A survey by the Roper Organization found that 45% of the people who were offended by a television program would change the channel, while 15% would turn off their television sets. The survey further stated that the margin of error is 3 percentage points, and 4000 adults were interviewed.

Several questions arise:

1.How do these estimates compare with the true population percentages?

2.What is meant by a margin of error of 3 percentage points?

3.Is the sample of 4000 large enough to represent the population of all adults who watch television in the United States?

See Statistics Today—Revisited at the end of the chapter for the answers.

After reading this chapter, you will be able to answer these questions, since this chapter explains how statisticians can use statistics to make estimates of parameters.

Source: The Associated Press.

Introduction

One aspect of inferential statistics is estimation, which is the process of estimating the value of a parameter from information obtained from a sample. For example, The Book of Odds, by Michael D. Shook and Robert L. Shook (New York: Penguin Putnam, Inc.), contains the following statements:

“One out of 4 Americans is currently dieting.” (Calorie Control Council)

“Seventy-two percent of Americans have flown on commercial airlines.” (“The Bristol Meyers Report: Medicine in the Next Century”)

“The average kindergarten student has seen more than 5000 hours of television.” (U.S. Department of Education)

“The average school nurse makes $32,786 a year.” (National Association of School Nurses)

“The average amount of life insurance is $108,000 per household with life insurance.” (American Council of Life Insurance)

Page 355

Since the populations from which these values were obtained are large, these values are only estimates of the true parameters and are derived from data collected from samples.

The statistical procedures for estimating the population mean, proportion, variance, and standard deviation will be explained in this chapter.

An important question in estimation is that of sample size. How large should the sample be in order to make an accurate estimate? This question is not easy to answer since the size of the sample depends on several factors, such as the accuracy desired and the probability of making a correct estimate. The question of sample size will be explained in this chapter also.

Objective 1

Find the confidence interval for the mean when σ is known.

7–1Confidence Intervals for the Mean When σ Is Known and Sample Size

Suppose a college president wishes to estimate the average age of students attending classes this semester. The president could select a random sample of 100 students and find the average age of these students, say, 22.3 years. From the sample mean, the president could infer that the average age of all the students is 22.3 years. This type of estimate is called a point estimate .

A point estimate is a specific numerical value estimate of a parameter. The best point estimate of the population mean µ is the sample mean .

You might ask why other measures of central tendency, such as the median and mode, are not used to estimate the population mean. The reason is that the means of samples vary less than other statistics (such as medians and modes) when many samples are selected from the same population. Therefore, the sample mean is the best estimate of the population mean.

Sample measures (i.e., statistics) are used to estimate population measures (i.e., parameters). These statistics are called estimators. As previously stated, the sample mean is a better estimator of the population mean than the sample median or sample mode.

A good estimator should satisfy the three properties described now.

Three Properties of a Good Estimator

1.The estimator should be an unbiased estimator. That is, the expected value or the mean of the estimates obtained from samples of a given size is equal to the parameter being estimated.

2.The estimator should be consistent. For a consistent estimator, as sample size increases, the value of the estimator approaches the value of the parameter estimated.

3.The estimator should be a relatively efficient estimator. That is, of all the statistics that can be used to estimate a parameter, the relatively efficient estimator has the smallest variance.

Page 356

Confidence Intervals

As stated in Chapter 6, the sample mean will be, for the most part, somewhat different from the population mean due to sampling error. Therefore, you might ask a second question: How good is a point estimate? The answer is that there is no way of knowing how close a particular point estimate is to the population mean.

This answer places some doubt on the accuracy of point estimates. For this reason, statisticians prefer another type of estimate, called an interval estimate .

An interval estimate of a parameter is an interval or a range of values used to estimate the parameter. This estimate may or may not contain the value of the parameter being estimated.

In an interval estimate, the parameter is specified as being between two values. For example, an interval estimate for the average age of all students might be 26.9 < µ < 27.7, or 27.3 ± 0.4 years.

Either the interval contains the parameter or it does not. A degree of confidence (usually a percent) can be assigned before an interval estimate is made. For instance, you may wish to be 95% confident that the interval contains the true population mean. Another question then arises. Why 95%? Why not 99 or 99.5%?

If you desire to be more confident, such as 99 or 99.5% confident, then you must make the interval larger. For example, a 99% confidence interval for the mean age of college students might be 26.7 < µ < 27.9, or 27.3 ± 0.6. Hence, a tradeoff occurs. To be more confident that the interval contains the true population mean, you must make the interval wider.

Historical Notes

Point and interval estimates were known as long ago as the late 1700s. However, it wasn’t until 1937 that a mathematician, J. Neyman, formulated practical applications for them.

The confidence level of an interval estimate of a parameter is the probability that the interval estimate will contain the parameter, assuming that a large number of samples are selected and that the estimation process on the same parameter is repeated.

A confidence interval is a specific interval estimate of a parameter determined by using data obtained from a sample and by using the specific confidence level of the estimate.

Intervals constructed in this way are called confidence intervals. Three common confidence intervals are used: the 90, the 95, and the 99% confidence intervals.

The algebraic derivation of the formula for determining a confidence interval for a mean will be shown later. A brief intuitive explanation will be given first.

The central limit theorem states that when the sample size is large, approximately 95% of the sample means taken from a population and same sample size will fall within ±1.96 standard errors of the population mean, that is,

Now, if a specific sample mean is selected, say, , there is a 95% probability that the interval µ ± 1.96(σ/) contains . Likewise, there is a 95% probability that the interval specified by

will contain µ, as will be shown later. Stated another way,

Interesting Fact

A postal worker who delivers mail walks on average 5.2 miles per day.

Page 357

Hence, you can be 95% confident that the population mean is contained within that interval when the values of the variable are normally distributed in the population.

The value used for the 95% confidence interval, 1.96, is obtained from Table E in Appendix C. For a 99% confidence interval, the value 2.58 is used instead of 1.96 in the formula. This value is also obtained from Table E and is based on the standard normal distribution. Since other confidence intervals are used in statistics, the symbol zα/2 (read “zee sub alpha over two”) is used in the general formula for confidence intervals. The Greek letter α (alpha) represents the total area in both tails of the standard normal distribution curve, and α/2 represents the area in each one of the tails. More will be said after Examples 7–1 and 7–2 about finding other values for zα/2.

The relationship between a and the confidence level is that the stated confidence level is the percentage equivalent to the decimal value of 1 – α, and vice versa. When the 95% confidence interval is to be found, α = 0.05, since 1 – 0.05 = 0.95, or 95%. When α = 0.01, then 1 – α = 1 – 0.01 = 0.99, and the 99% confidence interval is being calculated.

Formula for the Confidence Interval of the Mean for a Specific α

For a 90% confidence interval, zα/2 = 1.65; for a 95% confidence interval, zα/2 = 1.96; and for a 99% confidence interval, zα/2 = 2.58.

The term zα/2(α/) is called the maximum error of the estimate (also called the margin of error). For a specific value, say, α = 0.05, 95% of the sample means will fall within this error value on either side of the population mean, as previously explained. See Figure 7–1.

Figure 7–1

95% Confidence Interval

When n ≥ 30, s can be substituted for σ, but a different distribution is used.

The maximum error of the estimate is the maximum likely difference between the point estimate of a parameter and the actual value of the parameter.

A more detailed explanation of the maximum error of the estimate follows Examples 7–1 and 7–2, which illustrate the computation of confidence intervals.

Rounding Rule for a Confidence Interval for a Mean When you are computing a confidence interval for a population mean by using raw data, round off to one more decimal place than the number of decimal places in the original data. When you are computing a confidence interval for a population mean by using a sample mean and a standard deviation, round off to the same number of decimal places as given for the mean.

Page 358

Example 7–1

Days It Takes to Sell an Aveo

A researcher wishes to estimate the number of days it takes an automobile dealer to sell a Chevrolet Aveo. A sample of 50 cars had a mean time on the dealer’s lot of 54 days. Assume the population standard deviation to be 6.0 days. Find the best point estimate of the population mean and the 95% confidence interval of the population mean.

Source: Based on information obtained from Power Information Network.

Solution

The best point estimate of the mean is 54 days. For the 95% confidence interval use z = 1.96.

Hence one can say with 95% confidence that the interval between 52.3 and 55.7 days does contain the population mean, based on a sample of 50 automobiles.

Example 7–2

Ages of Automobiles

A survey of 30 adults found that the mean age of a person’s primary vehicle is 5.6 years. Assuming the standard deviation of the population is 0.8 year, find the best point estimate of the population mean and the 99% confidence interval of the population mean.

Source: Based on information in USA TODAY.

Solution

The best point estimate of the population mean is 5.6 years.

Hence, one can be 99% confident that the mean age of all primary vehicles is between 5.2 and 6.0 years, based on 30 vehicles.

Another way of looking at a confidence interval is shown in Figure 7–2. According to the central limit theorem, approximately 95% of the sample means fall within 1.96 standard deviations of the population mean if the sample size is 30 or more or if σ is known when n is less than 30 and the population is normally distributed. If it were possible to build a confidence interval about each sample mean, as was done in Examples 7–1 and 7–2 for µ, 95% of these intervals would contain the population mean, as shown in Figure 7–3. Hence, you can be 95% confident that an interval built around a specific sample mean would contain the population mean. If you desire to be 99% confident, you must enlarge the confidence intervals so that 99 out of every 100 intervals contain the population mean.

Page 359

Figure 7–2

95% Confidence Interval for Sample Means

Figure 7–3

95% Confidence Intervals for Each Sample Mean

Since other confidence intervals (besides 90, 95, and 99%) are sometimes used in statistics, an explanation of how to find the values for zα/2 is necessary. As stated previously, the Greek letter α represents the total of the areas in both tails of the normal distribution. The value for a is found by subtracting the decimal equivalent for the desired confidence level from 1. For example, if you wanted to find the 98% confidence interval, you would change 98% to 0.98 and find α = 1 – 0.98, or 0.02. Then α/2 is obtained by dividing α by 2. So α/2 is 0.02/2, or 0.01. Finally, z0.01 is the z value that will give an area of 0.01 in the right tail of the standard normal distribution curve. See Figure 7–4.

Page 360

Figure 7–4

Finding α/2 for a 98% Confidence Interval

Figure 7–5

Finding zα/2 for a 98% Confidence Interval

Once α/2 is determined, the corresponding zα/2 value can be found by using the procedure shown in Chapter 6, which is reviewed here. To get the zα/2 value for a 98% confidence interval, subtract 0.01 from 1.0000 to get 0.9900. Next, locate the area that is closest to 0.9900 (in this case, 0.9901) in Table E, and then find the corresponding z value. In this example, it is 2.33. See Figure 7–5.

For confidence intervals, only the positive z value is used in the formula.

When the original variable is normally distributed and σ is known, the standard normal distribution can be used to find confidence intervals regardless of the size of the sample. When n ≥ 30, the distribution of means will be approximately normal even if the original distribution of the variable departs from normality.

When σ is unknown, s can be used as an estimate of σ, but a different distribution is used for the critical values. This method is explained in Section 7–2.

Example 7–3

Credit Union Assets

The following data represent a sample of the assets (in millions of dollars) of 30 credit unions in southwestern Pennsylvania. Find the 90% confidence interval of the mean.

12.23	16.56	4.39
2.89	1.24	2.17
13.19	9.16	1.42
73.25	1.91	14.64
11.59	6.69	1.06
8.74	3.17	18.13
7.92	4.78	16.85
40.22	2.42	21.58
5.01	1.47	12.24
2.27	12.77	2.76

Source: Pittsburgh Post Gazette.

Solution

Step 1Find the mean and standard deviation for the data. Use the formulas shown in Chapter 3 or your calculator. The mean = 11.091. Assume the standard deviation of the population is 14.405.

Step 2Find α/2. Since the 90% confidence interval is to be used, α = 1 – 0.90 = 0.10, and

Step 3Find zα/2. Subtract 0.05 from 1.000 to get 0.9500. The corresponding z value obtained from Table E is 1.65. (Note: This value is found by using the z value for an area between 0.9495 and 0.9505. A more precise z value obtained mathematically is 1.645 and is sometimes used; however, 1.65 will be used in this textbook.)

Step 4Substitute in the formula

Hence, one can be 90% confident that the population mean of the assets of all credit unions is between $6.752 million and $15.430 million, based on a sample of 30 credit unions.

Comment to Computer and Statistical Calculator Users

This chapter and subsequent chapters include examples using raw data. If you are using computer or calculator programs to find the solutions, the answers you get may vary somewhat from the ones given in the textbook. This is so because computers and calculators do not round the answers in the intermediate steps and can use 12 or more decimal places for computation. Also, they use more exact values than those given in the tables in the back of this book. These discrepancies are part and parcel of statistics.

Objective 2

Determine the minimum sample size for finding a confidence interval for the mean.

Sample Size

Sample size determination is closely related to statistical estimation. Quite often, you ask, How large a sample is necessary to make an accurate estimate? The answer is not simple, since it depends on three things: the maximum error of the estimate, the population standard deviation, and the degree of confidence. For example, how close to the true mean do you want to be (2 units, 5 units, etc.), and how confident do you wish to be (90, 95, 99%, etc.)? For the purpose of this chapter, it will be assumed that the population standard deviation of the variable is known or has been estimated from a previous study.

Page 362

The formula for sample size is derived from the maximum error of the estimate formula

and this formula is solved for n as follows:

Hence,

Formula for the Minimum Sample Size Needed for an Interval Estimate of the Population Mean

where E is the maximum error of estimate. If necessary, round the answer up to obtain a whole number. That is, if there is any fraction or decimal portion in the answer, use the next whole number for sample size n.

Example 7–4

Depth of a River

A scientist wishes to estimate the average depth of a river. He wants to be 99% confident that the estimate is accurate within 2 feet. From a previous study, the standard deviation of the depths measured was 4.38 feet.

Solution

Since α = 0.01 (or 1 – 0.99), zα/2 = 2.58 and E = 2. Substituting in the formula,

Round the value 31.92 up to 32. Therefore, to be 99% confident that the estimate is within 2 feet of the true mean depth, the scientist needs at least a sample of 32 measurements. (Always round n up to the next whole number. For example, if n = 31.2, round it up to 32.)

Interesting Fact

It has been estimated that the amount of pizza consumed every day in the United States would cover a farm consisting of 75 acres.

Notice that when you are finding the sample size, the size of the population is irrelevant when the population is large or infinite or when sampling is done with replacement. In other cases, an adjustment is made in the formula for computing sample size. This adjustment is beyond the scope of this book.

The formula for determining sample size requires the use of the population standard deviation. What happens when σ is unknown? In this case, an attempt is made to estimate σ. One such way is to use the standard deviation s obtained from a sample taken previously as an estimate for σ. The standard deviation can also be estimated by dividing the range by 4.

Sometimes, interval estimates rather than point estimates are reported. For instance, you may read a statement: “On the basis of a sample of 200 families, the survey estimates that an American family of two spends an average of $84 per week for groceries. One can be 95% confident that this estimate is accurate within $3 of the true mean.” This statement means that the 95% confidence interval of the true mean is

$84 – $3 < μ < $84 + $3

$81 < μ < $87

Page 363

The algebraic derivation of the formula for a confidence interval is shown next. As explained in Chapter 6, the sampling distribution of the mean is approximately normal when large samples (n ≥ 30) are taken from a population. Also,

Furthermore, there is a probability of 1 – α that a z will have a value between –zα/2 and +zα/2. Hence,

By using algebra, the formula can be rewritten as

Subtracting from both sides and from the middle gives

Multiplying by – 1 gives

Reversing the inequality yields the formula for the confidence interval:

Applying the Concepts 7–1

Making Decisions with Confidence Intervals

Assume you work for Kimberly Clark Corporation, the makers of Kleenex. The job you are presently working on requires you to decide how many Kleenexes are to be put in the new automobile glove compartment boxes. Complete the following.

1.How will you decide on a reasonable number of Kleenexes to put in the boxes?

2.When do people usually need Kleenexes?

3.What type of data collection technique would you use?

4.Assume you found out that from your sample of 85 people, on average about 57 Kleenexes are used throughout the duration of a cold, with a population standard deviation of 15. Use a confidence interval to help you decide how many Kleenexes will go in the boxes.

5.Explain how you decided how many Kleenexes will go in the boxes.

See page 396 for the answers.

Page 364

Exercises 7–1

1.What is the difference between a point estimate and an interval estimate of a parameter? Which is better? Why?

2.What information is necessary to calculate a confidence interval?

3.What is the maximum error of the estimate?

4.What is meant by the 95% confidence interval of the mean?

5.What are three properties of a good estimator?

6.What statistic best estimates μ?

7.What is necessary to determine the sample size?

8.In determining the sample size for a confidence interval, is the size of the population relevant?

9.Find each.

a.zα/2 for the 99% confidence interval

b.zα/2 for the 98% confidence interval

c.zα/2 for the 95% confidence interval

d.zα/2 for the 90% confidence interval

e.zα/2 for the 94% confidence interval

10. Number of Faculty The numbers of faculty at 32 randomly selected state-controlled colleges and universities with enrollment under 12,000 students are shown below. Use these data to estimate the mean number of faculty at all state-controlled colleges and universities with enrollment under 12,000 with 92% confidence. Assume σ = 165.1.

Source: World Almanac.

11.Reading Scores A sample of the reading scores of 35 fifth-graders has a mean of 82. The standard deviation of the population is 15.

a.Find the best point estimate of the mean.

b.Find the 95% confidence interval of the mean reading scores of all fifth-graders.

c.Find the 99% confidence interval of the mean reading scores of all fifth-graders.

d.Which interval is larger? Explain why.

12.Freshmen’s GPA First-semester GPAs for a random selection of freshmen at a large university are shown below. Estimate the true mean GPA of the freshman class with 99% confidence. Assume σ = 0.62.

13.Workers’ Distractions A recent study showed that the modern working person experiences an average of 2.1 hours per day of distractions (phone calls, e-mails, impromptu visits, etc.). A random sample of 50 workers for a large corporation found that these workers were distracted an average of 1.8 hours per day and the population standard deviation was 20 minutes. Estimate the true mean population distraction time with 90% confidence, and compare your answer to the results of the study.

Source: Time Almanac.

14.Golf Averages A study of 35 golfers showed that their average score on a particular course was 92. The standard deviation of the population is 5.

a.Find the best point estimate of the mean.

b.Find the 95% confidence interval of the mean score for all golfers.

c.Find the 95% confidence interval of the mean score if a sample of 60 golfers is used instead of a sample of 35.

d.Which interval is smaller? Explain why.

15.Actuary Exams A survey of 35 individuals who passed the seven exams and obtained the rank of Fellow in the actuarial field finds the average salary to be $150,000. If the standard deviation for the population is $15,000, construct a 95% confidence interval for all Fellows.

Source: www.BeAnActuary.org

16. Number of Farms A random sample of the number of farms (in thousands) in various states follows. Estimate the mean number of farms per state with 90% confidence. Assume σ = 31.

Source: New York Times Almanac.

17.Television Viewing A study of 415 kindergarten students showed that they have seen on average 5000 hours of television. If the sample standard deviation of the population is 900, find the 95% confidence level of the mean for all students. If a parent claimed that his children watched 4000 hours, would the claim be believable?

Source: U.S. Department of Education.

Page 365

18.Day Care Tuition A random sample of 50 four-year-olds attending day care centers provided a yearly tuition average of $3987 and the population standard deviation of $630. Find the 90% confidence interval of the true mean. If a day care center were starting up and wanted to keep tuition low, what would be a reasonable amount to charge?

19.Hospital Noise Levels Noise levels at various area urban hospitals were measured in decibels. The mean of the noise levels in 84 corridors was 61.2 decibels, and the standard deviation of the population was 7.9. Find the 95% confidence interval of the true mean.

Source: M. Bayo, A. Garcia, and A. Garcia, “Noise Levels in an Urban Hospital and Workers’ Subjective Responses,” Archives of Environmental Health 50, no. 3, p. 249 (May–June 1995). Reprinted with permission of the Helen Dwight Reid Educational Foundation. Published by Heldref Publications, 1319 Eighteenth St. N.W., Washington, D.C. 20036-1802. Copyright © 1995.

20.Length of Growing Seasons The growing seasons for a random sample of 35 U.S. cities were recorded, yielding a sample mean of 190.7 days and the population standard deviation of 54.2 days. Estimate the true mean population of the growing season with 95% confidence.

Source: The Old Farmer’s Almanac.

21.Time on Homework A university dean of students wishes to estimate the average number of hours students spend doing homework per week. The standard deviation from a previous study is 6.2 hours. How large a sample must be selected if he wants to be 99% confident of finding whether the true mean differs from the sample mean by 1.5 hours?

22.In the hospital study cited in Exercise 19, the mean noise level in the 171 ward areas was 58.0 decibels, and the population standard deviation is 4.8. Find the 90% confidence interval of the true mean.

23.Chocolate Chips per Cookie It is desired to estimate the mean number of chocolate chips per cookie for a large national brand. How many cookies would have to be sampled to estimate the true mean number of chips per cookie within 2 chips with 98% confidence? Assume that σ = 10.1 chips.

24.Cost of Pizzas A pizza shop owner wishes to find the 95% confidence interval of the true mean cost of a large plain pizza. How large should the sample be if she wishes to be accurate to within $0.15? A previous study showed that the standard deviation of the price was $0.26.

25.National Accounting Examination If the variance of a national accounting examination is 900, how large a sample is needed to estimate the true mean score within 5 points with 99% confidence?

26.Commuting Times in New York The 90% confidence interval for the mean one-way commuting time in New York City is 37.8 < µ < 38.8 minutes. Construct a 95% confidence interval based on the same data. Which interval provides more information?

Source: www.census.gov

Technology Step by Step

MINITAB

Step by Step

Finding a z Confidence Interval for the Mean

For Example 7–3, find the 90% confidence interval estimate for the mean amount of assets for credit unions in southwestern Pennsylvania.

1.Maximize the worksheet, then enter the data into C1 of a MINITAB worksheet. If sigma is known, skip to step 3.

2.Calculate the standard deviation for the sample. It will be used as an estimate for sigma.

a)Select Calc>Column statistics.

b)Click the option for Standard deviation.

c)Enter C1 Assets for the Input variable and s for Store in:.

Page 366

3.Select Stat>Basic Statistics>1-Sample Z.

4.Select C1 Assets for the Samples in Columns.

5.Click in the box for Standard Deviation and enter s. Leave the box for Test mean empty.

6.Click the [Options] button. In the dialog box make sure the Confidence Level is 90 and the Alternative is not equal.

7.Optional: Click [Graphs], then select Boxplot of data. The boxplot of these data would clearly show the outliers!

8.Click [OK] twice. The results will be displayed in the session window.

One-Sample Z: Assets

TI-83 Plus or TI-84 Plus

Step by Step

Finding a z Confidence Interval for the Mean (Data)

1.Enter the data into L1.

2.Press STAT and move the cursor to TESTS.

3.Press 7 for ZInterval.

4.Move the cursor to Data and press ENTER.

5.Type in the appropriate values.

6.Move the cursor to Calculate and press ENTER.

Example TI7–1

This is Example 7–3 from the text. Find the 90% confidence interval for the population mean, given the data values.

The population standard deviation σ is unknown. Since the sample size is n = 30, you can use the sample standard deviation s as an approximation for σ. After the data values are entered in L1 (step 1 above), press STAT, move the cursor to CALC, press 1 for 1-Var Stats, then press ENTER. The sample standard deviation of 14.40544747 will be one of the statistics listed. Then continue with step 2. At step 5 on the line for σ, press VARS for variables, press 5 for Statistics, press 3 for Sx.

Page 367

The 90% confidence interval is 6.765 < μ < 15.417. The difference between these limits and the ones in Example 7–3 is due to rounding.

Finding a z Confidence Interval for the Mean (Statistics)

1.Press STAT and move the cursor to TESTS.

2.Press 7 for ZInterval.

3.Move the cursor to Stats and press ENTER.

4.Type in the appropriate values.

5.Move the cursor to Calculate and press ENTER.

Example TI7–2

Find the 95% confidence interval for the population mean, given σ = 2, = 23.2, and n = 50.

The 95% confidence interval is 22.6 < μ < 23.8.

Excel

Step by Step

Finding a z Confidence Interval for the Mean

Excel has a procedure to compute the maximum error of the estimate. But it does not compute confidence intervals. However, you may determine confidence intervals for the mean by using the MegaStat Add-in available on your CD. If you have not installed this add-in, do so following the instructions from the Chapter 1 Excel Step by Step.

Example XL7–1

Find the 95% confidence interval for the mean if σ = 11, using this sample:

1.Enter the data into an Excel worksheet.

2.From the toolbar, select Add-Ins, MegaStat>Confidence Intervals/Sample Size.

Note: You may need to open MegaStat from the MegaStat.xls file on your computer’s hard drive.

3.Enter the mean of the data, 32.03.

4.Select z for the standard normal distribution.

5.Enter 11 for the standard deviation and 30 for n, the sample size.

6.Either type in or scroll to 95% for the Confidence Level, then click [OK].

The result of the procedure is shown next.

Confidence interval—mean

95% Confidence level

32.03 Mean

11 Standard deviation

30 n

1.960 z

3.936 Half-width

35.966 Upper confidence limit

28.094 Lower confidence limit

Page 368

Objective 3

Find the confidence interval for the mean when σ is unknown.

7–2Confidence Intervals for the Mean When σ Is Unknown

When σ is known and the sample size is 30 or more, or the population is normally distributed if sample size is less than 30, the confidence interval for the mean can be found by using the z distribution as shown in Section 7–1. However, most of the time, the value of σ is not known, so it must be estimated by using s, namely, the standard deviation of the sample. When s is used, especially when the sample size is small, critical values greater than the values for zα/2 are used in confidence intervals in order to keep the interval at a given level, such as the 95%. These values are taken from the Student t distribution, most often called the t distribution.

To use this method, the samples must be simple random samples, and the population from which the samples were taken must be normally or approximately normally distributed, or the sample size must be 30 or more.

Some important characteristics of the t distribution are described now.

Historical Note

The t distribution was formulated in 1908 by an Irish brewing employee named W. S. Gosset. Gosset was involved in researching new methods of manufacturing ale. Because brewing employees were not allowed to publish results, Gosset published his finding using the pseudonym Student; hence, the t distribution is sometimes called Student’s t distribution.

Characteristics of the t Distribution

The t distribution shares some characteristics of the normal distribution and differs from it in others. The t distribution is similar to the standard normal distribution in these ways:

1.It is bell-shaped.

2.It is symmetric about the mean.

3.The mean, median, and mode are equal to 0 and are located at the center of the distribution.

4.The curve never touches the x axis.

The t distribution differs from the standard normal distribution in the following ways:

1.The variance is greater than 1.

2.The t distribution is actually a family of curves based on the concept of degrees of freedom, which is related to sample size.

3.As the sample size increases, the t distribution approaches the standard normal distribution. See Figure 7–6.

Many statistical distributions use the concept of degrees of freedom, and the formulas for finding the degrees of freedom vary for different statistical tests. The degrees of freedom are the number of values that are free to vary after a sample statistic has been computed, and they tell the researcher which specific curve to use when a distribution consists of a family of curves.

For example, if the mean of 5 values is 10, then 4 of the 5 values are free to vary. But once 4 values are selected, the fifth value must be a specific number to get a sum of 50, since 50 ÷ 5 = 10. Hence, the degrees of freedom are 5 – 1 = 4, and this value tells the researcher which t curve to use.

Figure 7–6

The t Family of Curves

Page 369

The symbol d.f. will be used for degrees of freedom. The degrees of freedom for a confidence interval for the mean are found by subtracting 1 from the sample size. That is, d.f. = n – 1. Note: For some statistical tests used later in this book, the degrees of freedom are not equal to n – 1.

The formula for finding a confidence interval about the mean by using the t distribution is given now.

Formula for a Specific Confidence Interval for the Mean When σ Is Unknown and n < 30

The degrees of freedom are n – 1.

The values for tα/2 are found in Table F in Appendix C. The top row of Table F, labeled Confidence Intervals, is used to get these values. The other two rows, labeled One tail and Two tails, will be explained in Chapter 8 and should not be used here.

Example 7–5 shows how to find the value in Table F for tα/2.

Example 7–5

Find the tα/2 value for a 95% confidence interval when the sample size is 22.

Solution

The d.f. = 22 – 1, or 21. Find 21 in the left column and 95% in the row labeled Confidence Intervals. The intersection where the two meet gives the value for tα/2, which is 2.080. See Figure 7–7.

Figure 7–7

Finding tα/2 for Example 7–5

When d.f. is greater than 30, it may fall between two table values. For example, if d.f. = 68, it falls between 65 and 70. Many textbooks say to use the closest value; for example, 68 is closer to 70 than 65; however, in this textbook a conservative approach is used. In this case, always round down to the nearest table value. In this case, 68 rounds down to 65.

Note: At the bottom of Table F where d.f. is large or ∞, the zα/2 values can be found for specific confidence intervals. The reason is that as the degrees of freedom increase, the t distribution approaches the standard normal distribution.

Examples 7–6 and 7–7 show how to find the confidence interval when you are using the t distribution.

Page 370

Example 7–6

Sleeping Time

Ten randomly selected people were asked how long they slept at night. The mean time was 7.1 hours, and the standard deviation was 0.78 hour. Find the 95% confidence interval of the mean time. Assume the variable is normally distributed.

Source: Based on information in Number Freaking.

Solution

Since σ is unknown and s must replace it, the t distribution (Table F) must be used for the confidence interval. Hence, with 9 degrees of freedom tα/2 = 2.262. The 95% confidence interval can be found by substituting in the formula.

Therefore, one can be 95% confident that the population mean is between 6.54 and 7.66 inches.

Example 7–7

Home Fires Started by Candles

The data represent a sample of the number of home fires started by candles for the past several years. (Data are from the National Fire Protection Association.) Find the 99% confidence interval for the mean number of home fires started by candles each year.

Page 361

Solution

Step 1Find the mean and standard deviation for the data. Use the formulas in Chapter 3 or your calculator. The mean = 7041.4. The standard deviation s = 1610.3.

Step 2Find tα/2 in Table F. Use the 99% confidence interval with d.f. = 6. It is 3.707.

Step 3Substitute in the formula and solve.

One can be 99% confident that the population mean number of home fires started by candles each year is between 4785.2 and 9297.6, based on a sample of home fires occurring over a period of 7 years.

Page 371

Figure 7–8

When to Use the z or t Distribution

Students sometimes have difficulty deciding whether to use zα/2 or tα/2 values when finding confidence intervals for the mean. As stated previously, when σ is known, zα/2 values can be used no matter what the sample size is, as long as the variable is normally distributed or n ≥ 30. When σ is unknown and n ≥ 30, then s can be used in the formula and tα/2 values can be used. Finally, when σ is unknown and n < 30, s is used in the formula and tα/2 values are used, as long as the variable is approximately normally distributed. These rules are summarized in Figure 7–8.

Applying the Concepts 7–2

Sport Drink Decision

Assume you get a new job as a coach for a sports team, and one of your first decisions is to choose the sports drink that the team will use during practices and games. You obtain a Sports Report magazine so you can use your statistical background to help you make the best decision. The following table lists the most popular sports drinks and some important information about each of them. Answer the following questions about the table.

1.Would this be considered a small sample?

2.Compute the mean cost per container, and create a 90% confidence interval about that mean. Do all the costs per container fall inside the confidence interval? If not, which ones do not?

3.Are there any you would consider outliers?

4.How many degrees of freedom are there?

5.If cost is a major factor influencing your decision, would you consider cost per container or cost per serving?

6.List which drink you would recommend and why.

See page 396 for the answers.

Page 372

Exercises 7–2

1.What are the properties of the t distribution?

2.What is meant by degrees of freedom?

3.When should the t distribution be used to find a confidence interval for the mean?

4.(ans) Find the values for each.

a.tα/2 and n = 18 for the 99% confidence interval for the mean

b.tα/2 and n = 23 for the 95% confidence interval for the mean

c.tα/2 and n = 15 for the 98% confidence interval for the mean

d.tα/2 and n = 10 for the 90% confidence interval for the mean

e.tα/2 and n = 20 for the 95% confidence interval for the mean

For Exercises 5 through 20, assume that all variables are approximately normally distributed.

5.Hemoglobin The average hemoglobin reading for a sample of 20 teachers was 16 grams per 100 milliliters, with a sample standard deviation of 2 grams. Find the 99% confidence interval of the true mean.

6. Digital Camera Prices The prices (in dollars) for a particular model of digital camera with 6.0 megapixels and an optical 3X zoom lens are shown below for 10 online retailers. Estimate the true mean price for this particular model with 95% confidence.

7. Women Representatives in State Legislature A state representative wishes to estimate the mean number of women representatives per state legislature. A random sample of 17 states is selected, and the number of women representatives is shown. Based on the sample, what is the point estimate of the mean? Find the 90% confidence interval of the mean population. (Note: The population mean is actually 31.72, or about 32.) Compare this value to the point estimate and the confidence interval. There is something unusual about the data. Describe it and state how it would affect the confidence interval.

8. State Gasoline Taxes A random sample of state gasoline taxes (in cents) is shown here for 12 states. Use the data to estimate the true population mean gasoline tax with 90% confidence. Does your interval contain the national average of 21 cents?

Source: World Almanac.

9.College Wrestler Weights A sample of six college wrestlers had an average weight of 276 pounds with a sample standard deviation of 12 pounds. Find the 90% confidence interval of the true mean weight of all college wrestlers. If a coach claimed that the average weight of the wrestlers on the team was 310, would the claim be believable?

10. Dance Company Students The number of students who belong to the dance company at each of several randomly selected small universities is shown below. Estimate the true population mean size of a university dance company with 99% confidence.

11.Distance Traveled to Work A recent study of 28 employees of XYZ company showed that the mean of the distance they traveled to work was 14.3 miles. The standard deviation of the sample mean was 2 miles. Find the 95% confidence interval of the true mean. If a manager wanted to be sure that most of his employees would not be late, how much time would he suggest they allow for the commute if the average speed were 30 miles per hour?

12.Thunderstorm Speeds A meteorologist who sampled 13 thunderstorms found that the average speed at which they traveled across a certain state was 15 miles per hour. The standard deviation of the sample was 1.7 miles per hour. Find the 99% confidence interval of the mean. If a meteorologist wanted to use the highest speed to predict the times it would take storms to travel across the state in order to issue warnings, what figure would she likely use?

13.Students per Teacher in U.S. Public Schools The national average for the number of students per teacher for all U.S. public schools is 15.9. A random sample of 12 school districts from a moderately populated area showed that the mean number of students per teacher was 19.2 with a variance of 4.41. Estimate the true mean number of students per teacher with 95% confidence. How does your estimate compare with the national average?

Source: World Almanac.

14.Stress Test For a group of 10 men subjected to a stress situation, the mean number of heartbeats per minute was 126, and the standard deviation was 4. Find the 95% confidence interval of the true mean.

15.For the stress test described in Exercise 14, six women had an average heart rate of 115 beats per minute. The standard deviation of the sample was 6 beats. Find the 95% confidence interval of the true mean for the women.

Page 373

16.Hospital Noise Levels For a sample of 24 operating rooms taken in the hospital study mentioned in Exercise 19 in Section 7–1, the mean noise level was 41.6 decibels, and the standard deviation was 7.5. Find the 95% confidence interval of the true mean of the noise levels in the operating rooms.

17. Costs for a 30-Second Spot on Cable Television The approximate costs for a 30-second spot for various cable networks in a random selection of cities are shown below. Estimate the true population mean cost for a 30-second advertisement on cable network with 90% confidence.

Source: www.spotrunner.com

18.Football Player Heart Rates For a group of 22 college football players, the mean heart rate after a morning workout session was 86 beats per minute, and the standard deviation was 5. Find the 90% confidence interval of the true mean for all college football players after a workout session. If a coach did not want to work his team beyond its capacity, what maximum value should he use for the mean number of heartbeats per minute?

19.Grooming Times for Men and Women It has been reported that 20- to 24-year-old men spend an average of 37 minutes per day grooming and 20- to 24-year-old women spend an average of 49 minutes per day grooming. Ask your classmates for their individual grooming time per day (unless you’re an 8:00 A.M. class), and use the data to estimate the true mean grooming time for your school with 95% confidence.

Source: Time magazine, Oct. 2006.

20. Unhealthy Days in Cities The number of unhealthy days based on the AQI (Air Quality Index) for a random sample of metropolitan areas is shown. Construct a 98% confidence interval based on the data.

Source: New York Times Almanac.

Extending the Concepts

21. A one-sided confidence interval can be found for a mean by using

where tα is the value found under the row labeled One tail. Find two one-sided 95% confidence intervals of the population mean for the data shown, and interpret the answers. The data represent the daily revenues in dollars from 20 parking meters in a small municipality.

Technology Step by Step

MINITAB

Step by Step

Find a t Interval for the Mean

For Example 7–7, find the 99% confidence interval for the mean number of home fires started by candles each year.

1.Type the data into C1 of a MINITAB worksheet. Name the column HomeFires.

2.Select Stat>Basic Statistics>1-Sample t.

Page 374

3.Double-click C1 HomeFires for the Samples in Columns.

4.Click on [Options] and be sure the Confidence Level is 99 and the Alternative is not equal.

5.Click [OK] twice.

6.Check for normality:

a)Select Graph>Probability Plot, then Single.

b)Select C1 HomeFires for the variable. The normal plot is concave, a skewed distribution.

In the session window you will see the results. The 99% confidence interval estimate for μ is between 4784.99 and 9784.99. The sample size, mean, standard deviation, and standard error of the mean are also shown.

However, this small sample appears to have a nonnormal population. The interval is less likely to contain the true mean.

One-Sample T: HomeFires

TI-83 Plus or TI-84 Plus

Step by Step

Finding a t Confidence Interval for the Mean (Data)

1.Enter the data into L1.

2.Press STAT and move the cursor to TESTS.

3.Press 8 for TInterval.

4.Move the cursor to Data and press ENTER.

5.Type in the appropriate values.

6.Move the cursor to Calculate and press ENTER.

Finding a t Confidence Interval for the Mean (Statistics)

1.Press STAT and move the cursor to TESTS.

2.Press 8 for TInterval.

3.Move the cursor to Stats and press ENTER.

4.Type in the appropriate values.

5.Move the cursor to Calculate and press ENTER.

Excel

Step by Step

Finding a t Confidence Interval for the Mean

Example XL7–2

Find the 95% confidence interval, using these sample data:

1.Enter the data into an Excel worksheet.

2.From the toolbar, select Add-Ins, MegaStat>Confidence Intervals/Sample Size. Note: You may need to open MegaStat from the MegaStat.xls file on your computer’s hard drive.

3.Enter the mean of the data, 563.2.

Page 375

4.Select t for the t distribution.

5.Enter 87.9 for the standard deviation and 10 for n, the sample size.

6.Either type in or scroll to 95% for the Confidence Level, then click [OK].

The result of the procedure is shown next.

Confidence interval—mean

95% Confidence level

563.2 Mean

87.9 Standard deviation

10 n

2.262 t (d.f. = 9)

62.880 Half-width

626.080 Upper confidence limit

500.320 Lower confidence limit

Objective 4

Find the confidence interval for a proportion.

7–3Confidence Intervals and Sample Size for Proportions

A USA TODAY Snapshots feature stated that 12% of the pleasure boats in the United States were named Serenity. The parameter 12% is called a proportion. It means that of all the pleasure boats in the United States, 12 out of every 100 are named Serenity. A proportion represents a part of a whole. It can be expressed as a fraction, decimal, or percentage. In this case, 12% = 0.12 = or . Proportions can also represent probabilities. In this case, if a pleasure boat is selected at random, the probability that it is called Serenity is 0.12.

Proportions can be obtained from samples or populations. The following symbols will be used.

Symbols Used in Proportion Notation

p = population proportion

(read “p hat”) = sample proportion

For a sample proportion,

where X = number of sample units that possess the characteristics of interest and n = sample size.

For example, in a study, 200 people were asked if they were satisfied with their job or profession; 162 said that they were. In this case, n = 200, X = 162, and = X/n = 162/200 = 0.81. It can be said that for this sample, 0.81, or 81%, of those surveyed were satisfied with their job or profession. The sample proportion is = 0.81.

The proportion of people who did not respond favorably when asked if they were satisfied with their job or profession constituted , where = (n – X)/n. For this survey, = (200 – 162)/200 = 38/200, or 0.19, or 19%.

When and are given in decimals or fractions, + = 1. When and are given in percentages, + = 100%. It follows, then, that = 1 – , or = 1 – , when and are in decimal or fraction form. For the sample survey on job satisfaction, can also be found by using = 1 – , or 1 – 0.81 = 0.19.

Similar reasoning applies to population proportions; that is, p = 1 – q, q = 1 – p, and p + q = 1, when p and q are expressed in decimal or fraction form. When p and q are expressed as percentages, p + q = 100%, p = 100% – q, and q = 100% – p.

Page 376

Example 7–8

Air Conditioned Households

In a recent survey of 150 households, 54 had central air conditioning. Find and , where is the proportion of households that have central air conditioning.

Solution

Since X = 54 and n = 150,

You can also find by using the formula = 1 – . In this case, = 1 – 0.36 = 0.64.

As with means, the statistician, given the sample proportion, tries to estimate the population proportion. Point and interval estimates for a population proportion can be made by using the sample proportion. For a point estimate of p (the population proportion), (the sample proportion) is used. On the basis of the three properties of a good estimator, is unbiased, consistent, and relatively efficient. But as with means, one is not able to decide how good the point estimate of p is. Therefore, statisticians also use an interval estimate for a proportion, and they can assign a probability that the interval will contain the population proportion.

The confidence interval for a particular p is based on the sampling distribution of . When the sample size n is no more than 5% of the population size, the sampling distribution of is approximately normal with a mean of p and a standard deviation of , where q = 1 – p.

Confidence Intervals

To construct a confidence interval about a proportion, you must use the maximum error of the estimate, which is

Confidence intervals about proportions must meet the criteria that n ≥ 5 and n ≥ 5.

Formula for a Specific Confidence Interval for a Proportion

when n and n are each greater than or equal to 5.

Rounding Rule for a Confidence Interval for a Proportion Round off to three decimal places.

Example 7–9

Male Nurses

A sample of 500 nursing applications included 60 from men. Find the 90% confidence interval of the true proportion of men who applied to the nursing program.

Page 377

Solution

Since α = 1 – 0.90 = 0.10 and zα/2 = 1.65, substituting in the formula

when = 60/500 = 0.12 and = 1 – 0.12 = 0.88, you get

Hence, you can be 90% confident that the percentage of applicants who are men is between 9.6 and 14.4%.

When a specific percentage is given, the percentage becomes when it is changed to a decimal. For example, if the problem states that 12% of the applicants were men, then = 0.12.

Example 7–10

Religious Books

A survey of 1721 people found that 15.9% of individuals purchase religious books at a Christian bookstore. Find the 95% confidence interval of the true proportion of people who purchase their religious books at a Christian bookstore.

Source: Baylor University.

Solution

Here = 0.159 (i.e., 15.9%), and = 1 – 0.159 = 0.841. For the 95% confidence interval zα/2 = 1.96.

Hence, you can say with 95% confidence that the true percentage is between 14.2 and 17.6%.

Objective 5

Determine the minimum sample size for finding a confidence interval for a proportion.

Sample Size for Proportions

To find the sample size needed to determine a confidence interval about a proportion, use this formula:

Formula for Minimum Sample Size Needed for Interval Estimate of a Population Proportion

If necessary, round up to obtain a whole number.

Page 378

This formula can be found by solving the maximum error of the estimate value for n in the formula

There are two situations to consider. First, if some approximation of is known (e.g., from a previous study), that value can be used in the formula.

Second, if no approximation of is known, you should use = 0.5. This value will give a sample size sufficiently large to guarantee an accurate prediction, given the confidence interval and the error of estimate. The reason is that when and are each 0.5, the product is at maximum, as shown here.


0.1	0.9	0.09
0.2	0.8	0.16
0.3	0.7	0.21
0.4	0.6	0.24
0.5	0.5	0.25
0.6	0.4	0.24
0.7	0.3	0.21
0.8	0.2	0.16
0.9	0.1	0.09

Example 7–11

Home Computers

A researcher wishes to estimate, with 95% confidence, the proportion of people who own a home computer. A previous study shows that 40% of those interviewed had a computer at home. The researcher wishes to be accurate within 2% of the true proportion. Find the minimum sample size necessary.

Solution

Since zα/2 = 1.96, E = 0.02, = 0.40, and = 0.60, then

which, when rounded up, is 2305 people to interview.

Example 7–12

Car Phone Ownership

The same researcher wishes to estimate the proportion of executives who own a car phone. She wants to be 90% confident and be accurate within 5% of the true proportion. Find the minimum sample size necessary.

Solution

Since there is no prior knowledge of , statisticians assign the values = 0.5 and = 0.5. The sample size obtained by using these values will be large enough to ensure the specified degree of confidence. Hence,

which, when rounded up, is 273 executives to ask.

In determining the sample size, the size of the population is irrelevant. Only the degree of confidence and the maximum error are necessary to make the determination.

Page 379

Speaking of Statistics

Does Success Bring Happiness?

W. C. Fields said, “Start every day off with a smile and get it over with.”

Do you think people are happy because they are successful, or are they successful because they are just happy people? A recent survey conducted by Money magazine showed that 34% of the people surveyed said that they were happy because they were successful; however, 63% said that they were successful because they were happy individuals. The people surveyed had an average household income of $75,000 or more. The margin of error was ±2.5%. Based on the information in this article, what would be the confidence interval for each percent?

Applying the Concepts 7–3

Contracting Influenza

To answer the questions, use the following table describing the percentage of people who reported contracting influenza by gender and race/ethnicity.

Influenza
Characteristic	Percent	(95% CI)
Gender
Men	48.8	(47.1–50.5%)
Women	51.5	(50.2–52.8%)
Race/ethnicity
Caucasian	52.2	(51.1–53.3%)
African American	33.1	(29.5–36.7%)
Hispanic	47.6	(40.9–54.3%)
Other	39.7	(30.8–48.5%)
Total	50.4	(49.3–51.5%)

Forty-nine states and the District of Columbia participated in the study. Weighted means were used. The sample size was 19,774. There were 12,774 women and 7000 men.

1.Explain what (95% CI) means.

2.How large is the error for men reporting influenza?

3.What is the sample size?

4.How does sample size affect the size of the confidence interval?

5.Would the confidence intervals be larger or smaller for a 90% CI, using the same data?

6.Where does the 51.5% under influenza for women fit into its associated 95% CI?

See page 396 for the answers.

Page 380

Exercises 7–3

1.In each case, find and .

a.n = 80 and X = 40

b.n = 200 and X = 90

c.n = 130 and X = 60

d.n = 60 and X = 35

e.n = 95 and X = 43

2.(ans) Find and for each percentage. (Use each percentage for .)

a.15%

b.37%

c.71%

d.51%

e.79%

3.Vacations A U.S. Travel Data Center survey conducted for Better Homes and Gardens of 1500 adults found that 39% said that they would take more vacations this year than last year. Find the 95% confidence interval for the true proportion of adults who said that they will travel more this year.

Source: USA TODAY.

4.Regular Voters in America Thirty-five percent of adult Americans are regular voters. A random sample of 250 adults in a medium-size college town were surveyed, and it was found that 110 were regular voters. Estimate the true proportion of regular voters with 90% confidence and comment on your results.

Source: Time magazine, Oct. 2006.

5.Private Schools The proportion of students in private schools is around 11%. A random sample of 450 students from a wide geographic area indicated that 55 attended private schools. Estimate the true proportion of students attending private schools with 95% confidence. How does your estimate compare to 11%?

Source: National Center for Education Statistics (www.nces.ed.gov).

6.Belief in Haunted Places A random sample of 205 college students was asked if they believed that places could be haunted, and 65 of them responded yes. Estimate the true proportion of college students who believe in the possibility of haunted places with 99% confidence. According to Time magazine, 37% of Americans believe that places can be haunted.

Source: Time magazine, Oct. 2006.

7.Work Interruptions A survey found that out of 200 workers, 168 said they were interrupted three or more times an hour by phone messages, faxes, etc. Find the 90% confidence interval of the population proportion of workers who are interrupted three or more times an hour.

Source: Based on information from USA TODAY Snapshot.

8.Travel to Outer Space A CBS News/New York Times poll found that 329 out of 763 adults said they would travel to outer space in their lifetime, given the chance. Estimate the true proportion of adults who would like to travel to outer space with 92% confidence.

Source: www.pollingreport.com

9.High School Graduates Who Take the SAT The national average for the percentage of high school graduates taking the SAT is 49%, but the state averages vary from a low of 4% to a high of 92%. A random sample of 300 graduating high school seniors was polled across a particular tristate area, and it was found that 195 of them had taken the SAT. Estimate the true proportion of high school graduates in this region who take the SAT with 95% confidence.

Source: World Almanac.

10.Canoe Survey A survey of 50 first-time white-water canoers showed that 23 did not want to repeat the experience. Find the 90% confidence interval of the true proportion of canoers who did not wish to canoe the rapids a second time. If a rafting company wants to distribute brochures for repeat trips, what is the minimum number it should print?

11.DVD Players A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval of the true proportion of families who own at least one DVD player. If another survey in a different location found that the proportion of families who owned at least one DVD player was 0.52, would you consider that the proportion of families in this area was larger than in the area where the original survey was done?

12.Students Who Major in Business It has been reported that 20.4% of incoming freshmen indicate that they will major in business or a related field. A random sample of 400 incoming college freshmen was asked their preference, and 95 replied that they were considering business as a major. Estimate the true proportion of freshman business majors with 98% confidence. Does your interval contain 20.4?

Source: New York Times Almanac.

13.Financial Well-being In a Gallup Poll of 1005 individuals, 452 thought they were worse off financially than a year ago. Find the 95% confidence interval for the true proportion of individuals who feel they are worse off financially.

Source: Gallup Poll.

Page 381

14.Fighting U.S. Hunger In a poll of 1000 likely voters, 560 say that the United States spends too little on fighting hunger at home. Find a 95% confidence interval for the true proportion of voters who feel this way.

Source: Alliance to End Hunger.

15.Vitamins for Women A medical researcher wishes to determine the percentage of females who take vitamins. He wishes to be 99% confident that the estimate is within 2 percentage points of the true proportion. A recent study of 180 females showed that 25% took vitamins.

a.How large should the sample size be?

b.If no estimate of the sample proportion is available, how large should the sample be?

16.Widows A recent study indicated that 29% of the 100 women over age 55 in the study were widows.

a.How large a sample must you take to be 90% confident that the estimate is within 0.05 of the true proportion of women over age 55 who are widows?

b.If no estimate of the sample proportion is available, how large should the sample be?

17.Direct Satellite Television It is believed that 25% of U.S. homes have a direct satellite television receiver. How large a sample is necessary to estimate the true population of homes which do with 95% confidence and within 3 percentage points? How large a sample is necessary if nothing is known about the proportion?

Source: New York Times Almanac.

18.Obesity Obesity is defined as a body mass index (BMI) of 3 kg/m2 or more. A 95% confidence interval for the percentage of U.S. adults aged 20 years and over who were obese was found to be 22.4 to 23.5%. What was the sample size?

Source: National Center for Health Statistics (www.cdc.gov/nchs).

19.Unmarried Americans Nearly one-half of Americans aged 25 to 29 are unmarried. How large a sample is necessary to estimate the true proportion of unmarried Americans in this age group within 2½ percentage points with 90% confidence?

Source: Time magazine, Oct. 2006.

20.Diet Habits A federal report indicated that 27% of children ages 2 to 5 years had a good diet—an increase over previous years. How large a sample is needed to estimate the true proportion of children with good diets within 2% with 95% confidence?

Source: Federal Interagency Forum on Child and Family Statistics, Washington Observer-Reporter.

Extending the Concepts

21.Gun Control If a sample of 600 people is selected and the researcher decides to have a maximum error of the estimate of 4% on the specific proportion who favor gun control, find the degree of confidence. A recent study showed that 50% were in favor of some form of gun control.

22.Survey on Politics In a study, 68% of 1015 adults said that they believe the Republicans favor the rich. If the margin of error was 3 percentage points, what was the confidence interval used for the proportion?

Source: USA TODAY.

Technology Step by Step

MINITAB

Step by Step

Find a Confidence Interval for a Proportion

MINITAB will calculate a confidence interval, given the statistics from a sample or given the raw data. In Example 7–9, in a sample of 500 nursing applications 60 were from men. Find the 90% confidence interval estimate for the true proportion of male applicants.

1.Select Stat>Basic Statistics>1 Proportion.

2.Click on the button for Summarized data. No data will be entered in the worksheet.

3.Click in the box for Number of trials and enter 500.

4.In the Number of events box, enter 60.

5.Click on [Options].

6.Type 90 for the confidence level.

Page 382

7.Check the box for Use test and interval based on normal distribution.

8.Click [OK] twice.

The results for the confidence interval will be displayed in the session window.

Test and CI for One Proportion

TI-83 Plus or TI-84 Plus

Step by Step

Finding a Confidence Interval for a Proportion

1.Press STAT and move the cursor to TESTS.

2.Press A (ALPHA, MATH) for 1-PropZlnt.

3.Type in the appropriate values.

4.Move the cursor to Calculate and press ENTER.

Example TI7–3

Find the 95% confidence interval of p when X = 60 and n = 500, as in Example 7–9.

The 95% confidence level for p is 0.09152 < p < 0.14848. Also p is given.

Excel

Step by Step

Finding a Confidence Interval for a Proportion

Excel has a procedure to compute the maximum error of the estimate. But it does not compute confidence intervals. However, you may determine confidence intervals for a proportion by using the MegaStat Add-in available on your CD. If you have not installed this add-in, do so following the instructions from the Chapter 1 Excel Step by Step.

Example XL7–3

There were 500 nursing applications in a sample, including 60 from men. Find the 90% confidence interval for the true proportion of male applicants.

1.From the toolbar, select Add-Ins, MegaStat>Confidence Intervals/Sample Size. Note: You may need to open MegaStat from the MegaStat.xls file on your computer’s hard drive.

2.In the dialog box, select Confidence interval—p.

3.Enter 60 in the box labeled p; p will automatically change to x.

4.Enter 500 in the box labeled n.

5.Either type in or scroll to 90% for the Confidence Level, then click [OK].

The result of the procedure is shown next.

Confidence interval—proportion

90% Confidence level

0.12 Proportion

500 n

1.645 z

0.024 Half-width

0.144 Upper confidence limit

0.096 Lower confidence limit

Page 383

Speaking of Statistics

Here is a survey about college students’ credit card usage. Suggest several ways that the study could have been more meaningful if confidence intervals had been used.

Objective 6

Find a confidence interval for a variance and a standard deviation.

7–4Confidence Intervals for Variances and Standard Deviations

In Sections 7–1 through 7–3 confidence intervals were calculated for means and proportions. This section will explain how to find confidence intervals for variances and standard deviations. In statistics, the variance and standard deviation of a variable are as important as the mean. For example, when products that fit together (such as pipes) are manufactured, it is important to keep the variations of the diameters of the products as small as possible; otherwise, they will not fit together properly and will have to be scrapped. In the manufacture of medicines, the variance and standard deviation of the medication in the pills play an important role in making sure patients receive the proper dosage. For these reasons, confidence intervals for variances and standard deviations are necessary.

Page 384

Historical Note

The χ2 distribution with 2 degrees of freedom was formulated by a mathematician named Hershel in 1869 while he was studying the accuracy of shooting arrows at a target. Many other mathematicians have since contributed to its development.

To calculate these confidence intervals, a new statistical distribution is needed. It is called the chi-square distribution.

The chi-square variable is similar to the t variable in that its distribution is a family of curves based on the number of degrees of freedom. The symbol for chi-square is χ2 (Greek letter chi, pronounced “ki”). Several of the distributions are shown in Figure 7–9, along with the corresponding degrees of freedom. The chi-square distribution is obtained from the values of (n – 1)s2/σ2 when random samples are selected from a normally distributed population whose variance is σ2.

A chi-square variable cannot be negative, and the distributions are skewed to the right. At about 100 degrees of freedom, the chi-square distribution becomes somewhat symmetric. The area under each chi-square distribution is equal to 1.00, or 100%.

Table G in Appendix C gives the values for the chi-square distribution. These values are used in the denominators of the formulas for confidence intervals. Two different values are used in the formula because the distribution is not symmetric. One value is found on the left side of the table, and the other is on the right. See Figure 7–10. For example, to find the table values corresponding to the 95% confidence interval, you must first change 95% to a decimal and subtract it from 1 (1 – 0.95 = 0.05). Then divide the answer by 2 (α/2 = 0.05/2 = 0.025). This is the column on the right side of the table, used to get the values for χ2right. To get the value for χ2left, subtract the value of α/2 from 1 (1 – 0.05/2 = 0. 975). Finally, find the appropriate row corresponding to the degrees of freedom n – 1. A similar procedure is used to find the values for a 90 or 99% confidence interval.

Figure 7–9

The Chi-Square Family of Curves

Page 385

Figure 7–10

Chi-Square Distribution for d.f. = n – 1

Example 7–13

Find the values for χ2right and χ2left for a 90% confidence interval when n = 25.

Solution

To find χ2right, subtract 1 – 0.90 = 0.10 and divide by 2 to get 0.05.

To find χ2left, subtract 1 – 0.05 to get 0.95. Hence, use the 0.95 and 0.05 columns and the row corresponding to 24 d.f. See Figure 7–11.

Figure 7–11

χ2 Table for Example 7–13

The answers are

χ2right = 36415

χ2left = 13.848

See Figure 7–12.

Page 386

Figure 7–12

χ2 Distribution for Example 7–13

Useful estimates for σ2 and σ are s2 and s, respectively.

To find confidence intervals for variances and standard deviations, you must assume that the variable is normally distributed.

The formulas for the confidence intervals are shown here.

Formula for the Confidence Interval for a Variance

d.f. = n – 1

Formula for the Confidence Interval for a Standard Deviation

d.f. = n – 1

Recall that s2 is the symbol for the sample variance and s is the symbol for the sample standard deviation. If the problem gives the sample standard deviation s, be sure to square it when you are using the formula. But if the problem gives the sample variance s2, do not square it when you are using the formula, since the variance is already in square units.

Rounding Rule for a Confidence Interval for a Variance or Standard Deviation When you are computing a confidence interval for a population variance or standard deviation by using raw data, round off to one more decimal place than the number of decimal places in the original data.

When you are computing a confidence interval for a population variance or standard deviation by using a sample variance or standard deviation, round off to the same number of decimal places as given for the sample variance or standard deviation.

Example 7–14 shows how to find a confidence interval for a variance and standard deviation.

Page 387

Example 7–14

Nicotine Content

Find the 95% confidence interval for the variance and standard deviation of the nicotine content of cigarettes manufactured if a sample of 20 cigarettes has a standard deviation of 1.6 milligrams.

Solution

Since α = 0.05, the two critical values, respectively, for the 0.025 and 0.975 levels for 19 degrees of freedom are 32.852 and 8.907. The 95% confidence interval for the variance is found by substituting in the formula.

Hence, you can be 95% confident that the true variance for the nicotine content is between 1.5 and 5.5.

For the standard deviation, the confidence interval is

Hence, you can be 95% confident that the true standard deviation for the nicotine content of all cigarettes manufactured is between 1.2 and 2.3 milligrams based on a sample of 20 cigarettes.

Example 7–15

Cost of Ski Lift Tickets

Find the 90% confidence interval for the variance and standard deviation for the price in dollars of an adult single-day ski lift ticket. The data represent a selected sample of nationwide ski resorts. Assume the variable is normally distributed.

Source: USA TODAY.

Solution

Step 1Find the variance for the data. Use the formulas in Chapter 3 or your calculator. The variance s2 = 28.2.

Step 2Find χ2right and χ2left from Table G in Appendix C. Since α = 0.10, the two critical values are 3.325 and 16.919, using d.f. = 9 and 0.95 and 0.05.

Step 3Substitute in the formula and solve.

Page 388

For the standard deviation

Hence you can be 90% confident that the standard deviation for the price of all single-day ski lift tickets of the population is between $3.87 and $8.73 based on a sample of 10 nationwide ski resorts. (Two decimal places are used since the data are in dollars and cents.)

Note: If you are using the standard deviation instead (as in Example 7–14) of the variance, be sure to square the standard deviation when substituting in the formula.

Applying the Concepts 7–4

Confidence Interval for Standard Deviation

Shown are the ages (in years) of the Presidents at the time of their deaths.

1.Do the data represent a population or a sample?

2.Select a random sample of 12 ages and find the variance and standard deviation.

3.Find the 95% confidence interval of the standard deviation.

4.Find the standard deviation of all the data values.

5.Does the confidence interval calculated in question 3 contain the mean?

6.If it does not, give a reason why.

7.What assumption(s) must be considered for constructing the confidence interval in step 3?

See page 396 for the answers.

Exercises 7–4

1.What distribution must be used when computing confidence intervals for variances and standard deviations?

2.What assumption must be made when computing confidence intervals for variances and standard deviations?

3.Using Table G, find the values for χ2left and χ2right.

a.α = 0.05, n = 12

b.α = 0.10, n = 20

c.α = 0.05, n = 27

d.α = 0.01, n = 6

e.α = 0.10, n = 41

4.Lifetimes of Disposable Cameras Find the 90% confidence interval for the variance and standard deviation for the lifetime of disposable camera batteries if a sample of 16 disposable camera batteries has a standard deviation of 2.1 months. Assume the variable is normally distributed. Do you feel that the lifetimes of the batteries are relatively consistent?

5. Carbohydrates in Yogurt The number of carbohydrates (in grams) per 8-ounce serving of yogurt for each of a random selection of brands is listed below. Estimate the true population variance and standard deviation for the number of carbohydrates per 8-ounce serving of yogurt with 95% confidence.

6.Weights of Containers Find the 99% confidence interval for the variance and standard deviation of the weights of 25 one-gallon containers of motor oil if a sample of 14 containers has a variance of 3.2. The weights are given in ounces. Assume the variable is normally distributed.

Page 389

7.Cost of Knee Replacement Surgery U.S. insurers’ costs for knee replacement surgery range from $17,627 to $25,462. Estimate the population variance (standard deviation) in cost with 98% confidence based on a random sample of 10 persons who have had this surgery. The retail costs (for uninsured persons) for the same procedure range from $40,640 to $58,702. Estimate the population variance and standard deviation in cost with 98% confidence based on a sample of 10 persons, and compare your two intervals.

Source: Time Almanac.

8.Age of College Students Find the 90% confidence interval for the variance and standard deviation of the ages of seniors at Oak Park College if a sample of 24 students has a standard deviation of 2.3 years. Assume the variable is normally distributed.

9. New-Car Lease Fees A new-car dealer is leasing various brand-new models for the monthly rates (in dollars) listed below. Estimate the true population variance (and standard deviation) in leasing rates with 90% confidence.

10. Stock Prices A random sample of stock prices per share (in dollars) is shown. Find the 90% confidence interval for the variance and standard deviation for the prices. Assume the variable is normally distributed.

Source: Pittsburgh Tribune Review.

11.Oil Changes A service station advertises that customers will have to wait no more than 30 minutes for an oil change. A sample of 28 oil changes has a standard deviation of 5.2 minutes. Find the 95% confidence interval of the population standard deviation of the time spent waiting for an oil change.

12. Home Ownership Rates The percentage rates of home ownership for 8 randomly selected states are listed below. Estimate the population variance and standard deviation for the percentage rate of home ownership with 99% confidence.

Source: World Almanac.

Extending the Concepts

13.Calculator Battery Lifetimes A confidence interval for a standard deviation for large samples taken from a normally distributed population can be approximated by Find the 95% confidence interval for the population standard deviation of calculator batteries. A sample of 200 calculator batteries has a standard deviation of 18 months.

Technology Step by Step

TI-83 Plus or TI-84 Plus

Step by Step

The TI-83 Plus and TI-84 Plus do not have a built-in confidence interval for the variance or standard deviation. However, the downloadable program named SDINT is available on your CD and Online Learning Center. Follow the instructions with your CD for downloading the program.

Finding a Confidence Interval for the Variance and Standard Deviation (Data)

1.Enter the data values into L1

2.Press PRGM, move the cursor to the program named SDINT, and press ENTER twice.

3.Press 1 for Data.

4.Type L1 for the list and press ENTER.

5.Type the confidence level and press ENTER.

6.Press ENTER to clear the screen.

Page 390

Example TI7–4

This refers to Example 7–15 in the text. Find the 90% confidence interval for the variance and standard deviation for the data:

Finding a Confidence Interval for the Variance and Standard Deviation (Statistics)

1.Press PRGM, move the cursor to the program named SDINT, and press ENTER twice.

2.Press 2 for Stats.

3.Type the sample standard deviation and press ENTER.

4.Type the sample size and press ENTER.

5.Type the confidence level and press ENTER.

6.Press ENTER to clear the screen.

Example TI7–5

This refers to Example 7–14 in the text. Find the 95% confidence interval for the variance and standard deviation, given n = 20 and s = 1.6.

Summary

An important aspect of inferential statistics is estimation. Estimations of parameters of populations are accomplished by selecting a random sample from that population and choosing and computing a statistic that is the best estimator of the parameter. A good estimator must be unbiased, consistent, and relatively efficient. The best estimators of µ and p are and , respectively. The best estimators of σ2 and σ are s2 and s, respectively.

There are two types of estimates of a parameter: point estimates and interval estimates. A point estimate is a specific value. For example, if a researcher wishes to estimate the average length of a certain adult fish, a sample of the fish is selected and measured. The mean of this sample is computed, for example, 3.2 centimeters. From this sample mean, the researcher estimates the population mean to be 3.2 centimeters.

The problem with point estimates is that the accuracy of the estimate cannot be determined. For this reason, statisticians prefer to use the interval estimate. By computing an interval about the sample value, statisticians can be 95 or 99% (or some other percentage) confident that their estimate contains the true parameter. The confidence level is determined by the researcher. The higher the confidence level, the wider the interval of the estimate must be. For example, a 95% confidence interval of the true mean length of a certain species of fish might be

3.17 < µ < 3.23

Page 391

whereas the 99% confidence interval might be

3.15 < µ < 3.25

When the confidence interval of the mean is computed, the z or t values are used, depending on whether the population standard deviation is known. If σ is known, the z values can be used. If σ is not known, the t values are used. When the sample size is less than 30, the population should be normally distributed.

Closely related to computing confidence intervals is the determination of the sample size to make an estimate of the mean. This information is needed to determine the minimum sample size necessary.

1.The degree of confidence must be stated.

2.The population standard deviation must be known or be able to be estimated.

3.The maximum error of the estimate must be stated.

Confidence intervals and sample sizes can also be computed for proportions, using the normal distribution; and confidence intervals for variances and standard deviations can be computed, using the chi-square distribution. A confidence interval is given as point estimate ± the maximum error of the estimate.

Important Terms

chi-square distribution

maximum error of the estimate

point estimate

proportion

relatively efficient estimator

t distribution

unbiased estimator

Important Formulas

Formula for the confidence interval of the mean when σ is known (when n ≥ 30, s can be used if σ is unknown):

Formula for the sample size for means:

where E is the maximum error.

Formula for the confidence interval of the mean when σ is unknown:

Formula for the confidence interval for a proportion:

where = X/n and = 1 – .

Formula for the sample size for proportions:

Formula for the confidence interval for a variance:

Formula for confidence interval for a standard deviation:

Page 392

Review Exercises

1. Eight chemical elements do not have isotopes (different forms of the same element having the same atomic number but different atomic weights). A random sample of 30 of the elements which do have isotopes showed a mean number of 19.63 isotopes per element and the population a standard deviation of 18.73. Estimate the true mean number of isotopes for all elements with isotopes with 90% confidence.

Source: Time Almanac.

2.Presidential Travel In a survey of 1004 individuals, 442 felt that President George W. Bush spent too much time away from Washington. Find a 95% confidence interval for the true population proportion.

Source: USA TODAY/CNN/Gallup Poll.

3. Vacation Days A U.S. Travel Data Center survey reported that Americans stayed an average of 7.5 nights when they went on vacation. The sample size was 1500. Find a point estimate of the population mean. Find the 95% confidence interval of the true mean. Assume the population standard deviation was 0.8.

Source: USA TODAY.

4. Lengths of Children’s Animated Films The lengths (in minutes) of a random selection of popular children’s animated films are listed below. Estimate the true mean length of all children’s animated films with 95% confidence.

5. Dog Bites to Postal Workers For a certain urban area, in a sample of 5 months, on average 28 mail carriers were bitten by dogs each month. The standard deviation of the sample was 3. Find the 90% confidence interval of the true mean number of mail carriers who are bitten by dogs each month. Assume the variable is normally distributed.

6.Teachers’Salaries A researcher is interested in estimating the average salary of teachers in a large urban school district. She wants to be 95% confident that her estimate is correct. If the standard deviation is $1050, how large a sample is needed to be accurate within $200?

7. Postage Costs A researcher wishes to estimate, within $25, the true average amount of postage a community college spends each year. If she wishes to be 90% confident, how large a sample is necessary? The standard deviation is known to be $80.

8.Vacation Sites A U.S. Travel Data Center’s survey of 1500 adults found that 42% of respondents stated that they favor historical sites as vacations. Find the 95% confidence interval of the true proportion of all adults who favor visiting historical sites as vacations.

Source: USA TODAY.

9. Snow Removal Survey In a recent study of 75 people, 41 said they were dissatisfied with their community’s snow removal service. Find the 95% confidence interval of the true proportion of individuals who are dissatisfied with their community’s snow removal service. Based on the results, should the supervisor consider making improvements in the snow removal service?

10.A local county has a very active adult education venue. A random sample of the population showed that 189 out of 400 persons 16 years old or older participated in some type of formal adult education activities, such as basic skills training, apprenticeships, personal interest courses, and part-time college or university degree programs. Estimate the true proportion of adults participating in some kind of formal education program with 98% confidence.

11. Health Insurance Coverage for Children A federal report stated that 88% of children under age 18 were covered by health insurance in 2000. How large a sample is needed to estimate the true proportion of covered children with 90% confidence with a confidence interval 0.05 wide?

Source: Washington Observer-Reporter.

12.Child Care Programs A study found that 73% of prekindergarten children ages 3 to 5 whose mothers had a bachelor’s degree or higher were enrolled in center- based early childhood care and education programs. How large a sample is needed to estimate the true proportion within 3 percentage points with 95% confidence? How large a sample is needed if you had no prior knowledge of the proportion?

13. Baseball Diameters The standard deviation of the diameter of 18 baseballs was 0.29 cm. Find the 95% confidence interval of the true standard deviation of the diameters of the baseballs. Do you think the manufacturing process should be checked for inconsistency?

14.MPG for Lawn Mowers A random sample of 22 lawn mowers was selected, and the motors were tested to see how many miles per gallon of gasoline each one obtained. The variance of the measurements was 2.6. Find the 95% confidence interval of the true variance.

15. Lifetimes of Snowmobiles A random sample of 15 snowmobiles was selected, and the lifetime (in months) of the batteries was measured. The variance of the sample was 8.6. Find the 90% confidence interval of the true variance.

16.Length of Children’s Animated Films Use the data from Exercise 4 to estimate the population variance (standard deviation) in length of children’s animated films with 99% confidence.

Page 393

Statistics Today

Would You Change the Channel?—Revisited

The estimates given in the survey are point estimates. However, since the margin of error is stated to be 3 percentage points, an interval estimate can easily be obtained. For example, if 45% of the people changed the channel, then the confidence interval of the true percentages of people who changed channels would be 42% < p < 48%. The article fails to state whether a 90%, 95%, or some other percentage was used for the confidence interval.

Using the formula given in Section 7–3, a minimum sample size of 1068 would be needed to obtain a 95% confidence interval for p, as shown. Use and as 0.5, since no value is known for .

Data Projects

The Data Bank is found in Appendix D , or on the World Wide Web by following links from www.mhhe.com/bluman/.

1.From the Data Bank choose a variable, find the mean, and construct the 95 and 99% confidence intervals of the population mean. Use a sample of at least 30 subjects. Find the mean of the population, and determine whether it falls within the confidence interval.

2.Repeat Exercise 1, using a different variable and a sample of 15.

3.Repeat Exercise 1, using a proportion. For example, construct a confidence interval for the proportion of individuals who did not complete high school.

4.From Data Set III in Appendix D, select a sample of 30 values and construct the 95 and 99% confidence intervals of the mean length in miles of major North American rivers. Find the mean of all the values, and determine if the confidence intervals contain the mean.

5.From Data Set VI in Appendix D, select a sample of 20 values and find the 90% confidence interval of the mean of the number of acres. Find the mean of all the values, and determine if the confidence interval contains the mean.

6.Select a random sample of 20 of the record high temperatures in the United States, found in Data Set I in Appendix D. Find the proportion of temperatures below 110°. Construct a 95% confidence interval for this proportion. Then find the true proportion of temperatures below 110°, using all the data. Is the true proportion contained in the confidence interval? Explain.

Chapter Quiz

Determine whether each statement is true or false. If the statement is false, explain why.

1. Interval estimates are preferred over point estimates since a confidence level can be specified.

2. For a specific confidence interval, the larger the sample size, the smaller the maximum error of the estimate will be.

3. An estimator is consistent if, as the sample size decreases, the value of the estimator approaches the value of the parameter estimated.

4. To determine the sample size needed to estimate a parameter, you must know the maximum error of the estimate.

Select the best answer.

5. When a 99% confidence interval is calculated instead of a 95% confidence interval with n being the same, the maximum error of estimate will be

a.Smaller

b.Larger

c.The same

d.It cannot be determined.

Page 394

6. The best point estimate of the population mean is

a.The sample mean

b.The sample median

c.The sample mode

d.The sample midrange

7. When the population standard deviation is unknown and the sample size is less than 30, what table value should be used in computing a confidence interval for a mean?

a.z

b.t

c.Chi-square

d.None of the above

Complete the following statements with the best answer.

8. A good estimator should be _____, _____, and _____.

9. The maximum difference between the point estimate of a parameter and the actual value of the parameter is called _____.

10. The statement “The average height of an adult male is 5 feet 10 inches” is an example of a(n) _____ estimate.

11. The three confidence intervals used most often are the _____%, _____%, and _____%.

12. Shopping Survey A random sample of 49 shoppers showed that they spend an average of $23.45 per visit at the Saturday Mornings Bookstore. The standard deviation of the population is $2.80. Find a point estimate of the population mean. Find the 90% confidence interval of the true mean.

13. Doctor Visit Costs An irate patient complained that the cost of a doctor’s visit was too high. She randomly surveyed 20 other patients and found that the mean amount of money they spent on each doctor’s visit was $44.80. The standard deviation of the sample was $3.53. Find a point estimate of the population mean. Find the 95% confidence interval of the population mean. Assume the variable is normally distributed.

14. Weights of Minivans The average weight of 40 randomly selected minivans was 4150 pounds. The standard deviation was 480 pounds. Find a point estimate of the population mean. Find the 99% confidence interval of the true mean weight of the minivans.

15. Ages of Insurance Representatives In a study of 10 insurance sales representatives from a certain large city, the average age of the group was 48.6 years and the standard deviation was 4.1 years. Assume the variable is normally distributed. Find the 95% confidence interval of the population mean age of all insurance sales representatives in that city.

16. Patients Treated in Hospital Emergency Rooms In a hospital, a sample of 8 weeks was selected, and it was found that an average of 438 patients was treated in the emergency room each week. The standard deviation was 16. Find the 99% confidence interval of the true mean. Assume the variable is normally distributed.

17. Burglaries For a certain urban area, it was found that in a sample of 4 months, an average of 31 burglaries occurred each month. The standard deviation was 4. Assume the variable is normally distributed. Find the 90% confidence interval of the true mean number of burglaries each month.

18. Hours Spent Studying A university dean wishes to estimate the average number of hours that freshmen study each week. The standard deviation from a previous study is 2.6 hours. How large a sample must be selected if he wants to be 99% confident of finding whether the true mean differs from the sample mean by 0.5 hour?

19. Money Spent on Road Repairs A researcher wishes to estimate within $300 the true average amount of money a county spends on road repairs each year. If she wants to be 90% confident, how large a sample is necessary? The standard deviation is known to be $900.

20. Bus Ridership A recent study of 75 workers found that 53 people rode the bus to work each day. Find the 95% confidence interval of the proportion of all workers who rode the bus to work.

21. Emergency Room Accidents In a study of 150 accidents that required treatment in an emergency room, 36% involved children under 6 years of age. Find the 90% confidence interval of the true proportion of accidents that involve children under the age of 6.

22. Television Set Ownership A survey of 90 families showed that 40 owned at least one television set. Find the 95% confidence interval of the true proportion of families who own at least one television set.

23. Skipping Lunch A nutritionist wishes to determine, within 3%, the true proportion of adults who do not eat any lunch. If he wishes to be 95% confident that his estimate contains the population proportion, how large a sample will be necessary? A previous study found that 15% of the 125 people surveyed said they did not eat lunch.

24. Novel Pages A sample of 25 novels has a standard deviation of 9 pages. Find the 95% confidence interval of the population standard deviation.

25. Truck Safety Check Find the 90% confidence interval for the variance and standard deviation for the time it takes a state police inspector to check a truck for safety if a sample of 27 trucks has a standard deviation of 6.8 minutes. Assume the variable is normally distributed.

26. Automobile Pollution A sample of 20 automobiles has a pollution by-product release standard deviation of 2.3 ounces when 1 gallon of gasoline is used. Find the 90% confidence interval of the population standard deviation.

Page 395

Critical Thinking Challenges

A confidence interval for a median can be found by using these formulas

to define positions in the set of ordered data values.

Suppose a data set has 30 values, and you want to find the 95% confidence interval for the median. Substituting in the formulas, you get

when n = 30 and zα/2 = 1.96.

Arrange the data in order from smallest to largest, and then select the 10th and 21st values of the data array; hence, X10 < median < X21.

Find the 90% confidence interval for the median for the given data.

Data Projects

1.Business and Finance Use 30 stocks classified as the Dow Jones industrials as the sample. Note the amount each stock has gained or lost in the last quarter. Compute the mean and standard deviation for the data set. Compute the 95% confidence interval for the mean and the 95% confidence interval for the standard deviation. Compute the percentage of stocks that had a gain in the last quarter. Find a 95% confidence interval for the percentage of stocks with a gain.

2.Sports and Leisure Use the top home run hitter from each major league baseball team as the data set. Find the mean and the standard deviation for the number of home runs hit by the top hitter on each team. Find a 95% confidence interval for the mean number of home runs hit.

3.Technology Use the data collected in data project 3 of Chapter 2 regarding song lengths. Select a specific genre and compute the percentage of songs in the sample that are of that genre. Create a 95% confidence interval for the true percentage. Use the entire music library and find the population percentage of the library with that genre. Does the population percentage fall within the confidence interval?

4.Health and Wellness Use your class as the sample. Have each student take her or his temperature on a healthy day. Compute the mean and standard deviation for the sample. Create a 95% confidence interval for the mean temperature. Does the confidence interval obtained support the long-held belief that the average body temperature is 98.6°F?

5.Politics and Economics Select five political polls and note the margin of error, sample size, and percent favoring the candidate for each. For each poll, determine the level of confidence that must have been used to obtain the margin of error given, knowing the percent favoring the candidate and number of participants. Is there a pattern that emerges?

6.Your Class Have each student compute his or her body mass index (BMI) (703 times weight in pounds, divided by the quantity height in inches squared). Find the mean and standard deviation for the data set. Compute a 95% confidence interval for the mean BMI of a student. A BMI score over 30 is considered obese. Does the confidence interval indicate that the mean for BMI could be in the obese range?

Page 396

Answers to Applying the Concepts

Section 7–1 Making Decisions with Confidence Intervals

1.Answers will vary. One possible answer is to find out the average number of Kleenexes that a group of randomly selected individuals use in a 2-week period.

2.People usually need Kleenexes when they have a cold or when their allergies are acting up.

3.If we want to concentrate on the number of Kleenexes used when people have colds, we select a random sample of people with colds and have them keep a record of how many Kleenexes they use during their colds.

4.Answers may vary. I will use a 95% confidence interval:

I am 95% confident that the interval 53.8–60.2 contains the true mean number of Kleenexes used by people when they have colds. It seems reasonable to put 60 Kleenexes in the new automobile glove compartment boxes.

5.Answers will vary. Since I am 95% confident that the interval contains the true average, any number of Kleenexes between 54 and 60 would be reasonable. Sixty seemed to be the most reasonable answer, since it is close to 2 standard deviations above the sample mean.

Section 7–2 Sport Drink Decision

1.Answers will vary. One possible answer is that this is a small sample since we are only looking at seven popular sport drinks.

2.The mean cost per container is $1.25, with standard deviation of $0.39. The 90% confidence interval is or 0.96 < μ < 1.54

The 10-K, All Sport, Exceed, and Hydra Fuel all fall outside of the confidence interval.

3.None of the values appear to be outliers.

4.There are 7 – 1 = 6 degrees of freedom.

5.Cost per serving would impact my decision on purchasing a sport drink, since this would allow me to compare the costs on an equal scale.

6.Answers will vary.

Section 7–3 Contracting Influenza

1.(95% CI) means that these are the 95% confidence intervals constructed from the data.

2.The margin of error for men reporting influenza is .

3.The total sample size was 19,774.

4.The larger the sample size, the smaller the margin of error (all other things being held constant).

5.A 90% confidence interval would be narrower (smaller) than a 95% confidence interval, since we need to include fewer values in the interval.

6.The 51.5% is the middle of the confidence interval, since it is the point estimate for the confidence interval.

Section 7–4 Confidence Interval for Standard Deviation

1.The data represent a population, since we have the age at death for all deceased Presidents (at the time of the writing of this book).

2.Answers will vary. One possible sample is 56, 67, 53, 46, 63, 77, 63, 57, 71, 57, 80, 65, which results in a standard deviation of 9.9 years and a variance of 98.0.

3.Answers will vary. The 95% confidence interval for the standard deviation is to . In this case we have to , or 7.0 to 16.8 years.

4.The standard deviation for all the data values is 11.6 years.

5.Answers will vary. Yes, the confidence interval does contain the population standard deviation.

6.Answers will vary.

7.We need to assume that the distribution of ages at death is normal.