statistic final
alnmar
final_review_fall2014.docReview Sheet for Math 273 Final Exam
Chapter 6: Confidence intervals
1. The length of the work commute time for a sample of 32 people has a mean of 9.5. Construct and interpret the 99% confidence interval for the mean. Assume a population standard deviation of 7.10.
Type of Interval: (z or t)
(if appropriate) df:
Correct critical value: 2.575
Margin for error:
Confidence Interval:
2. In a random sample of 15 CD players brought in for repair, the average repair cost was $80. Construct a 90% confidence interval for (. Assume the repair costs are normally distributed. And that the population standard deviation is $14
Type of Interval: (z or t)
(if appropriate) df:
Correct critical value: 1.645
Margin for error:
Confidence Interval:
3. The waking times of 40 people who start work at 8:00 AM (in minutes past 5:00 AM) were collected. The mean of the sample is 103.5 and the standared deviation is 34.66. Construct and interpret the 95% confidence interval for the mean waking time.
Type of Interval: (z or t)
(if appropriate) df:
Correct critical value: 2.023
Margin for error:
Confidence Interval:
4. In a random sample of eight people with advanced degrees in biology, the mean monthly income was $4744 and the standard deviation was $580. Assume the monthly incomes are normally distributed and use a t-distribution to construct a 95% confidence interval for the population mean monthly income for people with advanced degrees in biology.
Type of Interval: (z or t)
(if appropriate) df:
Correct critical value: 2.365
Margin for error:
Confidence Interval:
5. A survey of 957 adults found that 668 said they approve of labor unions. Find a 95% confidence interval for the proportion of US adults who say that they approve labor unions.
Type of Interval: (z or t)
(if appropriate) df:
Correct critical value: 1.960
Margin for error:
Confidence Interval:
6. A survey of 1011 adults found that 597 said they follow professional football. Find a 90% confidence interval for the proportion of US adults who say that they follow professional football.
Type of Interval: (z or t)
(if appropriate) df:
Correct critical value: 1.645
Margin for error:
Confidence Interval:
Chapter 7: Hypothesis testing
7. Use a t-test to test the claim
12,700
m
>
at a significance level of0.05
a
=
. The sample statistics are12,804
x
=
,248
s
=
,21
n
=
.Hypotheses: H0
Ha Correct p-value = 0.0346
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
8. A fitness magazine advertises that the average monthly cost of joining a health club is $25. You work for a consumer advocacy group and are asked to test the claim that the advertisement is incorrect. You find that a random sample of 18 clubs has a mean monthly cost of $26.25 and a standard deviation of $3.23. At
0.10
a
=
, do you have enough evidence to support the claim.Hypotheses: H0
Ha Correct p-value = 0.1190
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
9. A nutritional health agency claims that a restaurant chain’s hamburgers have more than 10 grams of fat. You work for the nutritional health agency and are asked to test this claim. You find that a random sample of nine hamburgers has a mean fat content of 13.5 grams and a standard deviation of 5.8 grams. At
0.10
a
=
, do you have enough evidence to support the claim.Hypotheses: H0
Ha Correct p-value = 0.0539
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
10. Use a z-test to test the claim
0.70
p
<
, at a significance level of0.01
a
=
. The sample statistics areˆ
0.50
p
=
,68
n
=
.Hypotheses: H0
Ha Correct p-value = 0.0002
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
11. A polling agency claims that over 75% of the people in the United States prepare and file their income taxes before April 15th. In a random survey in 1036 people in the United States, 818 said they prepare and file their income taxes before April 15th. Test the agency’s claim at the
0.10
a
=
level. What can you conclude?Hypotheses: H0
Ha Correct p-value = 0.0016
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
Chapter 8: Hypothesis testing with two samples:
12. A claim for a hypothesis test using two samples is
12
mm
<
. Using a significance level of0.10
a
=
to test this hypothesis. The sample statistics are1
0.28
x
=
,1
0.11
s
=
,1
41
n
=
,2
0.33
x
=
,2
0.10
s
=
,2
34
n
=
. Assume the population variances are the sameHypotheses: H0
Ha Correct p-value = 0.0224
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
13. In a fast food study, a nutritionist finds that the mean calorie content of 36 randomly selected Long John Silver’s fish sandwiches is 430 calories with a standard deviation of 43 calories. The mean calorie content of 41 randomly selected McDonald’s fish sandwiches is 410 calories with a standard deviation of 57 calories. At
0.05
a
=
, is there enough evidence for the nutritionist to conclude that the McDonald’s sandwich has fewer calories than the Long John Silver’s sandwich? Assume the population variances are the sameHypotheses: H0
Ha Correct p-value = 0.0449
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
14. A real estate agent claims that there is a difference between the mean household incomes of two neighborhoods. The mean income of 12 randomly selected households from the first neighborhood was $18,250 with a standard deviation of $1200. In the second neighborhood, 10 randomly selected households had a mean income of $17,500 with a standard deviation of $950. Assume normal distributions and equal population variances. Test the claim at
0.05
a
=
.Hypotheses: H0
Ha Correct p-value = 0.1253
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
15. Test the claim
2
1
m
m
<
at the significance level0.01
a
=
. The sample statistics are1
25.6
x
=
,2
8.25
s
=
,1
15
n
=
,2
22.4
x
=
,2
7.85
s
=
,2
34
n
=
. Assume the population variances are not the sameHypotheses: H0
Ha Correct p-value = 0.1800
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
16. A study of fast food nutrition compared the caloric content of french-fries. Thirty-eight randomly selected servings of Burger King medium French fries had a mean 360 calories and a standard deviation of 50 calories, and 35 randomly selected servings of Wendy’s medium French fries had a mean of 390 calories and a standard deviation of 45 calories. At
0.10
a
=
, can you support the claim that the caloric contents of the two types of French fries are different? Assume the population variances are not the sameHypotheses: H0
Ha Correct p-value = 0.0108
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
17. A team of heart surgeons at Saint Ann’s Hospital knows that many patients who undergo corrective heart surgery have a dangerous buildup of anxiety before their scheduled operations. The staff psychiatrist at the hospital has started a new counseling program intended to reduce this anxiety. A test of anxiety is given to patients who know they must undergo heart surgery. Then each patient participates in a series of counseling sessions with the staff psychiatrist. At the end of the counseling sessions, each patient is retested to determine anxiety level. The results are given below. From the given data, can we conclude that the counseling sessions reduce anxiety? Use a 0.01 level of significance.
|
|
Score Before |
Score After |
|
|
Counseling |
Counseling |
|
Patient |
B |
A |
|
Jan |
121 |
76 |
|
Tom |
93 |
93 |
|
Diane |
105 |
64 |
|
Barbara |
115 |
117 |
|
Mike |
130 |
82 |
|
Bill |
98 |
80 |
|
Frank |
142 |
79 |
|
Carol |
118 |
67 |
|
Alice |
125 |
89 |
|
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|
|
|
t-Test: Paired Two Sample for Means |
|
|
|
|
|
|
|
|
B |
A |
|
Mean |
116.3333333 |
83 |
|
Variance |
244.5 |
248 |
|
Observations |
9 |
9 |
|
Pearson Correlation |
-0.06700677 |
|
|
Hypothesized Mean Difference |
0 |
|
|
df |
8 |
|
|
t Stat |
4.362281022 |
|
|
P(T<=t) one-tail |
0.001202634 |
|
|
t Critical one-tail |
1.859548033 |
|
|
P(T<=t) two-tail |
0.002405269 |
|
|
t Critical two-tail |
2.306004133 |
|
Hypotheses: H0
Ha
α =
Distribution:
(if appropriate) df:
Standardized test statistic:
Correct p-value
Conclusion:
Conclusion in words:
Chapter 10: Hypothesis testing with χ²- and F-distributions:
18. Education Level of Adults The Census Bureau of the U.S. government found that 13% of adults did not finish high school, 30% graduated from high school only, 29% had some college education but did not obtain a bachelor’s degree, and 28% were college graduates. To see if these proportions were consistent with those people who lived in the Lincoln County area, a local researcher selected a random sample of 300 adults and found that 43 did not finish high school, 76 were high school graduates only, 96 had some college education, and 85 were college graduates. At α = 0.10, test the claim that the proportions are the same for the adults in Lincoln County as those stated by the Census Bureau.
|
χ² Test for Goodness of Fit |
|||||
|
Category |
Hypothesized Distribution |
Observed Data |
Expected |
|
|
|
Some HS |
13.00% |
43 |
39 |
|
χ² - statistic |
|
HS grad |
30.00% |
76 |
90 |
|
3.530973433 |
|
Some College |
29.00% |
96 |
87 |
|
|
|
College Grad |
28.00% |
85 |
84 |
|
df |
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|
3 |
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p-value of the Test |
|
|
|
|
|
|
0.3167671334 |
Hypotheses: H0
Ha Correct p-value =
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
19. In a recent year, of 79,160 students taking the SAT I who were planning to major in education, 61,745 were female. That year, of the 50,829 students planning to major in biological sciences, 33,039 were female. At
0.05
a
=
, test the claim that gender and major are independent..|
χ² Test for Independence |
|
|
|||
|
Actual Data |
Categories for Columns Variable |
Row |
|||
|
|
|
Education |
Biology |
|
Totals |
|
Categories |
Female |
61745 |
33039 |
|
94784 |
|
for |
Male |
17415 |
17790 |
|
35205 |
|
Rows |
|
|
|
|
|
|
Variable |
|
|
|
|
|
|
Column |
Totals |
79160 |
50829 |
|
129989 |
|
|
|
|
|
Sample |
Size |
|
|
Expected Data |
Education |
Biology |
|
|
|
|
Female |
57721.049 |
37062.951 |
|
|
|
|
Male |
21438.951 |
13766.049 |
|
|
|
χ²-stat = |
2648.9176 |
df = |
1 |
p-value = |
0.0000000 |
Hypotheses: H0
Ha Correct p-value =
α = Conclusion:
Distribution:
(if appropriate) df: Conclusion in words:
Standardized test statistic:
20. Sociology: Ethnic Groups A sociologist studying New York City ethnic groups wants to determine if there is a difference in income for immigrants from four different countries during their first year in the city. She obtained the data in the following table from a random sample of immigrants from these countries (incomes in thousands of dollars). Use a 0.05 level of significance to test the claim that there is no difference in the earnings of immigrants from the four different countries.
|
Country I |
Country II |
Country III |
Country IV |
|
|
|
|
12.7 |
8.3 |
20.3 |
17.2 |
|
|
|
|
9.2 |
17.2 |
16.6 |
8.8 |
|
|
|
|
10.9 |
19.1 |
22.7 |
14.7 |
|
|
|
|
8.9 |
10.3 |
25.2 |
21.3 |
|
|
|
|
16.4 |
|
19.9 |
19.8 |
|
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Anova: Single Factor |
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SUMMARY |
|
|
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|
|
|
Groups |
Count |
Sum |
Average |
Variance |
|
|
|
Country I |
5 |
58.1 |
11.62 |
9.447 |
|
|
|
Country II |
4 |
54.9 |
13.725 |
27.37583 |
|
|
|
Country III |
5 |
104.7 |
20.94 |
10.393 |
|
|
|
Country IV |
5 |
81.8 |
16.36 |
24.213 |
|
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ANOVA |
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|
|
Source of Variation |
SS |
df |
MS |
F |
P-value |
F crit |
|
Between Groups |
238.22471 |
3 |
79.408237 |
4.610691 |
0.01772 |
3.287382 |
|
Within Groups |
258.3395 |
15 |
17.222633 |
|
|
|
|
|
|
|
|
|
|
|
|
Total |
496.56421 |
18 |
|
|
|
|
Hypotheses: H0
Ha Correct p-value =
α = Conclusion:
Distribution:
Conclusion in words:
Standardized test statistic:
