. Looking at the percentages in the table, the hyperkalemic group is generally (for all but diabetes) quite different from the other two groups; the large sample sizes for hypokalemic and normal groups make even small differences (like the difference for gender) statistically signiÞcant.

Hypo. Normal Hyper. Hypertension (yes)

966 3662 11 4639 873.51 3743.94 21.56

(no) 128 1027 16 1171 220.49 945.06 5.44

Heart failure (yes)

181 1158 15 1354 254.95 1092.75 6.29

(no) 913 3531 12 4456 839.05 3596.25 20.71

Diabetes (yes)

225 1196 8 1429 269.08 1153.28 6.64

(no) 869 3493 19 4381 824.92 3535.72 20.36

Female 793 3189 13 3995 752.24 3224.19 18.57

Male 301 1500 14 1815 341.76 1464.81 8.43

Totals 1094 4689 27 5810

9.34 The variation in the percentage of woman pharmacy students is so great that it is not surprising that the differences are signiÞcant: X 2 = 359.677, df = 8, P < 0.0005. The plot (below, left) is roughly linear; the regression line is ŷ = −4448+ 2.27x .

For 9.34.

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For 9.35.

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1985 1990 1995

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9.35 Plot above, right. Thepercentage ofwomenpharmacy students has gradually increased since 1987, from 60% to nearly 64%; by the end it seems to have nearly leveled out. The rate of increase is considerably less than that shown from 1970 to 1986 (note the very different vertical scales on the two graphs above). To summarize, women were a minority of pharmacy students in the early 1970s, but

the proportion of women steadily increased until the mid-1980s, and has increased less rapidly since then. Women became the majority in the early 1980s.

Solutions 207

Estimates for 2000 should probably be between 63% and 64%, assuming that there are no big changes between 1996 and 2000.

9.36 X 2 = 852.433, df = 1, P < 0.0005. Using z = −29.2, computed in 8.66(c), this equals z2 (up to rounding).

Mexican- American Other

Juror 339 531 870 688.25 181.75

Not a juror

143 272 37 393 180 665 142 922.75 37 742.25 143 611 37 924 181 535

9.37 For cats: X 2 = 8.460 (df = 4), which gives P = 0.077. We do not reject H0 this time; with the 2× 3 table, we had P = 0.037, so having more cells has ÒweakenedÓ the evidence. For dogs: X 2 = 33.208 (df = 4), which gives P < 0.0005. The conclusion is the same as before: we reject H0.

9.38 Note to instructors: The distinctions between themodels can be quite difÞcult tomake, since the difference between several populations might, in fact, involve classiÞcation by a categorical variable. In many ways, it comes down to how the data were collected. For example, to compare male and female athletes (as in Example 9.3 and following), we can either (a) select n1 male and n2 female athletes and classify them according to some characteristic (e.g., social comparison and mastery categories)Ñas was described in Example 9.3Ñor (b) select a sample of athletes, then classify each as male or female, and also according to that other characteristic. The former case would be a Òcomparison of populationsÓ (a.k.a. ÒhomogeneityÓ) model, while the latter is a test of independence. Of course, the difÞculty is that themethodof collectingdatamaynot alwaysbe apparent,

in which case we have to make an educated guess. One question we can ask to educate our guess is whether we have data that can be used to estimate the (population) marginal distributions. E.g., in Example 9.3 and following, the table gives us no information about the proportion of all athletes who are male or female (these would be the proportions along the bottom margin); we simply picked 67 of each gender. Furthermore, we would get a different marginal distribution for the sports goals if we had a different mix of men andwomenÑsay, twice as manymen as womenÑsowe do not know the true sports goals marginal distribution, either. In Example 9.8, on the other hand, we could get information about the percentages of current smokers, former smokers, and ÒneverÓ smokers in our sample (the right margin), and also about the SES distribution in our sample (the bottom margin). For some of these problems, either answer may be acceptable, provided a reasonable

explanation is given. In 9.1, we are testing for independence between age and termination. (We have data

to compute the marginals for both.) In 9.3, we have two populations: cats brought into the humane society (ÒcasesÓ), and those which were not (control). (We do not know, and are not interested in, what proportion of all cats are brought to the humane society.) In 9.6, we are comparing three populationsÑone for each intervention. In 9.12, we test for independence between county type and tetracycline prescriptions.

208 Chapter 9 Inference for Two-Way Tables

9.39 Before we had X 2 = 7.118; with the counts doubled, X 2 = 14.235 (df = 3), which gives P = 0.003. The proportions are the same, but the increased sample size makes the differences between the categories statistically signiÞcant.

Solutions 209

Chapter 10 Solutions

10.1 (a) Ignoring the (circled) outlier, there is a weak positive association. (b) The regression equation is ŷ = 43.4 + 0.0733x . The signiÞcance test for the slope yields t = 2.85 from a t distribution with df = 59 − 2 = 57. This is signiÞcantÑusing Table E, we can estimate P < 2(0.005) = 0.01; Minitab reports P = 0.006. We conclude that linear regression on LOS is useful for predicting wages. (c) With b1 = 0.0733, we can say that wages increase by 0.0733 per week of service. (Note: This is not $0.0733, since we donÕt know the units of ÒWages.Ó) (d) From software, SEb1 = 0.02571; we compute b1 ± t∗ SEb1 . Using t∗ = 2.009 (df = 50, from the table), the interval is 0.0216 to 0.1250. With t∗ = 2.0025 (df = 57, from software), the interval is 0.0218 to 0.1248.

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Output from Minitab: The regression equation is Wages = 43.4 + 0.0733 LOS

Predictor Coef Stdev t-ratio p Constant 43.383 2.248 19.30 0.000 LOS 0.07325 0.02571 2.85 0.006

s = 10.21 R-sq = 12.5% R-sq(adj) = 10.9%

10.2 The new regression line is ŷ = 44.2+ 0.0731x . The intercept (b0) is higher, since the outlier ÒpullsÓ the line up. The slope has not changed much; it now has t = 2.42 (P = 0.018)Ñstill signiÞcant, though not as much as before. The estimated standard deviation is higher (11.98 vs. 10.21) since the outlier suggests a greater amount of variability in the data.

Output from Minitab: The regression equation is Wages = 44.2 + 0.0731 LOS

Predictor Coef Stdev t-ratio p Constant 44.213 2.628 16.82 0.000 LOS 0.07310 0.03015 2.42 0.018

s = 11.98 R-sq = 9.2% R-sq(adj) = 7.6%

210 Chapter 10 Inference for Regression

10.3 (a) The plot shows a weak posi- tive association. (b) Regression gives ŷ = 51, 938+ 47.7x . The slope is signif- icantly different from 0 (t = 6.94, df = 59− 2 = 57, P < 0.0005). We conclude that linear regression on square footage is useful for predicting selling price.

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1000 2000 3000 4000

Pr ic

e ($

10 00

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Square feet

10.4 The new regression equation is ŷ = 101, 458+ 25.4x . The slope is still signiÞcantly different from 0 (t = 5.30, P < 0.0005), but the average increase in price for added ßoor space (i.e., the slope) is considerably less with the Þve outliers removed. Those Þve homes were more expensive than we would expect from the pattern of the rest of the points, so they had the effect of increasing the slope.

10.5 (a) There is a fairly strong positive relationship. There are no particular outliers or unusual observations, but one noteworthy feature is that the spread seems to increase over time. (b) The regression equation is ŷ = −3545 + 1.84x . The slope is signiÞcantly different from 0 (t = 13.06, with df = 38). Yield has increased at an average rate of 1.84 bushels/acre each year.

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1960 1970 1980 1990 2000

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Year

10.6 (a) ŷ = 1.23 + 0.202x . (b) t = 17.66 (df = 7), which has P < 0.0005. The slope is signiÞcantly different from 0. (c) t∗ = 2.365 and SEb1 = 0.01145, so the interval is 0.175 to 0.229 (hundred ft3 of gas/heating degree day per day). (d) SEb0 = 0.2860, so the interval is 0.554 to 1.906 hundred ft3 of gas.

Output from Minitab: The regression equation is Gas = 1.23 + 0.202 HeatDeg

Predictor Coef Stdev t-ratio p Constant 1.2324 0.2860 4.31 0.004 HeatDeg 0.20221 0.01145 17.66 0.000

s = 0.4345 R-sq = 97.8% R-sq(adj) = 97.5%

Solutions 211

10.7 (a) Powerboats registered is the explanatory variable, so it should be on the horizontal axis. The (positive) association appears to be a straight-line relationship. (b) ŷ = −41.4 + 0.125x . (c) H0: β1 = 0; Ha: β1 > 0. The test statistic is t = b1/SEb1 = 14.24, which is signiÞcant (df = 12, P < 0.0005); this is good evidence that manatee deaths increase with powerboat registrations. (d) Use x = 716: the equation gives y = 48.1, or about 48 manatee deaths. The mean number of manatee deaths for 1991Ð93 is 42Ñless than the 48 predicted. Evidence of ÒsuccessÓ is perhaps in the eye of the beholder: the nature of the relationship between the two variables does not seem to have changed (not that we would have any reason to expect this), but the increase in the number of powerboat registrations evident in previous years seems to have been curtailed.

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400 450 500 550 600 650 700 750

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Powerboats registered

10.8 (a) The trend appears linear. (b) ŷ = −61.1 + 9.32x . The regression explains r 2 = 98.8% of the variation in lean. (c) The rate we seek is the slope. For df = 11, t∗ = 2.201, so the interval is 9.32 ± (2.201)(0.3099) = 8.64 to 10.00 tenths of a millimeter/year.

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Year

Output from Minitab: The regression equation is Lean = - 61.1 + 9.32 Year

Predictor Coef Stdev t-ratio p Constant -61.12 25.13 -2.43 0.033 Year 9.3187 0.3099 30.07 0.000

s = 4.181 R-sq = 98.8% R-sq(adj) = 98.7%

10.9 (a) ŷ = −61.1+9.32(18) .= 107, for a prediction of 2.9107m. (b) This is an example of extrapolationÑtrying tomake a prediction outside the range of given x values. Minitab reports SEŷ = 19.56, so a 95% prediction interval for ŷ when x∗ = 18 is about 62.6 to 150.7. The width of the interval is an indication of how unreliable the prediction is.

10.10 (a) ŷ = −61.1 + 9.32(97) .= 843, for a prediction of 2.9843 m. (b) A prediction interval is appropriate, since we are interested in one future observation, not the mean of all future observations; in this situation, it does not make sense to talk of more than one future observation.

10.11 (a) β1 represents the increase in gas consumption (in hundreds of cubic feet) for each additional degree day per day. With df = 16, t∗ = 2.120, so the interval is

212 Chapter 10 Inference for Regression

0.26896 ± (2.120)(0.00815), or 0.2517 to 0.2862. (b) The margin of error would be smaller here, since for a Þxed conÞdence level, the critical value t∗ decreases as df increases. (Additionally, the standard error is slightly smaller here.) The margin of error for 10.6 was t∗ SEb1 = (2.365)(0.01145) = 0.027, while it is 0.017 here.

10.12 (a) As stated in Exercise 10.6(d), β0 is the natural gas consumed for nonheating usesÑcooking, hot water, etc. t∗ = 2.120 (as in 10.11), so the interval is 2.405 ± (2.120)(0.20351) = 1.974 to 2.836 hundred ft3 of gas. (b) This interval is 0.862 units wide, while the interval of 10.6 was 1.352 units wide. This interval is shorter since for a Þxed conÞdence level, the critical value t∗ decreases as df increases. Also, the standard error is slightly less than in 10.6.

10.13 (a) t = b1/SEb1 = 0.20221/0.01145 = 17.66. (b) With df = 7, we have t∗ = 1.895. We reject H0 at this level (or any reasonable level). (c) From the table, we report P < 0.0005. This is probably more readily understandable than the software value: P .= 2.3× 10−7 = 0.00000023.

10.14 t = b1/SEb1 = 0.82/0.38 = 2.158. Table E gives a P-value between 0.02 and 0.04; software gives P .= 0.035. There is fairly good evidence that β1 6= 0 (signiÞcant at α = 0.05, but not at α = 0.01).

10.15 (a) x = 13.07 and ∑(xi − x)2 = 443.201. (b) H0: β1 = 0; Ha: β1 > 0. SEb1 = s/

√∑ (xi − x)2 = 0.0835, so t = 0.902/0.0835 = 10.80. For any reasonable α,

this is signiÞcant; we conclude that the two variables are positively associated. (c)With df = 8, we have t∗ = 3.355: 0.902 ± (3.355)(0.0835) = 0.622 to 1.182. (d) ŷ = 1.031+0.902(15) = 14.56. SEŷ = 1.757

√ 1+ 110 + (15.0−13.07)

443.201 = 1.850 and t∗ = 1.860, so the prediction interval is 14.56± (1.860)(1.850), or 11.12 to 18.00.

Output from Minitab: Fit Stdev.Fit 90.0% C.I. 90.0% P.I.

14.561 0.578 ( 13.485, 15.637) ( 11.121, 18.001)

10.16 (a) x = 1327/26 .= 51.038 and ∑(xi − x)2 .= 7836.96. (Be sure to use only the 26 rural readings for which there is also a city reading.) (b) H0: β1 = 0; Ha: β1 6= 0. SEb1 = s/

√∑ (xi − x)2 = 0.05060, so t = 1.0935/0.05060 = 21.61. Then

P < 0.001, which is signiÞcant for any reasonable α; we conclude that the slope is different from 0. (c) Use a prediction interval: ŷ = −2.580 + 1.0935(43) .= 44.44, SEŷ = 4.4792

√ 1+ 126 + (43−51.038)

7836.96 = 4.5826 and t∗ = 2.064, so the interval is 44.44± (2.064)(4.5826) = 34.98 to 53.90.

Output from Minitab: Fit Stdev.Fit 95.0% C.I. 95.0% P.I.

44.441 0.968 ( 42.442, 46.439) ( 34.980, 53.901)

Solutions 213

10.17 (a) The plot reveals no outliers or unusual points. (b) The regression equation is ŷ = −0.06485+1.184x , so we estimate 1/R = b1 = 1.184. SEb1 = 0.07790 and t∗ = 3.182, so the conÞdence interval is 0.936 to 1.432. (c) R .= 1/b1 = 0.8446; the conÞdence interval is 0.698 to 1.068. (d) SEb0 = 0.1142, so t = −0.6485/0.1142 = −0.5679. From the table, we can estimate that P > 2(0.25) = 0.50; Minitab gives P = 0.61. We have little reason to doubt that β0 = 0.

0.5

1.5

2.5

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C ur

re nt

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Voltage (V)

10.18 The new line is ŷ = 1.1434x ; the slope has standard error 0.02646, and t∗ = 2.776, so the 95% conÞdence interval for 1/R is 1.0699 to 1.2169. Taking reciprocals gives the interval for R: 0.8218 to 0.9346.

Output from Minitab: The regression equation is Current = 1.14 Voltage

Predictor Coef Stdev t-ratio p Noconstant Voltage 1.14339 0.02646 43.21 0.000

10.19 (a) The plot reveals no outliers or unusual points. (b) The regression equation is ŷ = −2.80 + 0.0387x . (c) t = 16.10 (df = 17); since P < 0.0005, we reject H0 and conclude that linear regression on HR is useful for predicting VO2. (d) When x = 95, we have ŷ = 0.8676 and SEŷ = 0.1205

√ 1+ 119 + (95−107)

2518 = 0.1269, so the 95% prediction interval is 0.8676 ± (2.110)(0.1269), or 0.5998 to 1.1354. When x = 110, we have ŷ = 1.4474 and SEŷ = 0.1205

√ 1+ 119 + (110−107)

2518 = 0.1238, so the interval is 1.4474 ± (2.110)(0.1238) = 1.1861 to 1.7086. A portion of the Minitab output that shows these intervals is reproduced below. (e) It depends on how accurately they need to know VO2; the regression equation predicts only the subjectÕs mean VO2 for a given heart rate, and the intervals in (d) reveal that a particular observation may vary quite a bit from that mean.

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90 100 110 120 130

VO 2

Output from Minitab: Fit Stdev.Fit 95.0% C.I. 95.0% P.I.

0.8676 0.0399 ( 0.7834, 0.9518) ( 0.5998, 1.1354) 1.4474 0.0286 ( 1.3871, 1.5076) ( 1.1861, 1.7086)

10.20 t = 0.83/0.065 .= 12.77. The alternative could reasonably be either β1 6= 0 or β1 > 0; the latter makes the reasonable assumption that the association between the two

214 Chapter 10 Inference for Regression

measurements should be positive. In either case, the P-value (for df = 79) is very small: P < 0.001 for the two-sided alternative, P < 0.0005 for the one-sided alternative. In words, this study gives strong evidence that oscillometric measurements are useful for estimating intra-arterialmeasurements (though, as indicated by the parenthetic comments, these estimates are not clinically useful).

10.21 (a) H0: β1 = 0; Ha: β1 > 0. t = 0.00665/0.00182 = 3.654; with df = 16, we have 0.001 < P < 0.0025 (software gives 0.0011). We reject H0 and conclude that greater airßow increases evaporation. (b) A 95% conÞdence interval for β1 is 0.00665± (2.120)(0.00182), or 0.00279 to 0.01051.

10.22 The plot shows a fairly strong positive relationship, with a hint of an upward curve at the high end. There are six unusually low observations in the middle of the plot. As with the corn yield plot, the spread seems to increase over time. Regression gives ŷ = −659 + 0.348x ; this line is shown on the plot. The slope is

signiÞcantly different from 0 (t = 11.03, df = 38, P < 0.0005). Yield has increased at an average rate of 0.348 bushels/acre each year. A normal quantile plot of the residuals (below) suggests deviation from normality. A

plot of residuals vs. year (not shown) again suggests that variability is higher in later years. Linear regression may not be appropriate for this data set.

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1960 1970 1980 1990 2000

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-3 -2 -1 0 1 2 3

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z score Output from Minitab:

The regression equation is Soybeans = - 659 + 0.348 Year

Predictor Coef Stdev t-ratio p Constant -659.47 62.39 -10.57 0.000 Year 0.34827 0.03156 11.03 0.000

s = 2.304 R-sq = 76.2% R-sq(adj) = 75.6%

10.23 (a) The prediction interval is about 123 to 168 bushels/acre: ŷ = 145.60, and SEŷ = 10.28

√ 1+ 140 + (2006−1976.5)

5330 = 11.21, so the 95% prediction interval is 145.60± (2.042)(11.21) = 122.71 to 168.49 (using the table value for df = 30), or 145.60 ± (2.024)(11.21) = 122.91 to 168.29 (using the software critical value for df = 38). (b) The centers are similar (150.25 vs. 145.60), but this interval is narrower. (c) The margin of error in Example 10.14 was 47 bushels/acre, compared to 23 here. It is smaller

Solutions 215

here because the sample size is larger, which decreases t∗ (since df is larger), and decreases SEŷ (since 1/n and 1/

∑ (xi− x)2 are smaller). [The latter effect is slightly offset because

x = 1976.5 rather than 1981, so (x∗ − x)2 = (2006− x)2 is larger than before, but this change is overcome by the greater change in

∑ (xi − x)2: It was 500 in Example 10.14,

and it is 5330 here.]

Output from Minitab: Fit Stdev.Fit 95.0% C.I. 95.0% P.I.

145.60 4.46 ( 136.57, 154.64) ( 122.91, 168.30) X

10.24 (a) There is a fairly strong positive close-to-linear relationship. (b) r = 0.9334Ñ this should be fairly good measure of the relationship, except to the extent that it is not linear. (c) Regression gives ŷ = 12.2+ 0.183x . For the slope, we have t = 16.04, df = 38, P < 0.0005; we conclude that the slope (and correlation) is not 0. (d) The residuals show a curved relationship with time: Generally, the residuals are positive in the earlier and later years, and mostly negative from about 1960 to 1990.

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1960 1970 1980 1990 2000

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Year Output from Minitab:

The regression equation is Soybeans = 12.2 + 0.183 Corn

Predictor Coef Stdev t-ratio p Constant 12.175 1.077 11.31 0.000 Corn 0.18306 0.01142 16.04 0.000

s = 1.695 R-sq = 87.1% R-sq(adj) = 86.8%

216 Chapter 10 Inference for Regression

10.25 The log yield model (using common [base 10] logs) is ŷ = −16.3+ 0.00925x ; if natural logs are used instead, the equation is ŷ = −37.6 + 0.0213x . For either regression, r 2 = 81.7% and t = 13.02. By comparison, for the original model we had t = 13.06 and r 2 = 81.8%. The log model is not particularly better than the original; the plot still suggests that the spread increases as ÒyearÓ increasesÑthough a plot of residuals vs. year shows some improvement in this respectÑand the numerical measures are actually slightly smaller than those in the original.

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1960 1970 1980 1990 2000

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10.26 (a) Below (from Minitab). (b) H0: β1 = 0; this says that current is not linearly related to voltage. (c) If H0 is true, F has an F(1, 3) distribution; F = 231.21 has P < 0.001.

Output from Minitab: Analysis of Variance

SOURCE DF SS MS F p Regression 1 2.0932 2.0932 231.21 0.001 Error 3 0.0272 0.0091 Total 4 2.1203

10.27 (a)Below (fromMinitab). (b) H0: β1 = 0; this says that VO2 is not linearly related to HR. (c) If H0 is true, F has an F(1, 17) distribution; F = 259.27 has P < 0.001. (d)We found t = 16.10, and t2 = 259.21. (e) r 2 = SSM/SST = 3.7619/4.0085 = 93.8%.

Output from Minitab: Analysis of Variance

SOURCE DF SS MS F p Regression 1 3.7619 3.7619 259.27 0.000 Error 17 0.2467 0.0145 Total 18 4.0085

10.28 (a) t = 0.39√38/√1− 0.392 .= 2.611 (b) This is a positive association; use Ha: ρ > 0. (c) P = 0.0064 (or 0.005 < P < 0.01). We conclude that ρ > 0.

10.29 (a) t = −0.19√711/√1− (−0.19)2 = −5.160. (b) We have df = 711, with t = −5.16, P < 0.001; this is signiÞcant (for any reasonable α), so we conclude that ρ 6= 0.

Solutions 217

10.30 (a) The plot shows a strong positive linear pat- tern. (b) ŷ = −0.0333 + 1.02x , s = 0.01472. (c) r = 0.99645; r2 = 99.3% of T2Õs variability is explained by T1. (d) t = 81.96. The alternative Ha could reasonably be either β1 6= 0 or β1 > 0; the latter makes the reasonable assumption that the association between the two measurements should be positive. Either way, P is tiny. In plain language: We can predict T2 to a very high degree of accuracy by multiplying the T1 measurement by 1.02 and sub- tracting 0.0333. The regression gives very strong evidence that the slope is not 0. (e) They agree (up to rounding error): t2 = 6717.44, while F = 6717.94.

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1 1.25 1.5 1.75 2

Output from Minitab: The regression equation is T2 = - 0.0333 + 1.02 T1

Predictor Coef Stdev t-ratio p Constant -0.03328 0.02034 -1.64 0.108 T1 1.01760 0.01242 81.96 0.000

s = 0.01472 R-sq = 99.3% R-sq(adj) = 99.3%

Analysis of Variance

SOURCE DF SS MS F p Regression 1 1.4564 1.4564 6717.94 0.000 Error 48 0.0104 0.0002 Total 49 1.4668

10.31 (a) Table below. b0 and sb1 have changed the most. (b) The full-set ANOVA table is above, the odd-set table below. The most important difference is that Fodds is about half as big as Ffull (though both are quite signiÞcant). MSE is similar in both tables, reßecting the similarity in s in the full and reduced regressions. (c) See table. (d) The relationship is still strong even with half as many data points; most values are similar in both regressions. (e) Since these values did not change markedly for n = 25 vs. n = 50, it seems likely that they will be similar when n = 100.

b0 b1 s sb1 r Full −0.03328 1.01760 0.01472 0.01242 0.99645 Odds −0.05814 1.03111 0.01527 0.01701 0.99688

Output from Minitab: Analysis of Variance

SOURCE DF SS MS F p Regression 1 0.85681 0.85681 3673.71 0.000 Error 23 0.00536 0.00023 Total 24 0.86218

218 Chapter 10 Inference for Regression

10.32 (a) ŷ = 110 − 1.13x . (b) t = −3.63; P .= 0.001 (or 0.0005 < P < 0.001). (c) t∗ = 2.093 and SEb1 = 0.3102; the interval is −1.116 to −0.478. (d) r 2 = 41.0%. (e) s = 11.02. (f) The new equation is only slightly changed: ŷ = 108 − 1.05x . The slope is still signiÞcantly different from 0, though it is not as signiÞcant as before (t = −2.51, 0.01 < P < 0.02). The conÞdence interval is −1.0499 ± (2.110)(0.4186) = −1.933 to−0.167Ñconsiderably wider than before. r2 has decreased (to 27.0%), as has s (to 8.831). Removing Case 18 (high age/low score) makes the association less linear (hence the drop in r 2 and the rise in P). The absence of Case 19 (typical age/high score) lowers s, the estimated variation about the line.

Output from Minitab: The regression equation is Gesell = 110 - 1.13 Age

Predictor Coef Stdev t-ratio p Constant 109.874 5.068 21.68 0.000 Age -1.1270 0.3102 -3.63 0.002

s = 11.02 R-sq = 41.0% R-sq(adj) = 37.9%

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Without 18 and 19 Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð The regression equation is Gesell = 108 - 1.05 Age

Predictor Coef Stdev t-ratio p Constant 107.585 5.724 18.80 0.000 Age -1.0499 0.4186 -2.51 0.023

s = 8.831 R-sq = 27.0% R-sq(adj) = 22.7%

10.33 (a) ŷ = −9.1+1.09x . (b) SEb1 = 0.6529. (c) t = 1.66, which gives P between 0.10 and 0.20 (software gives P = 0.140)Ñnot signiÞcant. With the original data, t = 17.66, which is strong evidence against H0.

Output from Minitab: The regression equation is Gas = - 9.1 + 1.09 HeatDeg

Predictor Coef Stdev t-ratio p Constant -9.10 16.31 -0.56 0.594 HeatDeg 1.0857 0.6529 1.66 0.140

s = 24.78 R-sq = 28.3% R-sq(adj) = 18.1%

Solutions 219

10.34 (a) The plot suggests a linear relationship, so it is appropriate to use a correlation. Note that in this case, since both measurements are in centimeters, it is best if both axes have the same scale. Also note that either variable may be on the horizontal axis. (b) r = 0.99415, which gives t = 15.94 (df = 3), so the two-sided P-value is P = 0.00054. This correlation is different from (greater than) 0.

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10.35 (a) There is a moderate positive relation- ship; player 8Õs point is an outlier. Note: Either variable may be plotted on the horizontal axis, although perhaps Round 1 scores are the most logical choice for the explanatory variable. Ide- ally, both scales should be equal. (b) r = 0.687, so t = 0.687√10/√1− 0.6872 = 2.99 (df = 10); this gives two-sided P-value 0.0136 (or 0.01 < P < 0.02)Ñfairly strong evidence that ρ 6= 0. (c) r = 0.842, so t = 0.842√9/√1− 0.8422 = 4.68 (df = 9); this gives P = 0.0012 (or 0.001 < P < 0.002)Ñstronger evidence that ρ 6= 0. The outlier makes the plot less linear, and so decreases the correlation.

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10.36 (a) b1 = r sy/sx = 0.68× 20.3/17.2 .= 0.80256, and b0 = y − b1x .= −21.43. The equation is ŷ = −21.42 + 0.80256x . (b) t = 0.68√48/√1− 0.682 = 6.42 (df = 48); this gives P < 0.0005. We conclude that the slope is not 0.

10.37 With n = 20, t = 2.45 (df = 18, 0.02 < P < 0.04), while with n = 10, t = 1.63 (df = 8, 0.1 < P < 0.2). With the larger sample size, r should be a better estimate of ρ, so we are less likely to get r = 0.5 unless ρ is really not 0.

10.38 Most of the small banks have nega- tive residuals, while the large ones have mostly positive residuals. This means that, generally, wages at large banks are higher, and small bank wages are lower, than we would predict from the regression.

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220 Chapter 10 Inference for Regression

10.39 (a) There is a positive association between scores. The 47th pair of scores (circled) is an outlierÑthe ACT score (21) is higher than one would expect for the SAT score (420). Since this SAT score is so low, this point may be inßuential. No other points fall outside the pattern. (b) The regression equation is ŷ = 1.63 + 0.0214x ; t = 10.78 which gives P < 0.001 (df = 58). (c) r = 0.8167.

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Output from Minitab: The regression equation is ACT = 1.63 + 0.0214 SAT

Predictor Coef Stdev t-ratio p Constant 1.626 1.844 0.88 0.382 SAT 0.021374 0.001983 10.78 0.000

s = 2.744 R-sq = 66.7% R-sq(adj) = 66.1%

10.40 (a) The means are identical (21.133). (b) For the observed ACT scores, sy = 4.714; for the Þtted values, sŷ = 3.850. (c) For z = 1, the SAT score is x+sx = 912.7+180.1 = 1092.8. The predicted ACT score is ŷ .= 25 (Minitab reports 24.983), which gives a standard score of about 1 (using the standard deviation of the predicted ACT scores. (d) For z = −1, the SAT score is x − sx = 912.7− 180.1 = 732.6. The predicted ACT score is ŷ .= 17.3 (Minitab reports 17.285), which gives a standard score of about −1. (e) It appears that the standard score of the predicted value is the same as the standard score of the explanatory variable value. (See note below.) Notes: (a) This will always be true, since

∑ i ŷi = ∑i(b0 + b1xi) = n b0 + b1∑i xi =

n(y − b1x) + b1n x = n y. (b) The standard deviation of the predicted values will be sŷ = |r |sy; in this case, sŷ = (0.8167)(4.714). To see this, note that the variance of the predicted values is 1n−1

∑ i(ŷi−y)2 = 1n−1

∑ i(b1xi−b1x)2 = b21 s2x = r 2s2y . (e) For a given

standard score z, note that ŷ = b0+ b1(x + z sx) = y − b1x + b1x + b1z sx = y + z r sy . If r > 0, the standard score for ŷ equals z; if r < 0, the standard score is −z.

10.41 (a) SAT: x = 912.6 and sx = 180.1 points. ACT: y = 21.13 and sy = 4.714 points. So, a1

.= 0.02617 and a0 .= −2.756. (More accurate computation gives a0

.= −2.752.) (b) The new line is dashed. (c) For example, the Þrst prediction is −2.756+ (0.02617)(1000) = 23.42. Up to rounding error, the mean and standard deviation are the same.

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222 Chapter 11 Multiple Regression

Chapter 11 Solutions

11.1 (a) H0: β1 = β2 = · · · = β13 = 0 vs. Ha: at least one β j 6= 0. The degrees of freedom are 13 and 2215, and P < 0.001 (referring to an F(12, 1000) distribution). We have strong evidence that at least one of the β j is not 0. (b) The regression explains 29.7% of the variation. (c) Each t statistic tests H0: β j = 0 vs. Ha: β j 6= 0, and has df = 2215. The critical value is t∗ = 1.961. (d) The only three coefÞcients that are not signiÞcantly different from 0 are those for Òtotal payments,Ó Òmale borrower,Ó and Òmarried.Ó (e) Interest rates are lower for larger loans, for longer terms, with larger down payments, when there is a cosigner, when the loan is secured, when the borrower has a higher income, when the credit report is not considered Òbad,Ó for older borrowers, when the borrower owns a home, and for borrowers who have lived for a long time at their present address.

11.2 (a) H0: β1 = β2 = · · · = β13 = 0 vs. Ha: at least one β j 6= 0. The degrees of freedom are 13 and 5650, and P < 0.001 (referring to an F(12, 1000) distribution). We have strong evidence that at least one of the β j is not 0. (b) The regression explains 14.1% of the variationÑmuch less than for the direct loans. (c) Each t statistic tests H0: β j = 0 vs. Ha: β j 6= 0, and has df = 5650. The critical value is t∗ = 1.9604. (d) Only the coefÞcients of Òloan size,Ó Òlength of loan,Ó Òpercent down payment,Ó and Òunsecured loanÓ are signiÞcantly different from 0. (e) Interest rates are lower for larger loans, for longer terms, with larger down payments, and when the loan is secured.

11.3 In 11.1, we found that 10 factors have a signiÞcant effect on the interest rate for direct loans, while based on 11.2, only four of the factors examined have a signiÞcant impact on the interest rate for indirect loans. Furthermore, a greater proportion of the variation in interest rates is explained by the regression for direct loans than that for indirect.

11.4 (a) Between GPA and IQ, r = 0.634 (straight-line regression explains r 2 = 40.2% of the variation in GPA). Between GPA and self-concept, r = 0.542 (straight-line regression explains r 2 = 29.4% of the variation in GPA). Since gender is cate- gorical, the correlation between GPA and gender is not meaningful. (b) Model: µGPA = β0 + β1 IQ + β2 Self-Concept. (c) Regression gives the equation ĜPA = −3.88 + 0.0772 IQ + 0.0513 Self-Concept. Based on the reported value of R2, the re- gression explains 47.1% of the variation in GPA. (So the inclusion of self-concept only adds about 6.9% to the variation explained by the regression.) (d) We test H0: β2 = 0 vs. Ha: β2 6= 0. The test statistic t = 3.14 (df = 75) has P = 0.002; we conclude that the coefÞcient of Self-Concept is not 0.

Solutions 223

Output from Minitab: The regression equation is GPA = - 3.88 + 0.0772 IQ + 0.0513 SelfCcpt

Predictor Coef Stdev t-ratio p Constant -3.882 1.472 -2.64 0.010 IQ 0.07720 0.01539 5.02 0.000 SelfCcpt 0.05125 0.01633 3.14 0.002

s = 1.547 R-sq = 47.1% R-sq(adj) = 45.7%

11.5 (a)With the given values,µGPA = β0+9β1+8β2+7β3. (b)Weestimate ĜPA = 2.697. Among all computer science students with the given high school grades, we expect the mean college GPA after three semesters to be about 2.7.

11.6 (a)With the given values,µGPA = β0+6β1+7β2+8β3. (b)Weestimate ĜPA = 2.202. Among all computer science students with the given high school grades, we expect the mean college GPA after three semesters to be about 2.2.

11.7 The critical value for df = 220 is t∗ .= 1.9708. If using the table, take t∗ = 1.984. (a) b1 ± t∗ SEb1 = 0.0986 to 0.2385 (or 0.0982 to 0.2390). This coefÞcient gives the average increase in college GPA for each 1-point increase in high school math grade. (b) b3± t∗ SEb3 = −0.0312 to 0.1214 (or−0.0317 to 0.1219). This coefÞcient gives the average increase in college GPA for each 1-point increase in high school English grade.

11.8 The critical value for df = 221 is t∗ .= 1.9708. If using the table, take t∗ = 1.984. (a) b1 ± t∗ SEb1 = 0.1197 to 0.2456 (or 0.1193 to 0.2461). This coefÞcient gives the average increase in college GPA for each 1-point increase in high school math grade. (b) b2± t∗ SEb2 = −0.0078 to 0.1291 (or−0.0082 to 0.1296). This coefÞcient gives the average increase in college GPA for each 1-point increase in high school English grade. The coefÞcients (and standard errors) can change greatly when the model changes.

11.9 (a) ĜPA = 0.590+0.169HSM+0.034HSS+0.045HSE. (b) s = √MSE = 0.69984. (c) H0: β1 = β2 = β3 = 0; Ha: at least one β j 6= 0. In words, H0 says that none of the high school grade variables are predictors of college GPA (in the form given in the model); Ha says that at least one of them is. (d) Under H0, F has an F(3, 220) distribution. Since P = 0.0001, we reject H0. (e) The regression explains 20.46% of the variation in GPA.

11.10 (a) ĜPA = 1.289+0.002283SATM−0.00002456SATV. (b) s = √MSE = 0.75770. (c) H0: β1 = β2 = 0; Ha: at least one β j 6= 0. In words, H0 says that neither SAT score predicts college GPA (in the form given in the model); Ha says that at least one of them is a predictor. (d) Under H0, F has an F(2, 221) distribution. Since P = 0.0007, we reject H0. (e) The regression explains 6.34% of the variation in GPA.

224 Chapter 11 Multiple Regression

11.11 A 95% prediction interval is $2.136± (1.984)($0.013), or $2.1102 to $2.1618. The actual price falls in this interval (in fact, it is less than one standard error below the predicted value), so there is not enough evidence to reject H0, which in this situation would be Òthere was no manipulation.Ó

11.12 There are no clear, strong patterns. The GPA/SATM plot suggests a slight positive association, but it is weakened by the two low SATM scores which do not follow the pattern.

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11.13 All of the plots display lots of scatter; high school grades seem to be poor predictors of college GPA. Of the three, the math plot seems to most strongly suggest a positive association, although the association appears to be quite weak, and almost nonexistent for HSM < 5. We also observe that scores below 5 are unusual for all three high school variables, and could be considered outliers and inßuential.

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Solutions 225

11.14 The regression equation (given in the answer to Exercise 11.9 and Figure 11.4) is ĜPA = 0.590+0.169HSM+0.034HSS+ 0.045 HSE. Among other things, we note that most of the residuals associated with low HS grades, and (not coincidentally) with low predicted GPAs, are ÒlargeÓ (positive, or just slightly negative). Also, using this model, the predicted GPAs are all between 1.33 and 3.07.

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11.15 The regression equation (given in the answer to Exercise 11.10 and Figure 11.7) is ĜPA = 1.289+ 0.002283 SATM− 0.00002456 SATV. The residual plots show no striking patterns, but one notice- able feature is the similarity between the predicted GPA and SATM plotsÑwhich results from the fact that the coefÞcient of SATV is so small that predicted GPA is almost a linear function of SATM alone.

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11.16 (a) ĜPA = 0.666+0.193HSM+0.000610SATM. (b) H0: β1 = β2 = 0; Ha: at least one β j 6= 0. In words, H0 says that neither mathematics variable is a predictor of college GPA (in the form given in the model); Ha says that at least one of them is. The F statistics

226 Chapter 11 Multiple Regression

(with df 2 and 221) is 26.63; this has P < 0.0005, sowe reject H0. Minitab output follows. (c) The critical value for df = 221 is t∗ .= 1.9708. If using the table, take t∗ = 1.984. For the coefÞcient of HSM, SEb1 = 0.03222, so the interval is 0.1295 to 0.2565 (or 0.1291 to 0.2569). For the coefÞcient of SATM, SEb2 = 0.0006112, so the interval is −0.000594 to 0.001815 (or −0.000602 to 0.001823)Ñwhich contains 0. (d) HSM: t = 5.99, P < 0.0005. SATM: t = 1.00, P = 0.319. As the intervals indicated, the coefÞcient of SATM is not signiÞcantly different from 0. (e) s = √MSE = 0.7028. (f) The regression explains 19.4% of the variation in GPA.

Output from Minitab: The regression equation is GPA = 0.666 + 0.193 HSM +0.000610 SATM

Predictor Coef Stdev t-ratio p Constant 0.6657 0.3435 1.94 0.054 HSM 0.19300 0.03222 5.99 0.000 SATM 0.0006105 0.0006112 1.00 0.319

s = 0.7028 R-sq = 19.4% R-sq(adj) = 18.7%

Analysis of Variance

SOURCE DF SS MS F p Regression 2 26.303 13.151 26.63 0.000 Error 221 109.160 0.494 Total 223 135.463

11.17 The regression equation is ĜPA = 1.28 + 0.143 HSE + 0.000394 SATV, and R2 = 8.6%. The regression is signiÞcant (F = 10.34, with df 2 and 221); the t-tests reveal that the coefÞcient of SATV is not signiÞcantly different from 0 (t = 0.71, P = 0.481). For mathematics variables, wehad R2 = 19.4%Ñnot overwhelmingly large, but considerably more than that for verbal variables.

Output from Minitab: The regression equation is GPA = 1.28 + 0.143 HSE +0.000394 SATV

Predictor Coef Stdev t-ratio p Constant 1.2750 0.3474 3.67 0.000 HSE 0.14348 0.03428 4.19 0.000 SATV 0.0003942 0.0005582 0.71 0.481

s = 0.7487 R-sq = 8.6% R-sq(adj) = 7.7%

Analysis of Variance

SOURCE DF SS MS F p Regression 2 11.5936 5.7968 10.34 0.000 Error 221 123.8692 0.5605 Total 223 135.4628

11.18 For males, regression gives ĜPA = 0.582+0.155HSM+0.0502HSS+0.0445HSE, with R2 = 18.4%. The regression is signiÞcant (F = 10.62 with df 3 and 141; P <

Solutions 227

0.0005), but only the coefÞcient of HSM is signiÞcantly different from 0 (even the constant 0.582 has t = 1.54 and P = 0.125). Regression with HSM and HSS (excluding HSE since it has the largest P-value) gives the equation ĜPA = 0.705 + 0.159 HSM + 0.0738 HSS, and R2 = 18.0%. The P-values for the constant and the coefÞcient of HSS are smaller (although the latter is still not signiÞcantly different from 0). One might also regress on HSM alone; this has R2 = 16.3%. Minitab output for all three models follows. Residual plots (not shown) do not suggest

problems with any of the models. Comparing these results to Figures 11.4 and 11.6, note that with all students, we

excluded HSS (rather than HSE) in the second model.

Output from Minitab: The regression equation is GPAm = 0.582 + 0.155 HSMm + 0.0502 HSSm + 0.0445 HSEm

Predictor Coef Stdev t-ratio p Constant 0.5818 0.3767 1.54 0.125 HSMm 0.15502 0.04487 3.45 0.001 HSSm 0.05015 0.05070 0.99 0.324 HSEm 0.04446 0.05037 0.88 0.379

s = 0.7363 R-sq = 18.4% R-sq(adj) = 16.7%

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð SECOND MODEL Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð

The regression equation is GPAm = 0.705 + 0.159 HSMm + 0.0738 HSSm

Predictor Coef Stdev t-ratio p Constant 0.7053 0.3495 2.02 0.045 HSMm 0.15863 0.04465 3.55 0.001 HSSm 0.07383 0.04299 1.72 0.088

s = 0.7357 R-sq = 18.0% R-sq(adj) = 16.8%

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð THIRD MODEL Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð

The regression equation is GPAm = 0.962 + 0.200 HSMm

Predictor Coef Stdev t-ratio p Constant 0.9619 0.3181 3.02 0.003 HSMm 0.19987 0.03790 5.27 0.000

s = 0.7407 R-sq = 16.3% R-sq(adj) = 15.7%

11.19 For females, regression gives ĜPA = 0.648+0.205HSM+0.0018HSS+0.0324HSE, with R2 = 25.1%. In this equation, only the coefÞcient of HSM is signiÞcantly different from 0 (even the constant 0.648 has t = 1.17 and P = 0.247). Regression with HSM and HSE (excluding HSS since it has the largest P-value) gives the equation ĜPA = 0.648+ 0.206 HSM+ 0.0333 HSE, and R2 = 25.1%Ñbut the P-values for the constant and coefÞcient of HSE have changed very little. With HSMalone, the regression equation is ĜPA = 0.821+0.220HSM, R2 decreases only slightly to 24.9%, and both the constant

228 Chapter 11 Multiple Regression

and coefÞcient are signiÞcantly different from 0. Minitab output for all three models follows. Residual plots (not shown) do not suggest

problems with any of the models. Comparing the results to males, we see that both HSM and HSS were fairly useful for

men, but HSM was sufÞcient for womenÑbased on R2, HSM alone does a better job for women than all three variables for men.

Output from Minitab: The regression equation is GPAf = 0.648 + 0.205 HSMf + 0.0018 HSSf + 0.0324 HSEf

Predictor Coef Stdev t-ratio p Constant 0.6484 0.5551 1.17 0.247 HSMf 0.20512 0.06134 3.34 0.001 HSSf 0.00178 0.05873 0.03 0.976 HSEf 0.03243 0.08270 0.39 0.696

s = 0.6431 R-sq = 25.1% R-sq(adj) = 22.1%

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð SECOND MODEL Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð

The regression equation is GPAf = 0.648 + 0.206 HSMf + 0.0333 HSEf

Predictor Coef Stdev t-ratio p Constant 0.6483 0.5514 1.18 0.243 HSMf 0.20596 0.05430 3.79 0.000 HSEf 0.03328 0.07732 0.43 0.668

s = 0.6389 R-sq = 25.1% R-sq(adj) = 23.1%

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð THIRD MODEL Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð

The regression equation is GPAf = 0.821 + 0.220 HSMf

Predictor Coef Stdev t-ratio p Constant 0.8213 0.3755 2.19 0.032 HSMf 0.21984 0.04347 5.06 0.000

s = 0.6355 R-sq = 24.9% R-sq(adj) = 24.0%

11.20 The correlations are on the right. Of these, the correlation between GPA and IQ is largest in absolute value, so the relationship between them is closest to a straight line. About 40.2% of the variation in GPA is explained by the relationship with IQ.

IQ 0.634 C2 0.601 AGE −0.389 C3 0.495 SEX −0.097 C4 0.267 SC 0.542 C5 0.472 C1 0.441 C6 0.401

Solutions 229

11.21 (a) Regression gives ĜPA = −2.83 + 0.0822 IQ + 0.163 C3, and R2 = 45.9%. For the coefÞcient of C3, t = 2.83, which has P = 0.006ÑsigniÞcantly different from 0. C3 increases R2 by 5.7% = 45.9% − 40.2%. (b) Regression now gives ĜPA = −3.49+0.0761 IQ+0.0670C3+0.0369SC, and R2 = 47.5%. For the coefÞcient of C3, t = 0.78, which has P = 0.436Ñnot signiÞcantly different from 0. When self-concept (SC) is included in the model, C3 adds little. (If we regress on IQ and SC, R2 = 47.1%). (c) The values change because coefÞcients are quite sensitive to changes in the model, especially when the explanatory variables are highly correlated (the correlation between SC and C3 is about 0.80). In this case, the predictive information of SC and C3 overlap, so that the two of them together add little more than either one separately (with IQ).

Output from Minitab: The regression equation is GPA = - 2.83 + 0.0822 IQ + 0.163 C3

Predictor Coef Stdev t-ratio p Constant -2.829 1.507 -1.88 0.064 IQ 0.08220 0.01508 5.45 0.000 C3 0.16289 0.05752 2.83 0.006

s = 1.564 R-sq = 45.9% R-sq(adj) = 44.5%

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð SECOND MODEL Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð

The regression equation is GPA = - 3.49 + 0.0761 IQ + 0.0670 C3 + 0.0369 SC

Predictor Coef Stdev t-ratio p Constant -3.491 1.558 -2.24 0.028 IQ 0.07612 0.01549 4.91 0.000 C3 0.06701 0.08558 0.78 0.436 SC 0.03691 0.02456 1.50 0.137

s = 1.551 R-sq = 47.5% R-sq(adj) = 45.4%

230 Chapter 11 Multiple Regression

11.22 (a)Regression gives ĜPA = −4.94+0.0815IQ+0.183C1+0.142C5, R2 = 52.5%, and s = 1.475. With the given values of IQ, C1, and C5, ĜPA = 7.457. (b) GPA would increase by about 0.0815 per IQ point (the coefÞcient of IQ). SEb1 = 0.01367; with df = 74, t∗ = 1.9926 (or use t∗ = 2.000 from the table). Interval: 0.0543 to 0.1087 (or 0.0542 to 0.1088). (c) The residual plots are below. The residual for OBS = 55 stands out as being extraordinarily low; this student had the lowest GPA and, at 15 years old, was the oldest. (d) Regression now gives ĜPA = −4.68+ 0.0805 IQ+ 0.197 C1+ 0.109 C5, R2 = 57.4%, and s = 1.303. With the given values of IQ, C1, and C5, ĜPA = 7.534. Removing this observation did not greatly change the model or the prediction, although the coefÞcient of C5 is not quite signiÞcant under the new regression (seeMinitab output).

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Output from Minitab: The regression equation is GPA = - 4.94 + 0.0815 IQ + 0.183 C1 + 0.142 C5

Predictor Coef Stdev t-ratio p Constant -4.937 1.491 -3.31 0.001 IQ 0.08145 0.01367 5.96 0.000 C1 0.18308 0.06475 2.83 0.006 C5 0.14205 0.06663 2.13 0.036

s = 1.475 R-sq = 52.5% R-sq(adj) = 50.6%

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Without OBS 55 Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð

The regression equation is GPA = - 4.68 + 0.0805 IQ + 0.197 C1 + 0.109 C5

Predictor Coef Stdev t-ratio p Constant -4.678 1.318 -3.55 0.001 IQ 0.08050 0.01207 6.67 0.000 C1 0.19707 0.05724 3.44 0.001 C5 0.10950 0.05923 1.85 0.069

s = 1.303 R-sq = 57.4% R-sq(adj) = 55.7%

Solutions 231

11.23 In the table, two IQRs are given; those in parentheses are based on quartiles reported by Minitab, which computes quartiles in a slightly different way from this textÕs method. None of the variables show striking deviations from normality in the quantile plots

(not shown). Taste and H2S are slightly right-skewed, and Acetic has two peaks. There are no outliers.

x M s IQR Taste 24.53 20.95 16.26 23.9 (or 24.58) Acetic 5.498 5.425 0.571 0.656 (or 0.713) H2S 5.942 5.329 2.127 3.689 (or 3.766) Lactic 1.442 1.450 0.3035 0.430 (or 0.4625)

Taste 0 00 0 556 1 1234 1 55688 2 011 2 556 3 24 3 789 4 0 4 7 5 4 5 67

Acetic 4 455 4 67 4 8 5 1 5 2222333 5 444 5 677 5 888 6 0011 6 3 6 44

H2S 2 9 3 1268899 4 17799 5 024 6 11679 7 4699 8 7 9 025 10 1

Lactic 8 6 9 9 10 689 11 56 12 5599 13 013 14 469 15 2378 16 38 17 248 18 1 19 09 20 1

232 Chapter 11 Multiple Regression

11.24 The plots show positive associations between the variables. The correlations and P-values (in parentheses) are at the right; all are positive (as expected) and signiÞcantly different from 0. [Recall that the P-values are correct if the two variables are normally distributed, in which case t = r√n − 2/√1− r 2 has a t (n − 2) distribution if ρ = 0.]

Taste Acetic H2S Acetic 0.5495

(0.0017) H2S 0.7558 0.6180

(<0.0001) (0.0003) Lactic 0.7042 0.6038 0.6448

(<0.0001) (0.0004) (0.0001)

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Solutions 233

11.25 The regression equation is T̂aste = −61.5 + 15.6 Acetic; the coefÞcient of Acetic has t = 3.48, which is signiÞcantly different from 0 (P = 0.002). The regression explains r 2 = 30.2% of the variation in Taste. Based on stem- and quantile plots (not

shown), the residuals seem to have a normal distribution. Scatterplots (below) reveal positive associations between residuals and both H2S and Lactic. Further analysis of the residuals shows a stronger positive association between residuals and Taste, while the plot of residuals vs. Acetic suggests greater scatter in the residuals for large Acetic values.

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234 Chapter 11 Multiple Regression

11.26 Regression gives T̂aste = −9.79 + 5.78 H2S. The coefÞcient of H2S has t = 6.11 (P < 0.0005); it is signiÞcantly different from 0. The regression explains r2 = 57.1% of the variation in Taste. Based on stem- and quantile plots (not

shown), the residuals may be slightly skewed, but do not differ greatly from a normal distribution. Scatterplots (below) reveal weak positive associations between residuals and both Acetic and Lactic. Further analysis of the residuals shows a moderate positive association between residuals and Taste, while the plot of residuals vs. H2S suggests greater scatter in the residuals for large H2S values.

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Solutions 235

11.27 Regression gives T̂aste = −29.9 + 37.7 Lactic. The coefÞcient of Lactic has t = 5.25 (P < 0.0005); it is signiÞcantly different from 0. The regression explains r 2 = 49.6% of the variation in Taste. Based on stem- and quantile plots (not

shown), the residuals seem to have a normal distribution. Scatterplots reveal a moderately strong positive association between residuals and Taste, but no striking patterns for residuals vs. the other variables.

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11.28 All information is in the table at the right. The intercepts differ from one model to the next because they represent different thingsÑe.g., in the Þrst model, the intercept is the predicted value of Taste with Acetic = 0, etc.

x T̂aste = F P r2 s Acetic −61.5+ 15.6x 12.11 0.002 30.2% 13.82 H2S −9.79+ 5.78x 37.29 <0.0005 57.1% 10.83 Lactic −29.9+ 37.7x 27.55 <0.0005 49.6% 11.75

236 Chapter 11 Multiple Regression

11.29 The regression equation is T̂aste = −26.9 + 3.80 Acetic + 5.15 H2S. The model explains 58.2% of the variation in Taste. The t-value for the coefÞcient of Acetic is 0.84 (P = 0.406), indicating that it does not add signiÞcantly to the model when H2S is used, because Acetic and H2S are correlated (in fact, r = 0.618 for these two variables). This model does a better job than any of the three simple linear regression models, but it is not much better than the model with H2S alone (which explained 57.1% of the variation in Taste)Ñas we might expect from the t-test result.

11.30 The regression equation is T̂aste = −27.6 + 3.95 H2S + 19.9 Lactic. The model explains 65.2% of the variation in Taste, which is higher than for the two simple linear regressions. Both coefÞcients are signiÞcantly different from 0 (P = 0.002 for H2S, and P = 0.019 for Lactic).

11.31 The regression equation is T̂aste = −28.9 + 0.33 Acetic + 3.91 H2S + 19.7 Lactic. The model explains 65.2% of the variation in Taste (the same as for the model with only H2S and Lactic). Residuals of this regression are positively associated with Taste, but they appear to be normally distributed and show no patterns in scatterplots with other variables. The coefÞcient of Acetic is not signiÞcantly different from 0 (P = 0.942); there is no

gain in adding Acetic to the model with H2S and Lactic. It appears that the best model is the H2S/Lactic model of Exercise 11.30.

11.32 (a) Equation: Ĉorn = −3545 + 1.84 Year. The slope is signiÞcantly different from 0 (t = 13.06, P < 0.0005). r 2 = 81.8%. (b) Plot below, left. The residuals look reasonably close to normal (perhaps slightly left-skewed). (c) Plot below, right. The residuals show a weak positive association with soybean yield.

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Solutions 237

11.33 (a) Equation: Ĉorn = −46.2 + 4.76 Soybeans. The slope is signiÞcantly different from 0 (t = 16.04, P < 0.0005). r2 = 87.1%. (b) Plot below, left. The quantile plot is fairly close to linear; there is one high residual, but it is not so high that we would call it an outlier. (c) The plot (below, right) reveals a curved pattern: Generally, the residuals are negative in the earlier and later years, and mostly positive from 1964 to 1990.

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11.34 (a) H0: β1 = β2 = 0; Ha: Not all β j = 0. F = 176.05 (df 2 and 37) has P < 0.0005; we reject H0. (b) R2 = 90.5%Ñslightly better than the Soybeans model (with r 2 = 87.1%) and considerably better than the Yearmodel (r 2 = 81.8%). (c) Ĉorn = −1510+0.765Year+3.08Soybeans. The coefÞcients can change greatly when the model changes. (d) Year: t = 3.62, P = 0.001. Soybeans: t = 5.82, P < 0.0005. Both are signiÞcantly different from 0. (e)With df = 37, use t∗ = 2.0262 (or 2.042 from the table). For the coefÞcient of Year, SEb1 = 0.2114, so the interval is 0.3369 to 1.1935 (or 0.3335 to 1.1969). For the coefÞcient of Soybeans, SEb2 = 0.5300, so the interval is 2.0109 to 4.1587 (or 2.0025 to 4.1671). (f) The plot of residuals vs. soybean yield looks Þne, but the plot of residuals vs. year still shows a curved relationship.

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Output from Minitab: The regression equation is Corn = - 1510 + 0.765 Year + 3.08 Soybeans

Predictor Coef Stdev t-ratio p Constant -1510.3 404.6 -3.73 0.001 Year 0.7652 0.2114 3.62 0.001 Soybeans 3.0848 0.5300 5.82 0.000

238 Chapter 11 Multiple Regression

(Output continues) s = 7.529 R-sq = 90.5% R-sq(adj) = 90.0%

Analysis of Variance

SOURCE DF SS MS F p Regression 2 19957.9 9978.9 176.05 0.000 Error 37 2097.3 56.7 Total 39 22055.2

11.35 (a) The regression equation is Ĉorn = −964 + 0.480 Year − 0.0451 Year2 + 3.90 Soybeans. (b) H0: β1 = β2 = β3 = 0 vs. Ha: Not all β j = 0. F = 233.05 (df 3 and 36) has P < 0.0005, so we conclude that the regression is signiÞcant (at least one coefÞcient is not 0). (c) R2 = 95.1% (compared with 90.5% for the model without Year2). (d)All three coefÞcients are signiÞcantly different from 0Ñthe t values are 2.97, −5.82, and 9.51, all with df = 36; the largest P-value (for the Þrst of these) is 0.005, while the other two are less than 0.0005. (e) The residuals seem to be (close to) normal, and they have no apparent relationship with the explanatory or response variables.

Output from Minitab: The regression equation is Corn = - 964 + 0.480 Year - 0.0451 Year2 + 3.90 Soybeans

Predictor Coef Stdev t-ratio p Constant -964.1 308.9 -3.12 0.004 Year 0.4800 0.1614 2.97 0.005 Year2 -0.045083 0.007742 -5.82 0.000 Soybeans 3.9039 0.4104 9.51 0.000

s = 5.477 R-sq = 95.1% R-sq(adj) = 94.7%

Analysis of Variance

SOURCE DF SS MS F p Regression 3 20975.2 6991.7 233.05 0.000 Error 36 1080.0 30.0 Total 39 22055.2

Solutions 239

11.36 (a) The regression is signiÞ- cant (F = 89.03 with df 2 and 37, P < 0.0005). The t-statistics for Year and Year2 are 13.26 (P < 0.0005) and −1.48 (P = 0.148), respectively. (b) CoefÞcients can change greatly when the model changes; the Year2 term does not make a signiÞcant con- tribution to the model in the absence of Soybeans. (c) The regression func- tions are similar from about 1960 to 1990; the differences emerge in the earlier and later years from the data set.

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Output from Minitab: The regression equation is Corn = - 3542 + 1.84 Year - 0.0198 Year2

Predictor Coef Stdev t-ratio p Constant -3542.0 274.2 -12.92 0.000 Year 1.8396 0.1387 13.26 0.000 Year2 -0.01985 0.01345 -1.48 0.148

s = 10.13 R-sq = 82.8% R-sq(adj) = 81.9%

11.37 Portions of the Minitab output follow; see also the graph in Exercise 11.36. For the simple linear regression, the predicted yield is 145.6; the 95% prediction interval is 122.91 to 168.30. For the multiple regression, the predicted yield is 130.98; the 95% prediction interval is 100.91 to 161.04. The second prediction is lower because the quadratic (curved) model allows for the rate of change to decreaseÑwith the multiple regression model, corn yield grows less rapidly in later years than it did in the earlier years.

Output from Minitab: Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Linear model Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð MTB > Regress ’Corn’ 1 ’Year’; SUBC> predict 2006.

The regression equation is Corn = - 3545 + 1.84 Year ...

Fit Stdev.Fit 95.0% C.I. 95.0% P.I. 145.60 4.46 ( 136.57, 154.64) ( 122.91, 168.30) X

240 Chapter 11 Multiple Regression

(Output continues) Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Quadratic model Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð

MTB > Regress ’Corn’ 2 ’Year’ ’Year2’; SUBC> predict 2006 870.25.

The regression equation is Corn = - 3542 + 1.84 Year - 0.0198 Year2 ...

Fit Stdev.Fit 95.0% C.I. 95.0% P.I. 130.98 10.84 ( 109.01, 152.95) ( 100.91, 161.04) XX

X denotes a row with X values away from the center

11.38 Portions of the Minitab output are below; see also the graph in Exercise 11.36. For the simple linear regression, the predicted yield is 129.05; the 95% prediction interval is 107.17 to 150.92. For the multiple regression, the predicted yield is 123.35; the 95% prediction interval is 100.41 to 146.29. The second prediction is again lower, but not as much as before: Since 1997 is not so far from the years in the data set, the two models have not separated toomuch. This also accounts for the prediction intervals being smaller.

Output from Minitab: Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Linear model Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð MTB > Regress ’Corn’ 1 ’Year’; SUBC> predict 1997.

The regression equation is Corn = - 3545 + 1.84 Year ...

Fit Stdev.Fit 95.0% C.I. 95.0% P.I. 129.05 3.31 ( 122.34, 135.76) ( 107.17, 150.92)

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Quadratic model Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð MTB > Regress ’Corn’ 2 ’Year’ ’year2’; SUBC> predict 1997 420.25.

The regression equation is Corn = - 3542 + 1.84 Year - 0.0198 year2 ...

Fit Stdev.Fit 95.0% C.I. 95.0% P.I. 123.35 5.05 ( 113.11, 133.59) ( 100.41, 146.29) X

X denotes a row with X values away from the center

Solutions 241

11.39 The outlier is observation 15 (Wages = 97.6801). Without it, the regression equation is Ŵages = 43.4+0.0733LOS. For the coefÞcient of LOS, t = 2.85 (P = 0.006). While this is signiÞcant, the predictions are not too good: Only r2 = 12.5% of the variation in Wages is explained by the regression.

Output from Minitab: The regression equation is Wages = 43.4 + 0.0733 LOS

Predictor Coef Stdev t-ratio p Constant 43.383 2.248 19.30 0.000 LOS 0.07325 0.02571 2.85 0.006

s = 10.21 R-sq = 12.5% R-sq(adj) = 10.9%

11.40 (a) The regression equation is Ŵages = 44.0+ 7.93 Size. For testing H0: β1 = 0 vs. Ha: β1 6= 0, t = 2.96 with df = 57; P = 0.004, so the coefÞcient of Size is signiÞcantly different from 0. (b) Large banks: n1 = 34, x1 = 51.91, s1 = 10.67. Small banks: n2 = 25, x2 = 43.97, s2 = 9.41. The pooled standard deviation is sp = 10.16 (the same as

√ MSE); the t-statistic is

the same (up to rounding), and df = n1 + n2 − 2 = 57. The slope β1 represents the change in Wages per unit change in bank size, so it estimates the difference in the means between small (size 0) and large (size 1) banks. Testing β1 = 0 is therefore equivalent to testing µ0 = µ1. (c) The residuals are positively associated with LOS.

o o o

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242 Chapter 11 Multiple Regression

11.41 The regression equation is Ŵages = 37.6+0.0829LOS+8.92Size. Both coefÞcients are signiÞcantly different from 0 (t = 3.53 and t = 3.63, respectively); the regression explains 29.1% of the variation in Wages (compared to 12.5% for LOS alone, and 13.4% for Size alone). The residuals look normal, and do not seem to be associated with LOS. There may be a relationship between the residuals and the size; speciÞcally, the residuals for small banks have less scatter than do those for large banks (s0 = 7.0 vs. s1 = 10.5).

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10 15 20 25

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es id

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z scores

Output from Minitab: The regression equation is Wages = 37.6 + 0.0829 LOS + 8.92 SizeCode

Predictor Coef Stdev t-ratio p Constant 37.565 2.596 14.47 0.000 LOS 0.08289 0.02349 3.53 0.001 SizeCode 8.916 2.459 3.63 0.001

s = 9.273 R-sq = 29.1% R-sq(adj) = 26.6%

Solutions 243

Chapter 12 Solutions

12.1 (a) Below (x, s, sx in mg/100g). (b) H0: µ1 = µ2 = µ3 = µ4 = µ5 vs. Ha: not all µi are equal. F = 367.74 with 4 and 5 degrees of freedom; P < 0.0005, so we reject the null hypothesis. Minitab output below. (c) Plot below. We conclude that vitamin C content decreases over time.

Condition n x s sx Immediate 2 48.705 1.534 1.085 One day 2 41.955 2.128 1.505 Three days 2 21.795 0.771 0.545 Five days 2 12.415 1.082 0.765 Seven days 2 8.320 0.269 0.190

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Immed One Three Five Seven 0

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m in

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on te

nt (m

g/ 10

0g )

Condition Output from Minitab:

Analysis of Variance on VitC Source DF SS MS F p Days 4 2565.72 641.43 367.74 0.000 Error 5 8.72 1.74 Total 9 2574.44

12.2 Means, etc., at right (x, s, sx in mg/100g). Plots of means below. The hypotheses are H0: µ1 = · · · = µ5 vs. Ha: not all µi are equal. For vitamin A, F = 12.09 (df 4 and 5), so P = 0.009Ñ we reject H0 and conclude that vitamin A content changes over time (it appears to decrease, except for the rise at ÒFive daysÓ). For vitamin E, F = 0.69 (df 4 and 5), so P = 0.630Ñwe cannot reject the null hypothesis. Minitab output on page 244.

Vitamin A n x s sx Immediate 2 3.350 0.01414 0.010 One day 2 3.240 0.05657 0.040 Three days 2 3.210 0.07071 0.050 Five days 2 3.305 0.07778 0.055 Seven days 2 2.965 0.06364 0.045 Vitamin E Immediate 2 95.30 0.98995 0.700 One day 2 94.45 1.76777 1.250 Three days 2 95.85 2.19203 1.550 Five days 2 96.35 1.90919 1.350 Seven days 2 93.70 1.97990 1.400

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Immed One Three Five Seven 2.9

3.1

3.2

3.3

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Condition

Immed One Three Five Seven 93

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Condition

244 Chapter 12 One-Way Analysis of Variance

Output from Minitab: Analysis of Variance on VitA Source DF SS MS F p Days 4 0.17894 0.04473 12.09 0.009 Error 5 0.01850 0.00370 Total 9 0.19744 --------------------------------------------------- Analysis of Variance on VitE Source DF SS MS F p Days 4 9.09 2.27 0.69 0.630 Error 5 16.47 3.29 Total 9 25.56

12.3 (a) All four data sets appear to be reasonably close to normal, although Ò3 promotionsÓ seems to have a low outlier. Plots below. (b) At right (x, s, sx in dollars). (c) The ratio of largest to smallest standard deviations is about 1.58, so the assumption of equal standard deviations is reasonable. (d) H0: µ1 = · · · = µ4; Ha: not all µi are equal. Minitab output (page 245) gives F = 59.90 with df 3 and 156, and P < 0.0005, so we reject the null hypothesis. With more promotions, expected price decreases.

Promotions n x s sx One 40 4.2240 0.2734 0.0432 Three 40 4.0627 0.1742 0.0275 Five 40 3.7590 0.2526 0.0399 Seven 40 3.5487 0.2750 0.0435

One Promotion.

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Three Promotions.

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Five Promotions.

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Seven Promotions.

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Solutions 245

Output from Minitab: Analysis of Variance on ExpPrice Source DF SS MS F p NumPromo 3 10.9885 3.6628 59.90 0.000 Error 156 9.5388 0.0611 Total 159 20.5273

12.4 We have six comparisons to make, and df = 156, so the Bonferroni critical value with α = 0.05 is t∗∗ = 2.67. The pooled standard deviation is sp .= 0.2473, so the standard deviation of each difference is sp

√ 1/40+ 1/40 .= 0.05529. All six differences

are signiÞcant. [Note that because the means decrease, we could consider only the differences in consecutive means, i.e., x1 − x3, x3 − x5, and x5 − x7. Since these three differences are signiÞcant, it follows that the others must be, too. (These are the three smallest t-values.)]

x1 − x3 = 0.16125 t13 = 2.916 x1 − x5 = 0.46500 t15 = 8.410 x1 − x7 = 0.67525 t17 = 12.212 x3 − x5 = 0.30375 t35 = 5.493 x3 − x7 = 0.51400 t37 = 9.296 x5 − x7 = 0.21025 t57 = 3.802

One Three Five Seven 3.50

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12.5 (a) At right. (b) H0: µ1 = · · · = µ4; Ha: not all µi are equal. F = 9.24 with df 3 and 74; P < 0.0005, so we reject the null hypothesis. The type of lesson does affect the mean score change; in particular, it appears that students who take piano lessons had signiÞcantly higher scores than the other students.

Lesson n x s sx Piano 34 3.618 3.055 0.524 Singing 10 −0.300 1.494 0.473 Computer 20 0.450 2.212 0.495 None 14 0.786 3.191 0.853

Output from Minitab: Analysis of Variance on Scores Source DF SS MS F p LssnCode 3 207.28 69.09 9.24 0.000 Error 74 553.44 7.48 Total 77 760.72

246 Chapter 12 One-Way Analysis of Variance

12.6 We have six comparisons to make, and df = 74, so the Bonferroni critical value with α = 0.05 is t∗∗ = 2.71. The pooled standard deviation is sp

.= 2.7348. The Piano mean is signiÞcantly higher

than the other three, but the other three means are not signiÞcantly different.

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Piano Singing Computer None

M ea

n sc

or e

Lessons

DPS = 3.91765 SEPS = 0.98380 tPS = 3.982

DPC = 3.16765 SEPC = 0.77066 tPC = 4.110

DPN = 2.83193 SEPN = 0.86843 tPN = 3.261

DSC = −0.75000 SESC = 1.05917 tSC = −0.708

DSN = −1.08571 SESN = 1.13230 tSN = −0.959 DCN = −0.33571 SECN = 0.95297 tCN = −0.352

12.7 We test the hypothesis H0: ψ = µ1 − 13(µ2 + µ3 + µ4) = 0; the sample contrast is c = 3.618 − 13(−0.300 + 0.450 + 0.786) = 3.306. The pooled standard deviation estimate is sp = 2.735, so SEc = 2.735

√ 1/34+ 19/10+ 19/20+ 19/14

.= 0.6356. Then t = 3.306/0.6356 .= 5.20, with df = 74. This is enough evidence (P < 0.001) to reject H0 in favor of Ha: ψ > 0, so we conclude that mean score changes for piano students are greater than the average of the means for the other three groups.

12.8 (a) Response: Yield (in pounds). Populations: Varieties A, B, C, and D. I = 4, ni = 12 (i = 1, 2, 3, 4), N = 48. (b) Response: Attractiveness rating. Popu- lations: Packaging type. I = 5, ni = 40 (i = 1, 2, 3, 4, 5), N = 200. (c) Re- sponse: Weight loss. Populations: Dieters using the various weight-loss programs. I = 3, ni = 20 (i = 1, 2, 3), N = 60.

12.9 (a)Response: Typical number of hours of sleep. Populations: Nonsmokers, moderate smokers, heavy smokers. I = 3, ni = 100 (i = 1, 2, 3), N = 300. (b) Response: Strength of the concrete. Populations: Mixtures A, B, C, and D. I = 4, ni = 5 (i = 1, 2, 3, 4), N = 20. (c) Response: Scores on Þnal exam. Populations: Students using Methods A, B, and C. I = 3, ni = 20 (i = 1, 2, 3), N = 60.

Solutions 247

12.10 (a) The data suggest that the presence of too many nematodes reduces growth. Table at right; two versions of the plot below. (The second shows accurately the scale for the number of nematodes.) (b) H0: µ1 = · · · = µ4 vs. Ha: not all µi are equal. This ANOVA tests whether nematodes affect mean plant growth. (c) Minitab output below. F = 12.08 with df 3 and 12; P = 0.001, so we reject H0; it appears that somewhere between 1000 and 5000 nematodes, the worms hurt seedling growth. sp = √ 2.78 = 1.667 and R2 = 100.65/133.97 = 75.1%.

Nematodes x s 0 10.650 2.053

1000 10.425 1.486 5000 5.600 1.244 10000 5.450 1.771

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(c m

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Output from Minitab: Analysis of Variance on Growth Source DF SS MS F p Nematode 3 100.65 33.55 12.08 0.001 Error 12 33.33 2.78 Total 15 133.97

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev ------+---------+---------+---------+ 0 4 10.650 2.053 (-------*------)

1000 4 10.425 1.486 (-------*------) 5000 4 5.600 1.244 (------*-------) 10000 4 5.450 1.771 (------*------)

------+---------+---------+---------+ Pooled StDev = 1.667 5.0 7.5 10.0 12.5

12.11 (a) Below. (b) H0: µ1 = · · · = µ4; Ha: not all µi are equal. ANOVA tests if there are differences in the mean number of insects attracted to each color. (c) F = 30.55 with df 3 and 20; P < 0.0005, so we reject H0. The color of the board does affect the number of insects attracted; in particular, it appears that yellow draws the most, green is second, and white and blue draw the least. The pooled standard deviation is sp = 6.784, and R2 = 4218.5/5139.0 = 82.1%. [Note that the largest-to-smallest SD ratio is almost 3, so the use of ANOVA is questionable here.]

248 Chapter 12 One-Way Analysis of Variance

Color x s Lemon yellow 47.17 6.79 White 15.67 3.33 Green 31.50 9.91 Blue 14.83 5.34

Yellow White Green Blue 10

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Board color Output from Minitab:

Analysis of Variance on Insects Source DF SS MS F p ColCode 3 4218.5 1406.2 30.55 0.000 Error 20 920.5 46.0 Total 23 5139.0

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev ---+---------+---------+---------+--- 1 6 47.167 6.795 (----*----) 2 6 15.667 3.327 (----*----) 3 6 31.500 9.915 (----*----) 4 6 14.833 5.345 (---*----)

---+---------+---------+---------+--- Pooled StDev = 6.784 12 24 36 48

12.12 (a)ψ = µ1− 13(µ2+µ3+µ4). (b) H0: ψ = 0; Ha: ψ > 0. (c) The sample contrast is c = 3.49. SEc = 1.6665

√ 1/4+ 19/4+ 19/4+ 19/4

.= 0.9622, so t = 3.49/0.9622 .= 3.63, with df = 12. This is enough evidence (P = 0.002) to reject H0, so we conclude that mean seedling growth with no nematodes is greater than the average of the means for the other three groups. (d) ψ2 = µ1 − µ4. The estimated contrast is c2 = 5.2, with SEc2 = 1.178; the 95% conÞdence interval is 2.632 to 7.768.

12.13 If doing computations by hand, note that sp √ 1/ni + 1/n j .= 3.916. The t statistics

for the multiple comparisons are t12 .= 8.04, t13 .= 4.00, t14 .= 8.26, t23 .= −4.04, t24 .=

0.21, t34 .= 4.26. These indicate that the only nonsigniÞcant difference is between white

and blue boards; lemon yellow is the best.

12.14 (a) Plot below, left. Form 2 scores are typically about one point higher than form 1 scores; form 3 scores are about two points higher than form 2 scores. (b) F = 7.61 (df 2 and 238) with P = 0.0006, so we conclude that themeans are different. The comparisons reveal that the form 3 minus form 1 difference is the signiÞcant one.

Solutions 249

For 12.14.

Form 1 Form 2 Form 3 15

Sc or

For 12.15.

1976 1977 1978 1982 1987

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ra tio

12.15 (a) Plot above, right. The low observation for 1977 is an outlier, as are the maximum and minimum in 1987 (using the 1.5× IQR criterion). There is no strong suggestion of a trend in the medians. Note that Òside-by-sideÓ boxplots are somewhat misleading since the elapsed times between observations differ. (b) F = 5.75 is signiÞcant (P = 0.0002), indicating that the mean log-concentration does vary over the years. The t-tests for individual differences suggest that the mean in 1987 is signiÞcantly lower than the others.

12.16 Yes: The ratio of largest to smallest standard deviations is 10.1/5.2 .= 1.94 < 2. The pooled variance is s2p = (19)(5.2

2)+ (19)(8.92)+ (19)(10.12) 19+ 19+ 19 = 69.42, so sp

.= 8.33.

12.17 Yes: The ratio of largest to smallest standard deviations is 12.2/9.2 .= 1.33 < 2. The pooled variance is s2p = (91)(12.2

2)+ (33)(10.42)+ (34)(9.22)+ (23)(11.72) 91+ 33+ 34+ 23

.= 127.845, so sp

.= 11.3.

12.18 The degrees of freedom are in the table at the right. ÒGroupsÓ refers to variation between (a) the mean yields for the tomato varieties, (b) the mean attractiveness ratings for each of the Þve packaging types, and (c) the mean weight-losses for each of the three diet methods.

df Source (a) (b) (c) Groups 3 4 2 Error 44 195 57 Total 47 199 59

12.19 The degrees of freedom are in the table at the right. ÒGroupsÓ refers to variation between (a) the mean hours of sleep for nonsmokers, moderate smokers, and heavy smokers, (b) the mean strengths for each of the four concrete mixtures, and (c) the mean scores for each of the three teaching methods.

df Source (a) (b) (c) Groups 2 3 2 Error 297 16 57 Total 299 19 59

250 Chapter 12 One-Way Analysis of Variance

12.20 (a) H0: µ1 = µ2 = µ3; Ha: not all µi are equal. (b) The sources of variation are Òamong groupsÓ (that is, among the mean SATM scores for each of the three groups), with df = 2, and Òwithin groups,Ó with df = 253. The total variation has df = 255. (c) If H0 is true, F has an F(2, 253) distribution. (d) Referring to the F(2, 200) distribution, the critical value is 3.04.

12.21 (a) H0: µ1 = · · · = µ4; Ha: not all µi are equal. (b) The sources of variation are Òamong groupsÓ (that is, among the mean amounts spent on books for each of the four classes), with df = 3, and Òwithin groups,Ó with df = 196. The total variation has df = 199. (c) If H0 is true, F has an F(3, 196) distribution. (d) Referring to the F(3, 100) distribution, the critical value is 2.70.

12.22 (a) At right. (b) H0: µ1 = µ2 = µ3; Ha: not all µi are equal. (c) If H0 is true, F has an F(3, 32) distribution. Referring to the F(3, 30) distribution, P < 0.001Ñstrong evidence of a difference. (d) s2p = MSE = 2203.14, so sp

.= 46.94.

Source df SS MS F Groups 3 104 855.87 34 951.96 15.86 Error 32 70 500.59 2 203.14 Total 35 175 356.46

12.23 (a) At right. (b) H0: µ1 = · · · = µ4; Ha: not all µi are equal. (c) If H0 is true, F has an F(3, 32) distribution. Referring to the F(3, 30) distribution, P is between 0.05 and 0.10 (there is some evidence of a difference, but not what we would usually call signiÞcant). (d) s2p = MSE = 62.81, so sp .= 7.925.

Source df SS MS F Groups 3 476.88 158.96 2.531 Error 32 2009.92 62.81 Total 35 2486.80

12.24 (a) s2p = MSE .= 3.898 .= (45)(2.52)+ (110)(1.82)+ (51)(1.82)

45+ 110+ 51 . (b) At right. (c) H0: µ1 = µ2 = µ3; Ha: not all µi are equal. (d) If H0 is true, F has an F(2, 206) distribution. Referring to the F(2, 200) distribution, P > 0.10; we have no reason to reject H0. (e) R2 = 17.22/802.89 .= 0.021 = 2.1%.

Source df SS MS F Groups 2 17.22 8.61 2.21 Error 206 802.89 3.90 Total 208 820.11

12.25 (a) s2p = MSE .= 72 412 .= (87)(3272)+ (90)(1842)+ (53)(2852)

87+ 90+ 53 . (b) At right. (c) H0: µ1 = µ2 = µ3; Ha: not all µi are equal. (d) If H0 is true, F has an F(2, 230) distribution. Referring to the F(2, 200) distribution, P < 0.001; we conclude that the means are not all the same. (e) R2 = 6 572 551/23 227 339 .= 0.283 = 28.3%.

Source df SS MS F Groups 2 6 572 551 3 286 275.5 45.38 Error 230 16 654 788 72 412 Total 232 23 227 339

12.26 (a) ψ1 = 12(µ1 + µ2)− µ3 = 0.5µ1 + 0.5µ2 − µ3 (or, ψ1 = µ3 − 0.5µ1 − 0.5µ2.) (b) ψ2 = µ1 − µ2 (or, ψ2 = µ2 − µ1).

Solutions 251

12.27 (a) ψ1 = 12(µ1 + µ2) − 12(µ3 + µ4) = 0.5µ1 + 0.5µ2 − 0.5µ3 − 0.5µ4 (or, ψ1 = 0.5µ3+0.5µ4−0.5µ1−0.5µ2.) (b)ψ2 = µ1−µ2 (or,ψ2 = µ2−µ1). (c)ψ3 = µ3−µ4 (or, ψ3 = µ4 − µ3).

12.28 (a) For ψ1 = 12(µ1 + µ2) − µ3, H0: ψ1 = 0 vs. Ha: ψ1 > 0 (since we might expect that the science majors would have higher SATM scores). For ψ2 = µ1 − µ2, H0: ψ2 = 0 vs. Ha: ψ2 6= 0 (since we have no prior expectation of the direction of the difference). (b) c1 = 12(619+ 629)− 575 = 49 and c2 = 619− 629 = −10. (c) SEc1 = 82.5

√ 1 4/103+ 14/31+ 1/122

.= 11.28 and SEc2 = 82.5 √ 1/103+ 1/31+ 0/122 .=

16.90. (d) t1 = 49/11.28 .= 4.344 (df = 253, P < 0.0005)Ñwe conclude that science majors have higher mean SATM scores than other majors. t2 = −10/16.90 .= −0.5916 (df = 253, P > 0.25)Ñthe difference in mean SATM scores for computer science vs. other science students is not signiÞcant. (e) Use t∗ = 1.984 (for df = 100, from the table), or t∗ = 1.9694 (for df = 253). For ψ1, this gives 26.6 to 71.4, or 26.8 to 71.2. For ψ2, −43.5 to 23.5, or −43.3 to 23.3.

12.29 (a) For ψ1 = 12(µ1 + µ2)− µ3, H0: ψ1 = 0 vs. Ha: ψ1 > 0 (since we might expect that the science majors would have higher math scores). For ψ2 = µ1 −µ2, H0: ψ2 = 0 vs. Ha: ψ2 6= 0 (since we have no prior expectation of the direction of the difference). (b) c1 = 12(8.77 + 8.75) − 7.83 = 0.93 and c2 = 8.77 − 8.75 = 0.02. (c) SEc1 = 1.581

√ 1 4/90+ 14/28+ 1/106

.= 0.2299 and SEc2 = 1.581 √ 1/90+ 1/28+ 0/106 .=

0.3421. (d) t1 = 0.93/0.2299 .= 4.045 (df = 221, P < 0.0005)Ñwe conclude that science majors have higher mean HSmath grades than other majors. t2 = 0.02/0.3421 .= 0.0585 (df = 221, P > 0.25)Ñthe difference in mean HS math grades for computer science vs. other science students is not signiÞcant. (e) Use t∗ = 1.984 (for df = 100, from the table), or t∗ = 1.9708 (for df = 221). Forψ1, this gives 0.474 to 1.386, or 0.477 to 1.383. For ψ2, −0.659 to 0.699, or −0.654 to 0.694.

12.30 (a) ψ1 = µT − µC ; H0: ψ1 = 0 vs. Ha: ψ1 > 0. ψ2 = µT − 12(µC + µS); H0: ψ2 = 0 vs. Ha: ψ2 > 0. ψ3 = µJ − 13(µT + µC + µS); H0: ψ3 = 0 vs. Ha: ψ3 > 0. (b) First note sp

.= 46.9432 and df = 32. c1 = −17.06, SEc1 .= 25.71, and t1 .= −0.66, which has P > 0.25Ñnot signiÞcant. c2 = 24.39, SEc2 .= 19.64, and t2 .= 1.24, which has 0.10 < P < 0.15Ñnot signiÞcant. c3 = 91.22, SEc3 .= 17.27, and t3 .= 5.28, which has P < 0.0005Ñstrong evidence of a difference. The contrasts allow us to determinewhich differences between samplemeans represent

ÒtrueÓ differences in population means: T is not signiÞcantly better than C, nor is it better than the average of C and S. Joggers have higher Þtness scores than the average of the other three groups. (c)No: Although this seems like a logical connection, we cannot draw this conclusion,

since the treatment imposed by the study (the T group) did not produce a signiÞcantly lower result than the control group. The only signiÞcant contrast involved all four groups, including the joggers and sedentary personswhodid not have treatments imposed on them. In these cases, causation cannot be determined because of confounding or Òcommon

252 Chapter 12 One-Way Analysis of Variance

responseÓ issues; e.g., perhaps some people choose not to jog because they are less Þt to begin with.

12.31 (a) ψ1 = µT − µC ; H0: ψ1 = 0 vs. Ha: ψ1 < 0. ψ2 = µT − 12(µC + µS); H0: ψ2 = 0 vs. Ha: ψ2 < 0. ψ3 = µJ − 13(µT + µC + µS); H0: ψ3 = 0 vs. Ha: ψ3 < 0. (b) First note sp =

√ MSE .= 7.93 and df = 32. c1 = −5.5, SEc1 .= 4.343, and

t1 .= −1.27, which has 0.10 < P < 0.15Ñnot signiÞcant. c2 = −5.9, SEc2 .= 3.317,

and t2 .= −1.78, which has 0.025 < P < 0.05Ñfairly strong evidence of a difference.

c3 = −6.103, SEc3 .= 2.917, and t3 .= −2.09, which has 0.02 < P < 0.025Ñfairly strong evidence of a difference. The contrasts allow us to determinewhich differences between samplemeans represent

ÒtrueÓ differences in population means: T is not signiÞcantly better than C, but it is better than the average of C and S. Joggers have lower mean depression scores than the average of the other three groups. (c) No: The treatment imposed by the study (the T group) did not produce a signiÞ-

cantly lower result than the control group. The contrasts that were signiÞcant involved joggers and sedentary personsÑthe two groups that did not have treatments imposed on them. In these cases, causation cannot be determined because of confounding or Òcom- mon responseÓ issues; e.g., there may be personality factors that dispose a person to be depressed and also to be sedentary.

12.32 In this context, α = 0.05 means that for all three comparisons, there is a probability no more than 0.05 that we will falsely conclude that means are unequal. sp =

√ MSE .= √3.898 .= 1.974; the t statistics (and standard errors) for the

three differences are at the right. According to the Bonferroni criterion, none of the differences are signiÞcant.

t12 .= 1.73 t13 .= 2.00

SE12 .= 0.3462 SE13 .= 0.3996

t23 .= 0.603

SE23 .= 0.3318

12.33 In this context, α = 0.05 means that for all three comparisons, there is a probability no more than 0.05 that we will falsely conclude that means are unequal. sp =

√ MSE .= √72 412 .= 269.095; the t statistics (and standard errors) for

the three differences are at the right. The mean toddler food intake for Kenya is signiÞcantly different from (less than) the means for the other two countries.

t12 .= 9.27∗ t13 .= 2.11

SE12 .= 40.232 SE13 .= 46.517

t23 .= −5.95∗

SE23 .= 46.224

Solutions 253

12.34 sp = √ MSE .= 46.94; the t

statistics (and standard errors) for the six differences are at the right. Those marked with an asterisk (∗) are signiÞcantly different. The T group is signiÞcantly lower (worse) than the jogging group; the Þrst three groups (T, C, J) are all signiÞcantly higher (better) than the sedentary group.

tTC .= −0.66 tT J .= −3.65∗ tT S .= 3.14∗

SETC .= 25.71 SET J .= 20.51 SET S .= 20.99

tC J .= −2.29 tCS .= 3.22∗

SEC J .= 25.32 SECS .= 25.71

tJ S .= 6.86∗

SEJ S .= 20.51

12.35 sp = √ MSE .= 7.925; the

t statistics (and standard errors) for the six differences are at the right. None of the differences are signiÞcant.

tTC .= −1.27 tT J .= 0.63 tT S .= −1.78

SETC .= 4.343 SET J .= 3.465 SET S .= 3.546

tC J .= 1.79 tCS .= −0.18

SEC J .= 4.277 SECS .= 4.343

tJ S .= −2.44

SEJ S .= 3.465

12.36 Results may vary slightly based on software used. µ = 3.0 and

λ = n(0.5 2 + 02 + 0.52) 2.32

= n10.58 With a total sample size of 3n, the degrees of freedom are 2 and 3n−3. Choices of sample size might vary. As n gets bigger, the return (increased power)

for larger sample size is smaller and smaller; n between 150 and 200 is probably a reasonable choice.

n DFG DFE F∗ λ Power 50 2 147 3.0576 4.7259 0.4719 100 2 297 3.0262 9.4518 0.7876 150 2 447 3.0159 14.1777 0.9295 175 2 522 3.0130 16.5406 0.9614 200 2 597 3.0108 18.9036 0.9795

12.37 Results may vary slightly based on software used. µ = 3.0 and

λ = n(0.3 2 + 02 + 0.32) 2.32

= 18n529 With a total sample size of 3n, the degrees of freedom are 2 and 3n − 3. Choices of sample size might vary. ÒAt least 150Ó is a reasonable response; one

might wish to go higher than n = 200 (to get more power). [In fact, we need n .= 325 in order to get power .= 0.90.]

n DFG DFE F∗ λ Power 50 2 147 3.0576 1.7013 0.1940 100 2 297 3.0262 3.4026 0.3566 150 2 447 3.0159 5.1040 0.5096 175 2 522 3.0130 5.9546 0.5780 200 2 597 3.0108 6.8053 0.6399

12.38 (a) Below. (b) H0: µ1 = · · · = µ4; Ha: not all µi are equal. The F statistic, with df 3 and 351, is 967.82, which has P < 0.0005. Minitab output below. We conclude that the means are different; speciÞcally, the ÔPlaceboÕ mean is much higher than the other three means.

254 Chapter 12 One-Way Analysis of Variance

Shampoo n x s sx PyrI 112 17.393 1.142 0.108 PyrII 109 17.202 1.352 0.130 Keto 106 16.028 0.931 0.090 Placebo 28 29.393 1.595 0.301 o o

PyrI PyrII Keto Placebo 15

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Shampoo Output from Minitab:

Analysis of Variance on Flaking Source DF SS MS F p Code 3 4151.43 1383.81 967.82 0.000 Error 351 501.87 1.43 Total 354 4653.30

12.39 (a)The plot (below) shows granularity (which varies between groups), but that should not make us question independence; it is due to the fact that the scores are all integers. (b) The ratio of the largest to the smallest standard deviations is 1.595/0.931 .= 1.714Ñ less than 2. (c) Apart from the granularity, the quantile plots (below) are reasonably straight. (d) Again, apart from the granularity, the quantile plot looks pretty good.

For 12.39(a).

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For 12.39(c)ÐPyrI.

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Solutions 255

For 12.39(c)ÐKeto.

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For 12.39(c)ÐPlacebo.

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12.40 Wehave six comparisons tomake, and df = 351, so the Bonferroni critical valuewith α = 0.05 is t∗∗ = 2.65. The pooled standard deviation is sp =

√ MSE .= 1.1958; the

differences, standard errors, and t statistics are below. The only nonsigniÞcant difference is between the two Pyr treatments (meaning the second application of the shampoo is of little beneÞt). The Keto shampoo mean is the lowest; the placebo mean is by far the highest.

D12 = 0.19102 SE12 = 0.16088 t12 = 1.187

D13 = 1.36456 SE13 = 0.16203 t13 = 8.421

D14 = −12.0000 SE14 = 0.25265 t14 = −47.497

D23 = 1.17353 SE23 = 0.16312 t23 = 7.195

D24 = −12.1910 SE24 = 0.25334 t24 = −48.121 D34 = −13.3646 SE34 = 0.25407 t34 = −52.601

12.41 (a) ψ1 = 13µ1+ 13µ2+ 13µ3−µ4, ψ2 = 12µ1+ 12µ2−µ3, ψ3 = µ1−µ2. (b) The pooled standard deviation is sp =

√ MSE .= 1.1958. The

estimated contrasts and their standard errors are in the table. For example, SEc1 = sp √ 1 9/112+ 19/109+ 19/106+ 1/28

.= 0.2355. (c) We test H0: ψi = 0 vs. Ha: ψi 6= 0 for each contrast. The t and P values are given in the table. The Placebo mean is signiÞcantly higher than the average of the other three, while

the Keto mean is signiÞcantly lower than the average of the two Pyr means. The difference between the Pyr means is not signiÞcant (meaning the second application of the shampoo is of little beneÞt)Ñthis agrees with our conclusion from 12.40.

c1 = −12.51 c2 = 1.269 c3 = 0.191 SEc1

.= 0.2355 SEc2 .= 0.1413 SEc3 .= 0.1609 t1 = −53.17 t2 = 8.98 t3 = 1.19 P1 < 0.0005 P2 < 0.0005 P3

.= 0.2359

256 Chapter 12 One-Way Analysis of Variance

12.42 (a) At right. (b) Each new value (except for n) is simply

(old value)/64× 100% (c) The SS and MS entries differ from those of Exercise 12.1Ñby a factor of (100/64)2. However, everything else is the same: F = 367.74 with df 4 and 5; P < 0.0005, so we (again) reject H0 and conclude that vitamin C content decreases over time.

Condition n x s sx Immediate 2 76.10% 2.40% 1.70% One day 2 65.55% 3.33% 2.35% Three days 2 34.055% 1.204% 0.852% Five days 2 19.40% 1.69% 1.20% Seven days 2 13% 0.420% 0.297%

Output from Minitab: Analysis of Variance on VitCPct Source DF SS MS F p Days 4 6263.97 1565.99 367.74 0.000 Error 5 21.29 4.26 Total 9 6285.26

12.43 Transformed values for vitamin A are at right; each value is

(old value)/5× 100% The transformation has no effect on vitamin E, since the number of mil- ligrams remaining is also the percent- age of the original 100 mg. For vitamin A, the SS and MS entries differ from those of Exercise 12.2Ñby a

factor of (100/5)2 = 400. Everything else is the same: F = 12.09 with df 4 and 5; P = 0.009, so we (again) reject H0 and conclude that vitamin A content decreases over time. Since the vitamin E numbers are unchanged, the ANOVA table is unchanged, and

we again fail to reject H0 (F = 0.69 with df 4 and 5; P = 0.630). In summary, transforming to percents (or doing any linear transformation) has no

effect on the results of the ANOVA.

Condition n x s sx Immediate 2 67.0% 0.28284% 0.2% One day 2 64.8% 1.13137% 0.8% Three days 2 64.2% 1.41421% 1.0% Five days 2 66.1% 1.55563% 1.1% Seven days 2 59.3% 1.27279% 0.9%

Output from Minitab: Analysis of Variance on VitAPct Source DF SS MS F p Days 4 71.58 17.89 12.09 0.009 Error 5 7.40 1.48 Total 9 78.98

12.44 There is no effect on the test statistic, df, P-value, and conclusion. The degrees of freedom are not affected, since the number of groups and sample sizes are unchanged; meanwhile, the SS and MS values change (by a factor of b2), but this change does not affect F , since the factors of b2 cancel out in the ratio F = MSG/MSE. With the same F and df values, the P-value and conclusion are necessarily unchanged. Proof of these statements is not too difÞcult, but it requires knowledge of the SS

formulas. For most students, a demonstration with several choices of a and b would

Solutions 257

probably be more convincing than a proof. However, here is the basic idea: Using results of Chapter 1, we know that the means undergo the same transformation as the data (x∗i = a + bxi ), while the standard deviations are changed by a factor of |b|. Let xT be the average of all the data; note that x∗T = a + bxT . Now SSG = ∑Ii=1 ni(xi − xT )2, so SSG∗ = ∑i ni(x∗i − x∗T )2 = ∑i ni(b xi − b xT )2 =∑ i ni b

2(xi − xT )2 = b2SSG. Similarly, we can establish that SSE∗ = b2SSE and SST∗ = b2SST; for these formulas, consult a more advanced text. Since the MS values are merely SS values divided by the (unchanged) degrees of freedom, these also change by a factor of b2.

12.45 (a) Below. (b) Below. There are no marked deviations from normality, apart from the granularity of the scores. (c) 2.7634/1.8639 = 1.4826 < 2; ANOVA is reasonable. (d) H0: µB = µD = µS; Ha: at least one mean is different. F = 0.11 with df 2 and 63, so P = 0.895; there is no evidence against H0. (e) There is no reason to believe that the mean PRE2 scores differ between methods.

Method n x s

Basal 22 5.27 2.7634 DRTA 22 5.09 1.9978 Strat 22 4.954 1.8639

Basal DRTA Strat 4.9

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Output from Minitab: Analysis of Variance on Pre2 Source DF SS MS F p GrpCode 2 1.12 0.56 0.11 0.895 Error 63 317.14 5.03 Total 65 318.26

258 Chapter 12 One-Way Analysis of Variance

12.46 (a) The mean for Basal increases by 1; the mean for Strat decreases by 1. (b) Minitab output below; F = 5.87 with df 2 and 63; P = 0.005. (c) SSG increases by 58 (from 1.12 to 59.12), so MSG increases by 29 (half as much as SSG). The resulting F statistic is much larger. (d) The altered data changes the formerly small differences between means into large, statistically signiÞcant differences

Method n x s

Basal 22 6.27 2.7634 DRTA 22 5.09 1.9978 Strat 22 3.954 1.8639

Output from Minitab: Analysis of Variance on Pre2X Source DF SS MS F p GrpCode 2 59.12 29.56 5.87 0.005 Error 63 317.14 5.03 Total 65 376.26

12.47 (a) Below. (b) Below. There are no marked deviations from normality, apart from the granularity of the scores. (c) 3.9271/2.7244 = 1.4415 < 2; ANOVA is reasonable. (d) H0: µB = µD = µS; Ha: at least one mean is different. F = 5.32 with df 2 and 63, so P = 0.007; this is strong evidence that the means differ. (e) sp =

√ MSE = 3.18852.

For the contrast ψ = µB − 12µD − 12µS, we have c = −2.09, SEc = 0.8326, and t = −2.51 with df = 63. The one-sided P-value (for the alternative ψ < 0) is 0.0073; this is strong evidence that the Basal mean is less than the average of the other two means. The 95% conÞdence interval is −3.755 to −0.427. (f) For the contrast ψ = µD − µS, we have c = 2, SEc = 0.9614, and t = 2.0504 with df = 63. The two-sided P-value is 0.0415; this is fairly strong evidence that the DRTA and Strat means differ. The 95% conÞdence interval is 0.079 to 3.921. (g) Among POST1 scores, the order of means is Basal (lowest), Strat, DRTA. The differences are big enough that they are not likely to occur by chance.

Method n x s

Basal 22 6.681 2.7669 DRTA 22 9.772 2.7244 Strat 22 7.772 3.9271 o

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Solutions 259

Output from Minitab: Analysis of Variance on Post1 Source DF SS MS F p GrpCode 2 108.1 54.1 5.32 0.007 Error 63 640.5 10.2 Total 65 748.6

12.48 (a) Below. (b) Below. Basal and Strat look Þne, apart from the granularity of the scores; the DRTA scores show some nonnormality (speciÞcally, there were many 6Õs). (c) 3.9040/2.0407 = 1.913 < 2; ANOVA is reasonable. (d) H0: µB = µD = µS; Ha: at least one mean is different. F = 8.41 with df 2 and 63, so P = 0.001; this is strong evidence that the means differ. (e) sp =

√ MSE .= 2.3785. For the contrast

ψ = µB − 12µD − 12µS, we have c = −1.75, SEc = 0.6211, and t = −2.82 with df = 63. The one-sided P-value (for the alternative ψ < 0) is 0.0032; this is strong evidence that the Basal mean is less than the average of the other two means. The 95% conÞdence interval is −2.991 to −0.509. (f) For the contrast ψ = µD − µS, we have c = −2.136, SEc = 0.7172, and t = −2.98 with df = 63. The two-sided P-value is 0.0041; this is fairly strong evidence that the DRTA and Strat means differ. The 95% conÞdence interval is −3.569 to −0.703. (g) Among POST2 scores, the order of means is Basal (lowest), DRTA, Strat. The differences are big enough that they are not likely to occur by chance.

Method n x s

Basal 22 5.54 2.0407 DRTA 22 6.227 2.0915 Strat 22 8.36 3.9040 o

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Output from Minitab: Analysis of Variance on Post2 Source DF SS MS F p GrpCode 2 95.12 47.56 8.41 0.001 Error 63 356.41 5.66 Total 65 451.53

260 Chapter 12 One-Way Analysis of Variance

12.49 (a) F = 1.33 with df 3 and 12, giving P = 0.310; not enough evidence to stop believing that all four means are equal. (b)With the correct data, F = 12.08 with df 3 and 12, giving P = 0.001. This is fairly strong evidence that the means are not all the same. Though the outlier made the means more different, it also increased the variability (sp = 24.41, compared to 1.667 with the correct data), which makes the difference between the means less signiÞcant. (c) The table is in the Minitab output below (x and s in cm). The marked difference in the values for 0 nematodes would have caught our attention, especially the relatively large standard deviation (which, had it been correct, would have made ANOVA unreasonable, since 48.74/1.24 is a lot bigger than 2).

Output from Minitab: One-Way Analysis of Variance

Analysis of Variance on Growth Source DF SS MS F p Nematode 3 2381 794 1.33 0.310 Error 12 7148 596 Total 15 9530

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev ---------+---------+---------+------- 0 4 34.95 48.74 (----------*----------)

1000 4 10.43 1.49 (---------*----------) 5000 4 5.60 1.24 (---------*----------) 10000 4 5.45 1.77 (---------*----------)

---------+---------+---------+------- Pooled StDev = 24.41 0 25 50

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð CORRECT DATA Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð

Analysis of Variance on Growth Source DF SS MS F p Nematode 3 100.65 33.55 12.08 0.001 Error 12 33.33 2.78 Total 15 133.97

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev ------+---------+---------+---------+ 0 4 10.650 2.053 (-------*------)

1000 4 10.425 1.486 (-------*------) 5000 4 5.600 1.244 (------*-------) 10000 4 5.450 1.771 (------*------)

------+---------+---------+---------+ Pooled StDev = 1.667 5.0 7.5 10.0 12.5

Solutions 261

12.50 (a) F = 2.00 with df 3 and 20, giving P = 0.146; not enough evidence to stop believing that all four means are equal. (b) With the correct data, F = 30.55 with df 3 and 20, giving P < 0.0005. This is strong evidence that the means are not all the same. Though the outlier made the means more different, it also increased the variability (sp = 82.42, compared to 6.784 with the correct data), which makes the difference between the means less signiÞcant. (c) The table is in the Minitab output below. The marked difference in the values for lemon yellow (ÒLevel 1Ó) would have caught our attention, especially the relatively large standard deviation (which, had it been correct, would have made ANOVA unreasonable, since 164.42/3.33 is a lot bigger than 2).

Output from Minitab: Analysis of Variance on Insects Source DF SS MS F p ColCode 3 40820 13607 2.00 0.146 Error 20 135853 6793 Total 23 176673

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev --------+---------+---------+-------- 1 6 114.67 164.42 (---------*---------) 2 6 15.67 3.33 (---------*---------) 3 6 31.50 9.91 (----------*---------) 4 6 14.83 5.34 (---------*---------)

--------+---------+---------+-------- Pooled StDev = 82.42 0 70 140

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð CORRECT DATA Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð

Analysis of Variance on Insects Source DF SS MS F p ColCode 3 4218.5 1406.2 30.55 0.000 Error 20 920.5 46.0 Total 23 5139.0

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev ---+---------+---------+---------+--- 1 6 47.167 6.795 (----*----) 2 6 15.667 3.327 (----*----) 3 6 31.500 9.915 (----*----) 4 6 14.833 5.345 (---*----)

---+---------+---------+---------+--- Pooled StDev = 6.784 12 24 36 48

262 Chapter 12 One-Way Analysis of Variance

12.51 (a) Table at right; plot not shown (it is similar to that from Exercise 12.10). [Note that this trans- formation actually makes the largest-to-smallest standard deviation ratio larger than it had beenÑ 1.94 vs. 1.65.] (b) The hypotheses are the same as before (H0: µ0 = µ1000 = µ5000 = µ10000 vs. Ha: at least one mean is different), except that now µi rep- resents the mean logarithm of the growth for each group. (c) The new F is 10.39 (using either natural or common logs), with df 3 and 12; the P-value is 0.001. The conclusion is the same as with the original data (although the new F is slightly smaller than the old F , meaning P is slightly greater): It appears that somewhere between 1000 and 5000 nematodes, the worms hurt seedling growth. For the original data, sp = 1.667 and R2 = 75.1%. For the transformed

data, sp = 0.2389 (natural logs) or sp = 0.1038 (common logs), and R2 = 1.7792/2.4643 = 0.3356/0.4648 = 72.2%.

Natural (base e) logarithms: Nematodes x s

0 2.3524 0.1843 1000 2.3357 0.1546 5000 1.7058 0.2077 10000 1.6509 0.3569

Common (base 10) logarithms: 0 1.0217 0.0800

1000 1.0144 0.0671 5000 0.7408 0.0902 10000 0.7170 0.1550

12.52 (a) Table at right; plot not shown (it is sim- ilar to that from Exercise 12.11). [Note that the largest-to-smallest SD ratio is slightly betterÑ 2.32 vs. 2.98Ñbut ANOVA is still questionable.] (b) H0: µ1 = · · · = µ4; Ha: not all µi are equal. ANOVA tests if there are differences in the mean square root of the number of insects attracted to each color. (c) The new F is 27.00, with df 3 and 20; this has P < 0.0005. The conclusion is the same as before (although the new F is slightly smaller than the old F , meaning P is slightly greater): Board color does affect the (square root of the) number of insects attracted; in particular, it appears that yellow draws the most, green is second, and white and blue draw the least. For the original data, sp = 6.784 and R2 = 4218.5/5139.0 = 82.1%. For the

transformed data, sp = 0.6835 and R2 = 37.836/47.180 = 80.2%.

Color x s Lemon yellow 6.8533 0.4882 White 3.9400 0.4142 Green 5.5435 0.9613 Blue 3.7931 0.7312

Output from Minitab: Analysis of Variance on SqrtIns Source DF SS MS F p ColCode 3 37.836 12.612 27.00 0.000 Error 20 9.343 0.467 Total 23 47.180

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev ----+---------+---------+---------+-- 1 6 6.8533 0.4882 (----*----) 2 6 3.9400 0.4142 (----*----) 3 6 5.5435 0.9613 (----*----) 4 6 3.7931 0.7312 (----*---)

----+---------+---------+---------+-- Pooled StDev = 0.6835 3.6 4.8 6.0 7.2

Solutions 263

12.53 Results may vary slightly based on software used. (a) µ = (620 + 600 + 580 + 560)/4 = 590 and

λ = n(30 2 + 102 + 102 + 302)

902 = 20n81

With a total sample size of 4n, the degrees of freedom are 3 and 4n − 4. Answers will vary with the

choices of α and n. The table and plot show values for α = 0.05. (b) The power rises to about 0.90 for n = 60; it continues rising (getting closer to 1) after that, but much more slowly. (c) Choice of sample size will vary; be sure to consider the balance between increased power and the additional expense of a larger sample.

n DFG DFE F∗ λ Power 25 3 96 2.6994 6.1728 0.5128 50 3 196 2.6507 12.3457 0.8437 75 3 296 2.6351 18.5185 0.9618 100 3 396 2.6274 24.6914 0.9922

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12.54 Results may vary slightly based on software used. (a) µ = (610 + 600 + 590 + 580)/4 = 595 and

λ = n(15 2 + 52 + 52 + 152)

902 = 5n81

With a total sample size of 4n, the degrees of freedom are 3 and 4n − 4. Answers will vary with the

choices of α and n. The table and plot show values for α = 0.05. (b) The power rises to about 0.80 for n = 180, and 0.90 for n = 225; it continues rising (getting closer to 1) after that, but much more slowly. Since the alternative we want to detect is less extreme (closer to the null) than in 12.53, it is harder to detect, so the power increases much more slowly than in the previous exercise. (c) Choice of sample size will vary; be sure to consider the balance between increased power and the additional expense of a larger sample.

n DFG DFE F∗ λ Power 50 3 196 2.6507 3.0864 0.2767 100 3 396 2.6274 6.1728 0.5262 150 3 596 2.6199 9.2593 0.7215 200 3 796 2.6161 12.3457 0.8493

0.2

0.4

0.6

0.8

0 50 100 150 200 250 300 350 400

Po w

Sample size

264 Chapter 12 One-Way Analysis of Variance

12.55 The regression equation is ŷ = 4.36 − 0.116x (ŷ is the expected price, and x is the number of promotions). The regression is signiÞcant (i.e., the slope is signiÞcantly different from 0): t = −13.31 with df = 158, giving P < 0.0005. The regression on number of promotions explains r 2 = 52.9% of the variation in expected price. (This is similar to the ANOVA value: R2 = 53.5%.) The granularity of the ÒNumber of promotionsÓ observations makes interpreting

the plot a bit tricky. For 5 promotions, the residuals seem to be more likely to be negative (in fact, 26 of the 40 residuals are negative), while for 3 promotions, the residuals are weighted toward the positive side. (We also observe that in the plot of mean expected price vs. number of promotions [see 12.4], the mean for 3 promotions is not as small as one would predict from a line near the other three points.) This suggests that a linear model may not be appropriate.

¥¥

¥ ¥

¥¥

¥ ¥¥ ¥

¥ ¥

¥¥¥ ¥ ¥

¥¥

¥ ¥¥

¥ ¥

¥¥

¥¥¥ ¥¥

¥ ¥

¥ ¥ ¥

¥¥ ¥

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¥ ¥¥¥ ¥¥

¥ ¥

¥¥ ¥ ¥

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¥ ¥

¥ ¥¥ ¥ ¥ ¥

¥¥

¥ ¥

¥¥¥ ¥ ¥ ¥

¥ ¥

¥¥

¥ ¥

¥¥

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¥¥ ¥

-0.8

-0.6

-0.4

-0.2

0.2

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0.8

1 3 5 7

R es

id ua

Number of promotions

Output from Minitab: The regression equation is ExpPrice = 4.36 - 0.116 NumPromo

Predictor Coef Stdev t-ratio p Constant 4.36452 0.04009 108.87 0.000 NumPromo -0.116475 0.008748 -13.31 0.000

s = 0.2474 R-sq = 52.9% R-sq(adj) = 52.6%

Solutions 265

Chapter 13 Solutions

13.1 (a) Response variable: Yield (pounds of tomatoes/plant). Factors: Variety (I = 5) and fertilizer type (J = 2). N = 5× 2× 4 = 40. (b) Response variable: Attractiveness rating. Factors: Packaging type (I = 6) and city (J = 6). N = 6 × 6 × 50 = 1800. (c) Response variable: Weight loss. Factors: Weight-loss program (I = 4) and gender (J = 2). N = 4× 2× 10 = 80.

13.2 (a) Response variable: Typical number of hours of sleep. Factors: Smoking level (I = 3) and gender (J = 2). N = 3 × 2 × 120 = 720. (b) Response variable: Strength of the concrete. Factors: Mixture (I = 4) and number of freezing/thawing cycles (J = 3). N = 4 × 3 × 2 = 24. (c) Response variable: Scores on Þnal exam. Factors: Teaching method (I = 3) and studentÕs area (J = 2). N = 3× 2× 7 = 42.

13.3 (a) Variety (df = 4), Fertilizer type (df = 1), Variety/Fertilizer interaction (df = 4), and Error (df = 30). Total df = 39. (b) Packaging type (df = 5), City (df = 5), Packaging/City interaction (df = 25), and Error (df = 1765). Total df = 1799. (c)Weight-loss program (df = 3), Gender (df = 1), Program/Gender interaction (df = 3), and Error (df = 72). Total df = 79.

13.4 (a) Smoking level (df = 2), Gender (df = 1), Smoking/Gender interaction (df = 2), and Error (df = 714). Total df = 719. (b) Mixture (df = 3), Cycles (df = 2), Mixture/Cycles interaction (df = 6), and Error (df = 12). Total df = 23. (c) Teaching method (df = 2), Area (df = 1), Method/Area interaction (df = 2), and Error (df = 36). Total df = 41.

13.5 (a) Plot below, left. (b) Nonwhite means are all slightly higher than white means. Mean systolic BP rises with age. There does not seem to be any interaction; both plots rise in a similar fashion. (c) By race, the marginal means are 135.98 (White) and 137.82 (Nonwhite). By age, they are 131.65, 133.25, 136.2, 140.35, 143.05. The Nonwhite minus White means are 1.3, 1.9, 2, 1.9, and 2.1. The mean systolic BP rises about 2 to 4 points from one age group to the next; the Nonwhite means are generally about 2 points higher.

For 13.5.

o o

x x

35Ð39 40Ð44 45Ð49 50Ð54 55Ð59 130

135

140

145

B lo

od p

re ss

ur e

Age

W hite

N onwhite

For 13.6.

o o

o o ox

x x

35Ð39 40Ð44 45Ð49 50Ð54 55Ð59 88

B lo

od p

re ss

ur e

Age

W hite

N onwhite

266 Chapter 13 Two-Way Analysis of Variance

13.6 (a) Plot above, right. (b) Nonwhite means are all slightly higher than white means. Generally, mean diastolic BP rises with age, except for the last age group, where it seems to drop. There may be an interactionÑthe two plots have different appearance (the means for nonwhites Òjump aroundÓmore than those of whites). (c)By race, the marginal means are 90.7 (White) and 93.12 (Nonwhite). By age, they are 90.3, 91.65, 92.1, 93.05, and 92.45. The Nonwhite minus White means are 1.8, 2.9, 2.4, 2.9, and 2.1. The mean diastolic BP rises about 0.5 to 1.5 points from one age group to the next, except in the end when it drops 0.6 points. The Nonwhite means are generally 2 to 3 points higher.

13.7 (a) Plot below, left. (b) There seems to be a fairly large difference between the means based on how much the rats were allowed to eat, but not very much difference based on the chromium level. There may be an interaction: the NM mean is lower than the LM mean, while the NRmean is higher than the LRmean. (c) Lmean: 4.86. N mean: 4.871. M mean: 4.485. R mean: 5.246. LR minus LM: 0.63. NR minus NM: 0.892. Mean GITH levels are lower for M than for R; there is not much difference for L vs. N. The difference between M and R is greater among rats who had normal chromium levels in their diets (N).

For 13.7.

o o

L N 4.3

4.5

G IT

H le

ve l

Chromium

E at-M

E at-R

For 13.8.

x x

Males Females 24

C ha

pi n

SI s

co re

Gender

13.8 Plot above, right. PG students generally scored higher than LU students. PG females outscored PGmales, while LUmales had a higher mean than LU females (an interaction). Male mean: 26.45. Female mean: 27.095. PG mean: 28.405. LU mean: 25.14.

Solutions 267

13.9 The ÒOtherÓ category had the lowest mean SATM score for both genders; this is apparent from the graph (below, left) as well as from the marginal means (CS: 605, EO: 624.5, O: 566.) Males had higher mean scores in CS and O, while females are slightly higher in EO; this seems to be an interaction. Overall, the marginal means are 611.7 (males) and 585.3 (females).

For 13.9.

x x

Males Females 525

550

575

600

625

M ea

n SA

Gender

For 13.10.

Males Females 7.5

8.5

M ea

n H

S m

at h

gr ad

Gender

13.10 The ÒOtherÓ category had the lowest mean HS math grades for both genders; this is apparent from the graph (above, right) as well as from the marginal means (CS: 8.895, EO: 8.855, O: 7.845.) Females had higher mean grades; the female marginal mean is 8.836 compared to 8.226 for males. The female−male difference is similar for CS and O (about 0.5), but is about twice as big for EO (an interaction).

13.11 (a) At right. (b) Plot on page 268, left. Elasticity appears to differ between species, with a smaller effect by ßake size. There also seems to be an interaction (birch has the smallest mean for S1, but the largest mean for S2). (c) Minitab output is below. With A = Species and B = Flake size, FA = 0.59 with df 2 and 12; this has P = 0.569. FB = 0.27 with df 1 and 12; this has P = 0.613. FAB = 1.70 with df 2 and 12; this has P = 0.224. None of these statistics are signiÞcant; the differences we observed could easily be attributable to chance.

Size of ßakes Species S1 S2 Aspen x = 387.333 x = 335.667 362

s = 68.712 s = 60.136 Birch x = 292.667 x = 455.333 374

s = 121.829 s = 117.717 Maple x = 323.333 x = 293.667 308

s = 52.013 s = 183.919 334 362

Output from Minitab: Analysis of Variance for Elast

Source DF SS MS F P Species 2 14511 7256 0.59 0.569 Flake 1 3308 3308 0.27 0.613 Species*Flake 2 41707 20854 1.70 0.224 Error 12 147138 12262 Total 17 206664

268 Chapter 13 Two-Way Analysis of Variance

For 13.11.

o x

Aspen Birch Maple 275

300

350

400

450

M od

ul us

o f e

la st

ic ity

(p si

)

Species

S 2

S 1

For 13.12.

x x

Aspen Birch Maple 1100

1200

1300

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St re

ng th

(k si

)

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13.12 (a) At right. (b) Plot above, right. Strength appears to differ between species, and between ßake sizes. There also seems to be an interaction (birch has the smallest mean for S1, but the largest mean for S2). (c) Minitab output is below. With A = Species and B = Flake size, FA = 1.10 with df 2 and 12; this has P = 0.365. FB = 1.92 with df 1 and 12; this has P = 0.191. FAB = 1.88 with df 2 and 12; this has P = 0.194. None of these statistics are signiÞcant; the differences we observed could easily be attributable to chance.

Size of ßakes Species S1 S2 Aspen x = 1659.67 x = 1321.33 1491

s = 351.241 s = 234.628 Birch x = 1168.00 x = 1345.33 1257

s = 235.879 s = 95.757 Maple x = 1526.33 x = 1142.00 1334

s = 308.262 s = 355.848 1451 1270

Output from Minitab: Analysis of Variance for Strength

Source DF SS MS F P Species 2 170249 85124 1.10 0.365 Flake 1 148694 148694 1.92 0.191 Species*Flake 2 291749 145874 1.88 0.194 Error 12 929765 77480 Total 17 1540456

Solutions 269

13.13 (a) At right. (b) Plot below. Except for tool 1, mean diameter is highest at time 2. Tool 1 had the highest mean diameters, fol- lowed by tool 2, tool 4, tool 3, and tool 5.

+ +

Tool 1 Tool 2 Tool 3 Tool 4 Tool 5 24.99

25.01

25.02

25.03

D ia

m et

er (m

m )

o Time 1

x Time 2

+ Time 3

(c) Minitab output below. With A = Tool and B = Time, FA = 412.98 with df 4 and 30; this has P < 0.0005. FB = 43.61 with df 2 and 30; this has P < 0.0005. FAB = 7.65 with df 8 and 30; this has P < 0.0005. (d) There is strong evidence of a difference in mean diameter among the tools (A) and among the times (B). There is also an interaction (speciÞcally, tool 1Õs mean diameters changed differently over time compared to the other tools).

Tool Time x s 1 1 25.0307 0.0011541

2 25.0280 0 3 25.0260 0

2 1 25.0167 0.0011541 2 25.0200 0.0019999 3 25.0160 0

3 1 25.0063 0.0015275 2 25.0127 0.0011552 3 25.0093 0.0011552

4 1 25.0120 0 2 25.0193 0.0011552 3 25.0140 0.0039997

5 1 24.9973 0.0011541 2 25.0060 0 3 25.0003 0.0015277

Output from Minitab: Analysis of Variance for Diameter

Source DF SS MS F P Tool 4 0.00359714 0.00089928 412.98 0.000 Time 2 0.00018992 0.00009496 43.61 0.000 Tool*Time 8 0.00013324 0.00001665 7.65 0.000 Error 30 0.00006533 0.00000218 Total 44 0.00398562

13.14 All means and standard deviations will change by a factor of 0.04; the plot is identical to that in Exercise 13.13, except that the vertical scale is different. All SS and MS values change by a factor of 0.042 = 0.0016, but the F (and P) values are the same. (Or at least they should be; Minitab [see output below] does not carry out the computation because the MS values are too small.)

Output from Minitab: Analysis of Variance for Diameter

Source DF SS MS F P Tool 4 5.7556E-06 1.4389E-06 ** Time 2 3.0385E-07 1.5193E-07 ** Tool*Time 8 2.1312E-07 2.6640E-08 ** Error 30 1.0453E-07 3.4844E-09 Total 44 6.3771E-06

** Denominator of F-test is zero.

270 Chapter 13 Two-Way Analysis of Variance

13.15 (a) Table at right; plot below. The mean expected price decreases as percent discount increases, and also as the number of promo- tions increases.

o o

x x

+ +

¥ ¥

One Three Five Seven 3.70

4.00

4.50

5.00

Ex pe

ct ed

p ric

e ($

)

Number of promotions

10% 20%

30%

40%

(b) Minitab output below. With A = Number of promotions and B = Percent discount, FA = 47.73 with df 3 and 144; this has P < 0.0005. FB = 47.42 with df 3 and 144; this has P < 0.0005. FAB = 0.44 with df 9 and 144; this has P = 0.912. (c) There is strong evidence of a difference in mean expected price based on the number of promotions and the percent discount. SpeciÞcally, the two effects noted in (a) are signiÞcant: more promotions and higher discounts decrease the expected price. There is no evidence of an interaction.

Promos Discount x s 1 10% 4.423 0.1848

20% 4.225 0.3856 30% 4.689 0.2331 40% 4.920 0.1520

3 10% 4.284 0.2040 20% 4.097 0.2346 30% 4.524 0.2707 40% 4.756 0.2429

5 10% 4.058 0.1760 20% 3.890 0.1629 30% 4.251 0.2648 40% 4.393 0.2685

7 10% 3.780 0.2144 20% 3.760 0.2618 30% 4.094 0.2407 40% 4.269 0.2699

Output from Minitab: Analysis of Variance for ExpPrice

Source DF SS MS F P Promos 3 8.3605 2.7868 47.73 0.000 Discount 3 8.3069 2.7690 47.42 0.000 Promos*Discount 9 0.2306 0.0256 0.44 0.912 Error 144 8.4087 0.0584 Total 159 25.3067

13.16 Note: If your software allows it, generate a new Òsubscript columnÓ from the Promo- tions and Discount columns. For example, in Minitab, Òlet c6=c2+100*c3Ó (where c2=Promos and c3=Discount) places in c6 the numbers 110, 120, 130, . . . , 740Ñthe Þrst digit is the number of promotions, and the last two are the percent discount. Then ÒOneway c4 c6Ó will do the 16-treatment analysis. The F statistic (with df 15 and 144) is 19.29, which has P < 0.0005Ñthere is a

signiÞcant difference between the 16 means. Full analysis of all possible differences between means is not given here. (There are 120 such differences!) For each difference, we (or a computer) must Þnd ti j = (xi − x j)/0.1080. (The divisor 0.1080 is the value of sp √ 1 10 + 110 , where sp =

√ MSE .= 0.2416.) Compare this to the appropriate Bonferroni

critical value t∗∗; answers will differ based on the chosen signiÞcance level αÑe.g., for α = 0.05, t∗∗ = 3.61. In the Minitab output below, we can see the individual 95% conÞdence intervals,

Solutions 271

which give some indication of which pairs of means may be different. SpeciÞcally, the conÞdence interval for level 110 (one promotion, 10% discount) overlaps those for 120 and 130, but not the 140 interval, meaning that for some choice of α, the 110 and 140 means are different. The t statistic for this comparison is t .= −4.60, so that is easily (Bonferroni-) signiÞcant at α = 0.05. (In fact, this difference is signiÞcant if we choose any α greater than about 0.0011.) Similarly, the 110 interval overlaps those for 310, 320, and 330, but just misses the 340

interval, so these means are different (for some choice of α greater than the Þrst, since these two intervals are closer together). The t statistic for this comparison is t .= −3.08; this is not (Bonferroni-) signiÞcant unless we choose α to be about 0.30 or higher. To put this another way, two means are (Bonferroni-) signiÞcantly different if they

differ by 0.1080t∗∗. For α = 0.05, this means they must differ by about 0.39; thus the 110 mean differs from the 140, 520, 710, and 720 means at the 5% level.

Output from Minitab: One-Way Analysis of Variance

Analysis of Variance on ExpPrice Source DF SS MS F p PromDisc 15 16.8980 1.1265 19.29 0.000 Error 144 8.4087 0.0584 Total 159 25.3067

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev ----------+---------+---------+------ 110 10 4.4230 0.1848 (---*--) 120 10 4.2250 0.3856 (---*--) 130 10 4.6890 0.2331 (---*---) 140 10 4.9200 0.1520 (---*---) 310 10 4.2840 0.2040 (---*---) 320 10 4.0970 0.2346 (--*---) 330 10 4.5240 0.2707 (---*---) 340 10 4.7560 0.2429 (---*---) 510 10 4.0580 0.1760 (--*---) 520 10 3.8900 0.1629 (---*---) 530 10 4.2510 0.2648 (---*---) 540 10 4.3930 0.2685 (---*---) 710 10 3.7800 0.2144 (---*--) 720 10 3.7600 0.2618 (---*---) 730 10 4.0940 0.2407 (--*---) 740 10 4.2690 0.2699 (---*---)

----------+---------+---------+------ Pooled StDev = 0.2416 4.00 4.40 4.80

272 Chapter 13 Two-Way Analysis of Variance

13.17 (a) At right. (b) For testing interaction, FAB = 5.7159. If there is no interaction, this comes from an F(1, 36) distribution; 5.7159 gives 0.010 < P < 0.025 (or P = 0.0222). (c) For testing the main effect of Chromium, FA = 0.04030. If there is no effect, this comes from an F(1, 36) distribution; 0.04030 gives P > 0.1 (or P = 0.8420). For testing the main effect of Eat, FB = 192.89. If there is no effect, this comes from an F(1, 36) distribution; 192.89 gives P < 0.001. (d) s2p = MSE = 0.03002, so sp = 0.1733. (e) The observations made in 13.7 are supported by the analysis: The amount the rats were allowed to eat made a difference in mean GITH levels, but chromium levels had no (signiÞcant) effect by themselves, although there was a Chromium/Eat interaction.

Source df SS MS F A(Chromium) 1 0.00121 0.00121 0.04 B(Eat) 1 5.79121 5.79121 192.89 AB 1 0.17161 0.17161 5.72 Error 36 1.08084 0.03002 Total 39 7.04487

13.18 (a) At right. Note that N = 4 × 150 = 600. (b) For testing interaction, FAB = 7.16. If there is no interaction, this comes from an F(1, 596) distribution; 7.16 gives 0.001 < P < 0.010 (or P = 0.0077). (c) For testing the main effect of Gender, FA = 2.73. If there is no effect, this comes from an F(1, 596) distribution; 2.73 gives P .= 0.1 (or P = 0.099). For testing the main effect of Group, FB = 69.92. If there is no effect, this comes from an F(1, 596) distribution; 69.92 gives P < 0.001. (d) s2p = MSE = 22.87, so sp = 4.7823. (e) PG students scored (signiÞcantly) higher than LU students. Although the means differ by gender, the difference is not overwhelming. There is an interaction: PG females outscored PG males, while LU males had a higher mean than LU females.

Source df SS MS F A(Gender) 1 62.40 62.40 2.73 B(Group) 1 1599.03 1599.03 69.92 AB 1 163.80 163.80 7.16 Error 596 13633.29 22.87 Total 599 15458.52

13.19 (a) All three F values have df 1 and 945, the P values are < 0.001, < 0.001, and 0.1477. Gender and handedness both have signiÞcant effects on mean lifetime, but there is no signiÞcant interaction. (b) Women live about 6 years longer than men (on the average), while right-handed people average 9 more years of life than left-handed people. ÒThere is no interactionÓ means that handedness affects both genders in the same way, and vice versa.

13.20 (a)WithA=Series andB=Holder, FA = 7.02with df 3 and 61; this has P = 0.0004. FB = 1.96 with df 1 and 61; this has P = 0.1665. FAB = 1.24 with df 3 and 61; this has P = 0.3026. Only the series had a signiÞcant effect; the presence or absence of a holder and series/holder interaction did not signiÞcantly affect the mean radon reading. (b) Since the ANOVA indicates that these means are signiÞcantly different, we conclude that detectors produced in different production runs give different readings for the same radon levelÑthis inconsistency may indicate poor quality control in production.

Solutions 273

13.21 The table and plot of the means (at the right) suggest that students who stay in the sciences have higher mean SATV scores than those who end up in the ÒOtherÓ group. Female CS and EO students have higher scores than males in those majors, but males have the higher mean in the Other group. Normal quantile plots (below)

suggest some right-skewness in the ÒWomen in CSÓ group, and also some nonnormality in the tails of the ÒWomen in EOÓ group. Other groups look reasonably normal. In the ANOVA, only the effect of

major is signiÞcant (F = 9.32, df 2 and 228, P < 0.0005).

Major Gender CS EO Other Male n = 39 39 39

x = 526.949 507.846 487.564 s = 100.937 57.213 108.779

Female n = 39 39 39 x = 543.385 538.205 465.026 s = 77.654 102.209 82.184

x x

CS EO Other 450

475

500

525

550

SA TV

s co

re Major

Male

Female

Men in CS.

o o

oo o

250 300

400

500

600

700

800

-3 -2 -1 0 1 2 3

SA TV

s co

z score

Women in CS.

oo ooooo

o o

oooo

oo oo

400

450

500

550

600

650

700

750

-3 -2 -1 0 1 2 3

SA TV

s co

z score

Men in EO.

ooooo

o o

oo ooo

o o o

ooo oo

ooooo

375 400

450

500

550

600

-3 -2 -1 0 1 2 3

SA TV

s co

z score

Women in EO.

oo o o

ooo

oooo

o o oo ooo

ooo

oooooo

oooo

400

450

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550

600

650

700

-3 -2 -1 0 1 2 3

SA TV

s co

z score

274 Chapter 13 Two-Way Analysis of Variance

Men in Other.

o o

o oo

o o

o oo

o o

o 250 300

400

500

600

700

800

-3 -2 -1 0 1 2 3

SA TV

s co

z score

Women in Other.

o o

ooo

o o

o oo

o o

250 300 350 400 450 500 550 600 650

-3 -2 -1 0 1 2 3

SA TV

s co

z score

Output from Minitab: Analysis of Variance for SATV

Source DF SS MS F P Maj 2 150723 75362 9.32 0.000 Sex 1 3824 3824 0.47 0.492 Maj*Sex 2 29321 14661 1.81 0.166 Error 228 1843979 8088 Total 233 2027848

13.22 The table and plot of the means (at the right) suggest that, within a given gender, students who stay in the sciences have higher HSS grades than those who end up in the ÒOtherÓ group. Males have a slightly higher mean in the CS group, but females have the edge in the other two. Normal quantile plots (below) show

no great deviations from normality, apart from the granularity of the grades (most evident among Women in EO). In the ANOVA, sex, major, and

interaction are all signiÞcant: For the main effect of gender, F = 5.06, df 1 and 228, P = 0.025; for major, F = 8.69, df 2 and 228, P < 0.0005; for interaction, F = 4.86, df 2 and 228, P = 0.009.

Major Gender CS EO Other Male n = 39 39 39

x = 8.66667 7.92308 7.43590 s = 1.28418 2.05688 1.71364

Female n = 39 39 39 x = 8.38461 9.23077 7.82051 s = 1.66410 0.70567 1.80455

CS EO Other 7.3 7.5

8.0

8.5

9.0

H SS

g ra

Major

Male Female

Solutions 275

Men in CS.

ooo

oooo

o o

-3 -2 -1 0 1 2 3

H SS

g ra

z score

Women in CS.

ooooooo

oooo

o o

oooo

o o

-3 -2 -1 0 1 2 3

H SS

g ra

z score

Men in EO.

ooooo

oooo

ooo

o o o

ooo o o

-3 -2 -1 0 1 2 3

H SS

g ra

z score

Women in EO.

oooo

o o o

oooooooooooo

oooo

ooo

o o o

oooooooooo

-3 -2 -1 0 1 2 3

H SS

g ra

z score

Men in Other.

-3 -2 -1 0 1 2 3

H SS

g ra

z score

Women in Other.

ooo

-3 -2 -1 0 1 2 3

H SS

g ra

z score

Output from Minitab: Analysis of Variance for HSS

Source DF SS MS F P Sex 1 12.927 12.927 5.06 0.025 Maj 2 44.410 22.205 8.69 0.000 Sex*Maj 2 24.855 12.427 4.86 0.009 Error 228 582.923 2.557 Total 233 665.115

276 Chapter 13 Two-Way Analysis of Variance

13.23 The table and plot of the means (at the right) suggest that females have higher HSE grades than males. For a given gender, there is not too much difference among majors. Normal quantile plots (below) show

no great deviations from normality, apart from the granularity of the grades (most evident among Women in EO). In the ANOVA, only the effect of

gender is signiÞcant (F = 50.32, df 1 and 228, P < 0.0005).

Major Gender CS EO Other Male n = 39 39 39

x = 7.79487 7.48718 7.41026 s = 1.50752 2.15054 1.56807

Female n = 39 39 39 x = 8.84615 9.25641 8.61539 s = 1.13644 0.75107 1.16111

o o

CS EO Other 7.3 7.5

8.0

8.5

9.0

H SE

g ra

Major

Male

Female

Men in CS.

oooo

-3 -2 -1 0 1 2 3

H SE

g ra

z score

Women in CS.

ooo

oooo

o o

-3 -2 -1 0 1 2 3

H SE

g ra

z score

Men in EO.

ooo

o o

ooo

ooooo

ooo

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oo4

-3 -2 -1 0 1 2 3

H SE

g ra

z score

Women in EO.

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o o o

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-3 -2 -1 0 1 2 3

H SE

g ra

z score

Solutions 277

Men in Other.

oooooo

ooo

-3 -2 -1 0 1 2 3

H SE

g ra

z score

Women in Other.

ooo

-3 -2 -1 0 1 2 3

H SE

g ra

z score

Output from Minitab: Analysis of Variance for HSE

Source DF SS MS F P Sex 1 105.338 105.338 50.32 0.000 Maj 2 5.880 2.940 1.40 0.248 Sex*Maj 2 5.573 2.786 1.33 0.266 Error 228 477.282 2.093 Total 233 594.073

13.24 The table and plot of the means (at the right) suggest that students who stay in the sciences have higher mean GPAs than those who end up in the ÒOtherÓ group. Both genders have similar mean GPAs in the EO group, but in the other two groups, females come out on top. Normal quantile plots (below) show

no great deviations from normality, apart from a few low outliers in the two EO groups. In the ANOVA, sex and major are

signiÞcant, while there is some (not quite signiÞcant) evidence for the interaction. For the main effect of gender, F = 7.31, df 1 and 228, P = 0.007; for major, F = 31.42, df 2 and 228, P < 0.0005; for interaction, F = 2.77, df 2 and 228, P = 0.065.

Major Gender CS EO Other Male n = 39 39 39

x = 2.74744 3.09641 2.04769 s = 0.68399 0.51297 0.73041

Female n = 39 39 39 x = 2.97923 3.08077 2.52359 s = 0.53347 0.64813 0.76556

x x

CS EO Other

2.0

2.5

3.0

G PA

Major

Male

Female

278 Chapter 13 Two-Way Analysis of Variance

Men in CS.

o o

oo o

o o

-3 -2 -1 0 1 2 3

G PA

z score

Women in CS.

ooo

o o

oo oo

oooo

o o

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-3 -2 -1 0 1 2 3

G PA

z score

Men in EO.

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o o

oo oo

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-3 -2 -1 0 1 2 3

G PA

z score

Women in EO.

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-3 -2 -1 0 1 2 3

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z score

Men in Other.

oo o

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-3 -2 -1 0 1 2 3

G PA

z score

Women in Other.

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-3 -2 -1 0 1 2 3

G PA

z score

Output from Minitab: Analysis of Variance for GPA

Source DF SS MS F P Sex 1 3.1131 3.1131 7.31 0.007 Maj 2 26.7591 13.3795 31.42 0.000 Sex*Maj 2 2.3557 1.1779 2.77 0.065 Error 228 97.0986 0.4259 Total 233 129.3265

Solutions 279

Chapter 14 Solutions

Section 1: The Wilcoxon Rank Sum Test

14.1 (a) Normal quantile plots are not shown. The score 0.00 for child 8 seems to be a low outlier (although with only 5 observations, such judgments are questionable). (b) H0: µ1 = µ2 vs. Ha: µ1 > µ2. x1 = 0.676, x2 = 0.406, t = 2.059, which gives P = 0.0446 (df = 5.5). We have fairly strong evidence that high-progress readers have higher mean scores. (c)We test

H0: Scores for both groups are identically distributed vs. Ha: High-progress children systematically score higher

W = 36, P .= 0.0463; we have strong evidence against the hypothesis of identical distributions. This is equivalent to the conclusion reached in (b).

Output from Minitab: Mann-Whitney Confidence Interval and Test

HiProg1 N = 5 Median = 0.7000 LoProg1 N = 5 Median = 0.4000 Point estimate for ETA1-ETA2 is 0.2100 96.3 Percent C.I. for ETA1-ETA2 is (-0.0199,0.7001) W = 36.0 Test of ETA1 = ETA2 vs. ETA1 > ETA2 is significant at 0.0473 The test is significant at 0.0463 (adjusted for ties)

14.2 (a) Normal quantile plots are not shown. The score 0.54 for child 3 seems to be a low outlier. (b) H0: µ1 = µ2 vs. Ha: µ1 > µ2. x1 = 0.768, x2 = 0.516, t = 2.346, which gives P = 0.0258 (df = 6.9). We have fairly strong evidence that high-progress readers have higher mean scores. (c)We test

H0: Scores for both groups are identically distributed vs. Ha: High-progress children systematically score higher

W = 38, P .= 0.0184; we have strong evidence against the hypothesis of identical distributions. This is equivalent to the conclusion reached in (b).

Output from Minitab: Mann-Whitney Confidence Interval and Test

HiProg2 N = 5 Median = 0.8000 LoProg2 N = 5 Median = 0.4900 Point estimate for ETA1-ETA2 is 0.2600 96.3 Percent C.I. for ETA1-ETA2 is (0.0200,0.5199) W = 38.0 Test of ETA1 = ETA2 vs. ETA1 > ETA2 is significant at 0.0184

280 Chapter 14 Nonparametric Tests

14.3 (a) See table. (b) For Story 2, W = 8+ 9+ 4+ 7+ 10 = 38. Under H0,

µW = (5)(11)2 = 27.5

σW = √ (5)(5)(11)

12 .= 4.787

the continuity correction, we compute 37.5−27.54.787

.= 2.09, which gives P = P(Z > 2.09) = 0.0183. (d) See the table.

Story 1 Story 2 Child Progress Score Rank score Rank

1 high 0.55 4.5 0.80 8 2 high 0.57 6 0.82 9 3 high 0.72 8.5 0.54 4 4 high 0.70 7 0.79 7 5 high 0.84 10 0.89 10 6 low 0.40 3 0.77 6 7 low 0.72 8.5 0.49 3 8 low 0.00 1 0.66 5 9 low 0.36 2 0.28 1 10 low 0.55 4.5 0.38 2

14.4 (a) Testing H0: Yields are identically distributed vs. Ha: Yields are systematically higher with no weeds

we Þnd W = 26 and P .= 0.0152. We have strong evidence against the hypothesis of identical distributions. (b) We test H0: µ0 = µ9 vs. Ha: µ0 > µ9. x0 = 170.2, s0 = 5.42, x9 = 157.6, s9 = 10.1, t = 2.20, which gives P = 0.042 (df = 4.6). We have fairly strong evidence that the mean yield is higher with no weedsÑbut the evidence is not quite as strong as in (a). (c) Both tests still reach the same conclusion, so there is no Òpractically important impactÓ on our conclusions. The Wilcoxon evidence is slightly weaker: W = 22, P .= 0.0259. The t-test evidence is slightly stronger: t = 2.79, df = 3, P = 0.034. (The new statistics for the 9-weeds-per-meter group are x9 = 162.633 and s9 = 0.208; these are substantial changes for each value.)

14.5 (a) H0: Nerve response is unaffected by DDT; Ha: Nerve response is systematically different with DDT. (b) We Þnd W = 53 and P .= 0.0306. We have strong evidence that DDT affects nerve response. (c) The conclusions are essentially the same.

14.6 (a) W = 579, which has P = 0.0064; the evidence is slightly stronger with the Wilcoxon test. (b) For the t test, H0: µ1 = µ2 vs. Ha: µ1 > µ2. For the Wilcoxon test, H0: DRP scores are identically distributed for both groups vs. Ha: DRP score are systematically higher for those who had directed reading activities

14.7 (a)W = 106.5, which has P .= 0.16. In Example 7.20, P = 0.059, while in Exercise 7.69, P = 0.06 or 0.07 (depending on df used). In none of these tests did we conclude that the difference is signiÞcant, but the evidence was stronger using the t tests. (b) For the two t tests, we use H0: µ1 = µ2 vs. Ha: µ1 > µ2. For the Wilcoxon test,

H0: Both BP distributions are identical vs. Ha: BP is systematically lower in the calcium group.

(c) For the t tests, we assume we have SRSs from two normal populations (and equal variances, for the Þrst t test). For the Wilcoxon test, we assume only that we have SRSs from continuously distributed populations.

Solutions 281

14.8 Testing H0: Both score distributions are identical vs. Ha: Piano students have sys- tematically higher scores, we obtain W = 1787, which has P < 0.0001, so we reject H0.

14.9 For H0: Responses are identically distributed for both genders vs. Ha: WomenÕs responses are systematically higher, Minitab reports W = 32, 267.5 and a P-value of 0.0003. Women are also more concerned about food safety in restaurants.

14.10 Wedonot have independent samples from twopopulations; rather, wehavedependent samples (each person answered both questions).

14.11 (a) X 2 = 3.955 with df = 4, giving P = 0.413. There is little evidence to make us believe that there is a relationship between city and income. (b) Minitab reports W = 56, 370, with P .= 0.5; again, there is no evidence that incomes are systematically higher in one city.

Section 2: The Wilcoxon Signed Rank Test

14.12 The hypotheses are

H0: Pre- and posttest scores are identically distributed vs. Ha: Posttest scores are systematically higher

(One might also state a two-sided alternative, since the exercise suggests no direction for the difference, but an improvement in scores is a reasonable expectation.) The statistic is W+ = 138.5, and the reported P-value is 0.002Ñstrong evidence that posttest scores are higher.

Output from Minitab: TEST OF MEDIAN = 0.000000 VERSUS MEDIAN G.T. 0.000000

N FOR WILCOXON ESTIMATED N TEST STATISTIC P-VALUE MEDIAN

Diffs 20 17 138.5 0.002 3.000

14.13 (a) The hypotheses are

H0: Pre- and posttest scores are identically distributed vs. Ha: Posttest scores are systematically higher

(b) The Wilcoxon rank sum test requires two independent samples; we have dependent data. (c) xpre = 27.3 and xpost = 28.75 (an increase of 1.45), while the median changes from 29 to 30. The signed rank statistic is W+ = 154.5, and the reported P-value is 0.034Ñstrong evidence that posttest scores are higher.

Output from Minitab: TEST OF MEDIAN = 0.000000 VERSUS MEDIAN G.T. 0.000000

N FOR WILCOXON ESTIMATED N TEST STATISTIC P-VALUE MEDIAN

Diff 20 20 154.5 0.034 1.500

282 Chapter 14 Nonparametric Tests

14.14 There are 17 nonzero differences; only one is negative (the boldface 6 in the list below).

Diff: 1 1 2 2 2 3 3 3 3 3 3 6 6 6 6 6 6 Rank: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸ Value: 1.5 4 8.5 14.5

This gives W+ = 138.5. (Note that the only tie we really need to worry about is the last group; all other ties involve only positive differences.)

14.15 For the differences sfair− srest, x = 0.5149 (other measures may also be used). Ap- plying theWilcoxon signed rank test to these differences, with the one-sided alternativeÑ Òfood at fairs is systematically rated higher (less safe) than restaurant foodÓÑwe obtain W+ = 10, 850.5 (P < 0.0005), so we conclude that restaurant food is viewed as being safer.

Output from Minitab: TEST OF MEDIAN = 0.000000 VERSUS MEDIAN G.T. 0.000000

N FOR WILCOXON ESTIMATED N TEST STATISTIC P-VALUE MEDIAN

Diffs 303 157 10850.5 0.000 0.5000

14.16 For the differences sfair− sfast, x = 0.0693 (other measures may also be used). Ap- plying theWilcoxon signed rank test to these differences, with the one-sided alternativeÑ Òfood at fairs is systematically rated higher (less safe) than fast foodÓÑwe obtain W+ = 4, 730.5 (P = 0.103), so we conclude that the difference in safety ratings is not signiÞcant.

Output from Minitab: TEST OF MEDIAN = 0.000000 VERSUS MEDIAN G.T. 0.000000

N FOR WILCOXON ESTIMATED N TEST STATISTIC P-VALUE MEDIAN

Diffs 303 129 4730.5 0.103 0.000E+00

14.17 A stemplot of the differences is left-skewed, which suggests that a nonparametric test is appropriate. The mean difference is −5.71, and the median difference is −3. The Wilcoxon statistic is W+ = 22.5, with P-value 0.032Ñfairly strong evidence that the wounds healed faster with the natural electric Þeld.

−3 1 −2 −2 2 −1 −1 20 −0 7 −0 433311 0 34 0 1 0

Output from Minitab: TEST OF MEDIAN = 0.000000 VERSUS MEDIAN L.T. 0.000000

N FOR WILCOXON ESTIMATED N TEST STATISTIC P-VALUE MEDIAN

Diffs 14 14 22.5 0.032 -4.000

Solutions 283

14.18 For Before− After differences, we Þnd W+ = 15 (all Þve differences are positive), and P = 0.03; we conclude that vitamin C is lost in cooking.

Output from Minitab: TEST OF MEDIAN = 0.000000 VERSUS MEDIAN G.T. 0.000000

N FOR WILCOXON ESTIMATED N TEST STATISTIC P-VALUE MEDIAN

Diffs 5 5 15.0 0.030 54.50

14.19 The mean change is −5.33; the median is −6. The stemplot is somewhat left-skewed. The Wilcoxon statistic is W+ = 37 (P < 0.0005); the differences (drops in vitamin C content) are systemati- cally positive, so vitamin C content is lower in Haiti.

−1 4 −1 3322 −1 −0 9988 −0 7776666 −0 5444 −0 2 −0 1 0 1 0 33 0 4 0 0 8

Output from Minitab: TEST OF MEDIAN = 0.000000 VERSUS MEDIAN L.T. 0.000000

N FOR WILCOXON ESTIMATED N TEST STATISTIC P-VALUE MEDIAN

Change 27 27 37.0 0.000 -5.500

14.20 The mean and median (right-threaded − left-threaded) differ- ences are −13.32 and −12; the stemplot shows many negative dif- ferences, but it looks reasonably normal. Our hypotheses are ÒTimes have the same distribution for both directionsÓ and ÒClockwise times are systematically lower.Ó The test statistic is W+ = 56.5, which has P = 0.004, so we conclude that clockwise times are lower.

−5 2 −4 853 −3 511 −2 94 −1 66621 −0 74331 0 02 1 1 2 03 3 8Output from Minitab:

TEST OF MEDIAN = 0.000000 VERSUS MEDIAN L.T. 0.000000

N FOR WILCOXON ESTIMATED N TEST STATISTIC P-VALUE MEDIAN

RH-LH 25 24 56.5 0.004 -14.00

Section 3: The Kruskal-Wallis Test

14.21 (a) For ANOVA, H0: µ0 = µ1000 = µ5000 = µ10000 vs. Ha: Not all µi are equal. For Kruskal-Wallis,

H0: The distribution of growth is the same for all nematode counts vs. Ha: Growth is systematically larger for some counts

(b) The medians are 10, 11.1, 5.2, and 5.55 cmÑnoticeably lower for the latter two, suggesting that nematodes retard growth (after a point). The Kruskal-Wallis test statistic is H = 11.34, with df = 3; the P-value is 0.01, so we have strong evidence that growth is not the same for all nematode counts (that is, the difference we observed is statistically signiÞcant).

284 Chapter 14 Nonparametric Tests

Output from Minitab: Kruskal-Wallis Test

LEVEL NOBS MEDIAN AVE. RANK Z VALUE 0 4 10.000 12.3 1.82

1000 4 11.100 12.8 2.06 5000 4 5.200 4.2 -2.06 10000 4 5.550 4.7 -1.82 OVERALL 16 8.5

H = 11.34 d.f. = 3 p = 0.010 H = 11.35 d.f. = 3 p = 0.010 (adjusted for ties)

14.22 (a)Normal quantile plots (not shown) suggest that there may be outliers in the lemon yellow counts (38 is low, 59 is high). No other striking violations are evident (given the small sample sizes). (b) For ANOVA, H0: µ1 = µ2 = µ3 = µ4 vs. Ha: Not all µi are equal. For Kruskal-Wallis,

H0: The distribution of the trapped insect count is the same for all board colors vs. Ha: Insects trapped is systematically higher for some colors

(c) In the order given, the medians are 46.5, 15.5, 34.5, and 15 insects; it appears that yellow is most effective, green is in the middle, and white and blue are least effective. The Kruskal-Wallis test statistic is H = 16.95, with df = 3; the P-value is 0.001, so we have strong evidence that color affects the insect count (that is, the difference we observed is statistically signiÞcant).

Output from Minitab: LEVEL NOBS MEDIAN AVE. RANK Z VALUE

1 6 46.50 21.2 3.47 2 6 15.50 7.3 -2.07 3 6 34.50 14.8 0.93 4 6 15.00 6.7 -2.33

OVERALL 24 12.5

H = 16.95 d.f. = 3 p = 0.001 H = 16.98 d.f. = 3 p = 0.001 (adjusted for ties)

14.23 We test H0: All hot dogs have the same calorie distribution vs. Ha: Some type is systematically different (lower/higher) than some other. With H = 15.89, df = 2, and P < 0.0005, we conclude that there is a difference; speciÞcally, poultry hot dogs are lower than the other two types (which differ very little).

Min Q1 M Q3 Max Beef 111 140.0 152.5 178.5 190 Meat 107 138.5 153.0 180.5 195 Poultry 86 100.5 129.0 143.5 170

Output from Minitab: LEVEL NOBS MEDIAN AVE. RANK Z VALUE

1 20 152.5 33.1 2.02 2 17 153.0 33.5 1.89 3 17 129.0 14.9 -3.99

OVERALL 54 27.5

H = 15.89 d.f. = 2 p = 0.000 H = 15.90 d.f. = 2 p = 0.000 (adjusted for ties)

Solutions 285

14.24 (a) I = 4, ni = 6, N = 24. (b) The columns in the table at the right are rank, number of insects, and color. The Ri (rank sums) are Yellow 17+ 20+ 21+ 22+ 23+ 24 = 127 White 3+ 4+ 5.5+ 9.5+ 9.5+ 12.5 = 44 Green 7+ 14+ 15+ 16+ 18+ 19 = 89 Blue 1+ 2+ 5.5+ 8+ 11+ 12.5 = 40 (c) H = 12

24(25)

( 1272 + 442 + 892 + 402

) − 3(25)

= 91.953− 75 = 16.953. Under H0, this has approximately the chi-squared distribution with df = I − 1 = 3; comparing to this distribution tells us that 0.0005 < P < 0.001.

1 7 B 12.5 21 B 2 11 B 14 25 G 3 12 W 15 32 G 4 13 W 16 37 G 5.5 14 W 17 38 Y 5.5 14 B 18 39 G 7 15 G 19 41 G 8 16 B 20 45 Y 9.5 17 W 21 46 Y 9.5 17 W 22 47 Y 11 20 B 23 48 Y 12.5 21 W 24 59 Y

14.25 We test H0: All hot dogs have the same sodium distribution vs. Ha: Some type is systematically different (lower/higher) than some other. With H = 4.71, df = 2, and P = 0.095, we have some evidence of difference, but not enough to reject H0.

Min Q1 M Q3 Max Beef 253 320.5 380.5 478 645 Meat 144 379.0 405.0 501 545 Poultry 357 379.0 430.0 535 588

Output from Minitab: LEVEL NOBS MEDIAN AVE. RANK Z VALUE

1 20 380.5 22.0 -1.95 2 17 405.0 28.1 0.20 3 17 430.0 33.3 1.83

OVERALL 54 27.5

H = 4.71 d.f. = 2 p = 0.095 H = 4.71 d.f. = 2 p = 0.095 (adjusted for ties)

14.26 (a) The Þve-number summaries (right) suggest that the scores of piano students are higher; there is little difference among the other three (except in the extremes). (b) The normal quantile plots (not shown) show a low outlier (−4) for singing, and another (−6) for the no-lessons group. The others are reasonably normal (aside from granularity). (c) The test statistic is H = 21 (df = 3), which has P < 0.0005Ñstrong evidence against the null hypothesis (Òscores are identically distributed for all four groupsÓ). Some treatment (presumably piano lessons) is systematically different (higher) than other treatments.

Lessons Min Q1 M Q3 Max Piano −3 2 4 6 9 Singing −4 −1 0 1 1 Computer −3 −1 0.5 2 4 None −6 −1 0 2 7

286 Chapter 14 Nonparametric Tests

Output from Minitab: LEVEL NOBS MEDIAN AVE. RANK Z VALUE

1 34 4.00E+00 52.6 4.47 2 10 0.00E+00 23.6 -2.38 3 20 5.00E-01 29.9 -2.19 4 14 0.00E+00 32.8 -1.22

OVERALL 78 39.5

H = 21.00 d.f. = 3 p = 0.000 H = 21.25 d.f. = 3 p = 0.000 (adjusted for ties)

14.27 For the Kruskal-Wallis test, we need two or more independent samples. Since these data come from different questions being asked of the same people, the responses are not independent.

14.28 (a)Yes, the data support this statement: 68211 .= 32.2%of high-SES subjects have never

smoked, compared to 17.3% and 23.7% of middle- and low-SES subjects (respectively). Also, only 51211

.= 24.2% of high-SES subjects are current smokers, versus 42.3% and 46.2% of those in the middle- and low-SES groups. (b) X 2 = 18.510 with df = 4; this has P = 0.001. There is a signiÞcant relationship. (c) H = 12.72 with df = 2, so P = 0.002Ñor, after adjusting for ties, H = 14.43 and P = 0.001. The observed differences are signiÞcant; some SES groups smoke systematically more.

Output from Minitab: LEVEL NOBS MEDIAN AVE. RANK Z VALUE

1 211 2.000 162.4 -3.56 2 52 2.000 203.6 1.90 3 93 2.000 201.0 2.46

OVERALL 356 178.5

H = 12.72 d.f. = 2 p = 0.002 H = 14.43 d.f. = 2 p = 0.001 (adjusted for ties)

14.29 (a)We compare beef and meat, beef and poultry, and meat and poultry. (b)Minitab output (portions appear below) gives P = 0.9393, P = 0.0005, and P = 0.0007, respectively. (c) The latter two P-values are (quite a bit) less than 0.0167. Beef and meat are not signiÞcantly different; poultry is signiÞcantly lower in calories than both beef and meat hot dogs.

Output from Minitab: Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Beef Ð Meat Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð 95.1 Percent C.I. for ETA1-ETA2 is (-19.99,13.00) W = 377.0 Test of ETA1 = ETA2 vs. ETA1 ~= ETA2 is significant at 0.9393 The test is significant at 0.9392 (adjusted for ties)

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Beef Ð Poultry Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð 95.1 Percent C.I. for ETA1-ETA2 is (15.01,49.99) W = 495.5 Test of ETA1 = ETA2 vs. ETA1 ~= ETA2 is significant at 0.0005 The test is significant at 0.0005 (adjusted for ties)

Solutions 287

(Output continues) Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Meat Ð Poultry Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð 95.0 Percent C.I. for ETA1-ETA2 is (17.00,52.01) W = 396.0 Test of ETA1 = ETA2 vs. ETA1 ~= ETA2 is significant at 0.0007 The test is significant at 0.0007 (adjusted for ties)

14.30 The P-values (from Minitab; output below) are summarized in the table at the right. To be Bonferroni- signiÞcant, we must have P ≤ α/6 = 0.0083, so only the yellow/white and yellow/blue differences are signiÞcant. Green is (barely) not signiÞcantly different from the other colors.

Yellow Ð White 0.0051* Yellow Ð Green 0.0131 Yellow Ð Blue 0.0051* White Ð Green 0.0202 White Ð Blue 0.8102 Green Ð Blue 0.0202

Output from Minitab: Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Yellow Ð White Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð 95.5 Percent C.I. for ETA1-ETA2 is (25.00,38.00) W = 57.0 Test of ETA1 = ETA2 vs. ETA1 ~= ETA2 is significant at 0.0051 The test is significant at 0.0050 (adjusted for ties)

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Yellow Ð Green Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð 95.5 Percent C.I. for ETA1-ETA2 is (6.00,30.00) W = 55.0 Test of ETA1 = ETA2 vs. ETA1 ~= ETA2 is significant at 0.0131

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Yellow Ð Blue Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð 95.5 Percent C.I. for ETA1-ETA2 is (25.00,40.00) W = 57.0 Test of ETA1 = ETA2 vs. ETA1 ~= ETA2 is significant at 0.0051

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð White Ð Green Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð 95.5 Percent C.I. for ETA1-ETA2 is (-25.00,-3.00) W = 24.0 Test of ETA1 = ETA2 vs. ETA1 ~= ETA2 is significant at 0.0202 The test is significant at 0.0200 (adjusted for ties)

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð White Ð Blue Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð 95.5 Percent C.I. for ETA1-ETA2 is (-6.003,6.996) W = 41.0 Test of ETA1 = ETA2 vs. ETA1 ~= ETA2 is significant at 0.8102 The test is significant at 0.8092 (adjusted for ties)

Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Green Ð Blue Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð Ð 95.5 Percent C.I. for ETA1-ETA2 is (4.00,27.00) W = 54.0 Test of ETA1 = ETA2 vs. ETA1 ~= ETA2 is significant at 0.0202

288 Chapter 15 Logistic Regression

Chapter 15 Solutions

15.1 (a) For the high blood pressure group, p̂ = 553338 .= 0.01648, giving odds p̂1− p̂ = 553283

.= 0.01675, or about 1 to 60. (If students give odds in the form Òa to b,Ó their choices of a and b might be different.) (b) For the low blood pressure group, p̂ = 212676

.= 0.00785, giving odds p̂1− p̂ = 212655

.= 0.00791, or about 1 to 126 (or 125). (c) The odds ratio is about 2.118. Odds of death from cardiovascular disease are about 2.1 times greater in the high blood pressure group.

15.2 (a) For female references, p̂ = 4860 = 0.8, giving odds p̂1− p̂ = 4812 = 4 (Ò4 to 1Ó). (b) For male references, p̂ = 52132 = 0.39, giving odds p̂1− p̂ = 5280 = 0.65 (Ò13 to 20Ó). (c) The odds ratio is about 6.154. (The odds of a juvenile reference are more than six times greater for females.)

15.3 (a) Find b1 ± z∗SEb1 , using either z∗ = 2 or z∗ = 1.96. These give 0.2349 to 1.2661, or 0.2452 to 1.2558, respectively. (b) X 2 =

( 0.7505 0.2578

)2 .= 8.47. This gives a P-value between 0.0025 and 0.005. (c)We have strong evidence that there is a real (signiÞcant) difference in risk between the two groups.

15.4 (a) Find b1 ± z∗SEb1 , using either z∗ = 2 or z∗ = 1.96. These give 1.0799 to 2.5543, or 1.0946 to 2.5396, respectively. (b) X 2 =

( 1.8171 0.3686

)2 .= 24.3023. This gives P < 0.0005. (c)We have strong evidence that there is a real (signiÞcant) difference in juvenile references between male and female references.

15.5 (a) The estimated odds ratio is eb1 .= 2.118 (as we found in Exercise 15.1). Exponen- tiating the intervals for β1 in Exercise 15.3(a) gives odds-ratio intervals from about 1.26 to 3.55 (z∗ = 2), or 1.28 to 3.51 (z∗ = 1.96). (b) We are 95% conÞdent that the odds of death from cardiovascular disease are about 1.3 to 3.5 times greater in the high blood pressure group.

15.6 (c) The estimated odds ratio is eb1 .= 6.154 (as we found in Exercise 15.2). Exponen- tiating the intervals for β1 in Exercise 15.4(a) gives odds-ratio intervals from about 2.94 to 12.86 (z∗ = 2), or 2.99 to 12.67 (z∗ = 1.96). (b)We are 95% conÞdent that the odds of a juvenile reference are about 3 to 13 times greater among females.

15.7 (a) The model is log (

pi 1−pi

) = β0 + β1xi , where xi = 1 if the i th person is over 40,

and 0 if he/she is under 40. (b) pi is the probability that the i th person is terminated; this model assumes that the probability of termination depends on age (over/under 40). In this case, that seems to have been the case, but we might expect that other factors were taken into consideration. (c) The estimated odds ratio is eb1 .= 3.859. (Of course, we can also get this from 41/7657/504 .) We can also Þnd, e.g., a 95% conÞdence interval for b1: b1 ± 1.96SEb1 = 0.5409 to 2.1599. Exponentiating this translates to a 95%

Solutions 289

conÞdence interval for the odds: 1.7176 to 8.6701. The odds of being terminated are 1.7 to 8.7 times greater for those over 40. (d) Use a multiple logistic regression model, e.g., log

( pi

1−pi ) = β0 + β1xi + β2yi .

15.8 We show the steps for doing this by hand; if software is available, the results should

be the same. The model is log (

p 1−p

) = β0 + β1x . We make the arbitrary choice to take

x to be the indicator variable for ÒmaleÓÑi.e., x = 1 for men, 0 for women. (We could also choose to have x = 1 for women, and 0 for men.) Then

log (

pm 1− pm

) = β0 + β1, and log

( pf

1− pf ) = β0

With the given data, we estimate

log (

p̂m 1− p̂m

) = log

( 515 1005

) .= −0.6686 = b0 + b1 and

log (

p̂f 1− p̂f

) = log

( 27 164

) .= −1.8040 = b0

so we Þnd that b0 = −1.8040 and b1 = 1.1354. This gives an odds ratio of about eb1 .= 3.11; we estimate that the odds for a male testing positive are about three times those for a female. With software, we Þnd SEb1

.= 0.2146 and X 2 = 27.98 (P < 0.0001). The logistic regression is signiÞcant (i.e., we conclude that β1 6= 0). A 95% conÞdence interval for β1 is b1 ± 1.96SEb1 = 0.7148 to 1.5561, so we are 95% conÞdent that the odds ratio is between about 2.04 and 4.74.

15.9 For the model log (

p 1−p

) = β0 + β1x , we obtain the Þtted model log(ODDS) =

b0+ b1x = −7.2789+ 0.9399x . (Here p is the probability that the cheese is acceptable, and x is the value of H2S.) We have b1 = 0.9399 and SEb1 = 0.3443, so we estimate that the odds ratio increases by a factor of eb1 .= 2.56 for every unit increase in H2S. For testing β1 = 0, we Þnd X 2 = 7.45 (P = 0.0063), so we conclude that β1 6= 0. We are 95% conÞdent that β1 is in the interval b1±1.96SEb1 = 0.2651 to 1.6147; exponentiating this tells us that the odds ratio increases by a factor between 1.3035 and 5.0265 (with 95% conÞdence) for each unit increase in H2S.

15.10 For the model log (

p 1−p

) = β0 + β1x , we obtain the Þtted model log(ODDS) =

b0+b1x = −10.7799+6.3319x . (Here p is the probability that the cheese is acceptable, and x is the value of Lactic.) We have b1 = 6.3319 and SEb1 = 2.4532, so we estimate that the odds ratio increases by a factor of eb1 .= 562.22 for every unit increase in Lactic. For testing β1 = 0, we Þnd X 2 = 6.66 (P = 0.0098), so we conclude that β1 6= 0. We are 95% conÞdent that β1 is in the interval b1 ± 1.96SEb1 = 1.5236 to 11.1402; exponentiating this tells us that the odds ratio increases by a factor between 4.5889 and about 68,884 (with 95% conÞdence) for each unit increase in Lactic.

290 Chapter 15 Logistic Regression

15.11 The seven models are summarized below. The P-value in the right column is for the null hypothesis that all slopes equal 0 (i.e., the signiÞcance of the regression); all are signiÞcant. For the three new models (those with two predictors), all have only one coefÞcient

signiÞcantly different from 0 (in the last case, arguably neither coefÞcient is nonzero). The standard errors are given in parentheses below each coefÞcient; the six respective P-values are 0.4276, 0.0238; 0.3094, 0.0355; 0.0567, 0.1449. In summary, we might conclude that that H2S has the greatest effect: It had the

smallest P-value among the three single-predictor models, and in the three multiple logistic regression models in which it was used, it had the minimum P-value. (It was the closest to being signiÞcant in the last two models in the table below.)

Fitted Model P

log(ODDS) = −13.71 + 2.249 Acetic 0.0285 log(ODDS) = −7.279 + 0.9399 H2S 0.0063 log(ODDS) = −10.78 + 6.332 Lactic 0.0098 log(ODDS) = −12.85 + 1.096 Acetic + 0.8303 H2S 0.0008

(1.382) (0.3673)

log(ODDS) = −16.56 + 1.309 Acetic + 5.257 Lactic 0.0016 (1.288) (2.500)

log(ODDS) = −11.72 + 0.7346 H2S + 3.777 Lactic 0.0003 (0.3866) (2.596)

log(ODDS) = −14.26 + 0.584 Acetic + 0.6849 H2S + 3.468 Lactic 0.0010

15.12 Portions of SAS and GLMStat output are given below. (a) The X 2 statistic for testing this hypothesis is 33.65 (df = 3), which has P = 0.0001. We conclude that at least one coefÞcient is not 0. (b)Themodel is log(ODDS) = −6.053+0.3710HSM+0.2489HSS+ 0.03605HSE. The standard errors of the three coefÞcients are 0.1302, 0.1275, and 0.1253, giving respective 95% conÞdence intervals 0.1158 to 0.6262, −0.0010 to 0.4988, and −0.2095 to 0.2816. (c) Only the coefÞcient of HSM is signiÞcantly different from 0, though HSS may also be useful. (Only HSM was useful in the multiple linear regression model of GPA on high school grades.)

Output from SAS: Intercept

Intercept and Criterion Only Covariates Chi-Square for Covariates

AIC 297.340 269.691 . SC 300.751 283.338 . -2 LOG L 295.340 261.691 33.648 with 3 DF (p=0.0001) Score . . 29.672 with 3 DF (p=0.0001)

Solutions 291

(Output continues) Analysis of Maximum Likelihood Estimates

Parameter Standard Wald Pr > Standardized Variable DF Estimate Error Chi-Square Chi-Square Estimate

INTERCPT 1 -6.0528 1.1562 27.4050 0.0001 . HSM 1 0.3710 0.1302 8.1155 0.0044 0.335169 HSS 1 0.2489 0.1275 3.8100 0.0509 0.233265 HSE 1 0.0361 0.1253 0.0828 0.7736 0.029971

Output from GLMStat: estimate se(est) z ratio Prob>|z|

1 Constant -6.053 1.156 -5.236 <0.0001 2 HSM 0.3710 0.1302 2.849 0.0044 3 HSS 0.2489 0.1275 1.952 0.0509 4 HSE 3.605e-2 0.1253 0.2877 0.7736

15.13 Portions of SAS and GLMStat output are given below. (a) The X 2 statistic for testing this hypothesis is 14.2 (df = 2), which has P = 0.0008. We conclude that at least one coefÞcient is not 0. (b) The model is log(ODDS) = −4.543 + 0.003690 SATM + 0.003527 SATV. The standard errors of the two coefÞcients are 0.001913 and 0.001751, giving respective 95% conÞdence intervals −0.000059 to 0.007439, and 0.000095 to 0.006959. (The Þrst coefÞcient has a P-value of 0.0537, and the second has P = 0.0440.) (c)We (barely) cannot reject βSATM = 0Ñthough since 0 is just in the conÞdence interval, we are reluctant to discard SATM.Meanwhile, we conclude that βSATV 6= 0. (By contrast, with multiple linear regression of GPA on SAT scores, we found SATM useful but not SATV.)

Output from SAS: Intercept

Intercept and Criterion Only Covariates Chi-Square for Covariates

AIC 297.340 287.119 . SC 300.751 297.354 . -2 LOG L 295.340 281.119 14.220 with 2 DF (p=0.0008) Score . . 13.710 with 2 DF (p=0.0011)

Analysis of Maximum Likelihood Estimates

Parameter Standard Wald Pr > Standardized Variable DF Estimate Error Chi-Square Chi-Square Estimate

INTERCPT 1 -4.5429 1.1618 15.2909 0.0001 . SATM 1 0.00369 0.00191 3.7183 0.0538 0.175778 SATV 1 0.00353 0.00175 4.0535 0.0441 0.180087

Output from GLMStat: estimate se(est) z ratio Prob>|z|

1 Constant -4.543 1.161 -3.915 <0.0001 2 SATM 3.690e-3 1.913e-3 1.929 0.0537 3 SATV 3.527e-3 1.751e-3 2.014 0.0440

292 Chapter 15 Logistic Regression

15.14 The coefÞcients and standard errors for the Þtted model are below. (a) The X 2

statistic for testing this hypothesis is 23.0 (df = 3); since P < 0.0001, we reject H0 and conclude that high school grades add a signiÞcant amount to the model with SAT scores. (b) The X 2 statistic for testing this hypothesis is 3.6 (df = 2); since P = 0.1653, we cannot reject H0; SAT scores do not add signiÞcantly to the model with high school grades. (c) For modeling the odds of HIGPA, high school grades (speciÞcally HSM, and to a lesser extent HSS) are useful, while SAT scores are not.

Output from SAS: Analysis of Maximum Likelihood Estimates

Parameter Standard Wald Pr > Standardized Variable DF Estimate Error Chi-Square Chi-Square Estimate

INTERCPT 1 -7.3732 1.4768 24.9257 0.0001 . HSM 1 0.3427 0.1419 5.8344 0.0157 0.309668 HSS 1 0.2249 0.1286 3.0548 0.0805 0.210704 HSE 1 0.0190 0.1289 0.0217 0.8829 0.015784 SATM 1 0.000717 0.00220 0.1059 0.7448 0.034134 SATV 1 0.00289 0.00191 2.2796 0.1311 0.147566

Output from GLMStat: estimate se(est) z ratio Prob>|z|

1 Constant -7.373 1.477 -4.994 <0.0001 2 SATM 7.166e-4 2.201e-3 0.3255 0.7448 3 SATV 2.890e-3 1.914e-3 1.510 0.1311 4 HSM 0.3427 0.1419 2.416 0.0157 5 HSS 0.2249 0.1286 1.748 0.0805 6 HSE 1.899e-2 0.1289 0.1473 0.8829

15.15 (a) The Þtted model is log(ODDS) = −0.6124 + 0.0609 Gender; the coefÞcient of gender is not signiÞcantly different from 0 (SEbGender = 0.2889, P = 0.8331). (b) Now log(ODDS) = −5.214 + 0.3028 Gender + 0.004191 SATM + 0.003447 SATV. In this model, gender is still not signiÞcant (P = 0.3296). (c)Gender is not useful for modeling the odds of HIGPA.

Output from GLMStat: estimate se(est) z ratio Prob>|z|

1 Constant -5.214 1.362 -3.828 0.0001 2 Gender 0.3028 0.3105 0.9750 0.3296 3 SATM 4.191e-3 1.987e-3 2.109 0.0349 4 SATV 3.447e-3 1.760e-3 1.958 0.0502

Solutions 293

15.16 (a) The Þtted model is log(ODDS) = 3.4761 + 0.4157x , x = 0 for Hospital A and 1 for Hospital B. With b1

.= 0.4157 and SEb1 .= 0.2831, we Þnd that X 2 = 2.16 (P = 0.1420), so we do not have evidence to suggest that β1 is not 0. A 95% conÞdence interval for β1 is−0.1392 to 0.9706 (this interval includes 0). We estimate the odds ratio to be eb1 .= 1.52, with conÞdence interval 0.87 to 2.64 (this includes 1, since β1 might be 0). (a) The Þtted model is log(ODDS) = −6.930+ 1.009 Hospital− 0.09132 Condition; as before, use 0 for Hospital A and 1 for Hospital B, and 1 for good condition, and 0 for poor. Now we estimate the odds ratio to be eb1 .= 2.74, with conÞdence interval 0.30 to 25.12. (c) In neither case is the effect signiÞcant; SimpsonÕs paradox is seen in the increased width of the interval from part(a) to part (b).

Output from GLMStat: estimate se(est) z ratio Prob>|z|

1 Constant -6.930 0.7693 -9.009 <0.0001 2 Hosp 1.009 1.130 0.8928 0.3720 3 Cond -9.132e-2 1.130 8.080e-2 0.9356

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