- The χ2 value for a one-tailed (upper tail) hypothesis test at 95% confidence and a sample size of 20 is _____.
| a. 10.117 |
| b. 10.851 |
| c. 31.410 |
| d. 30.144 |
2. Which of the following rejection rules is proper?
| a. Reject H0 if p-value ≥ α/2. |
| b. Reject H0 if p-value ≤ Fα. |
| c. Reject H0 if F ≤ Fα/2. |
| d. Reject H0 if F ≥ Fα. |
3. Consider the following hypothesis problem.
| n = 23 | s2 = 60 | H0: σ2 ≤ 66 |
| Ha: σ2 > 66 |
If the test is to be performed at the 5% level, the critical value(s) from the chi-square distribution table is(are)
| a. 12.338 and 33.924. |
| b. 12.338. |
| c. 33.924. |
| d. 10.982 and 36.781. |
4. A manufacturer of carpentry screws claims that its production machine is very accurate and that the standard deviation of the length of all screws produced is .08 mm or less. A sample of 30 screws showed a standard deviation of .10. The test statistic to test the manufacturer’s claim is _____.
| a. 18.56 |
| b. 19.2 |
| c. 45.31 |
| d. 23.2 |
5. The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The test statistic is
| a. 1.2. |
| b. 30. |
| c. 500. |
| d. 31.2. |
6. You want to test whether or not the following sample of 30 observations follows a normal distribution. The mean of the sample equals 11.83 and the standard deviation equals 4.53.
| 2 | 3 | 5 | 5 | 7 | 8 | 8 | 9 | 9 | 10 |
| 11 | 11 | 12 | 12 | 12 | 12 | 13 | 13 | 13 | 14 |
| 15 | 15 | 15 | 16 | 16 | 17 | 17 | 18 | 18 | 19 |
The number of intervals or categories used to test the hypothesis for this problem is
| a. 10. |
| b. 4. |
| c. 6. |
| d. 5 |
7. A study was conducted to examine whether the proportion of females was the same for five groups (Groups A, B, C, D, and E). How many degrees of freedom would the χ2 test statistic have when testing the hypothesis that the proportions in each group are all equal?
| a. 0.20 |
| b. 4 |
| c. 5 |
| d. 1 |
8. You want to test whether or not the following sample of 30 observations follows a normal distribution. The mean of the sample equals 11.83 and the standard deviation equals 4.53.
| 2 | 3 | 5 | 5 | 7 | 8 | 8 | 9 | 9 | 10 |
| 11 | 11 | 12 | 12 | 12 | 12 | 13 | 13 | 13 | 14 |
| 15 | 15 | 15 | 16 | 16 | 17 | 17 | 18 | 18 | 19 |
The calculated value for the test statistic equals
| a. 6. |
| b. 1.67. |
| c. 2. |
| d. 0. |
9. When individuals in a sample of 150 were asked whether or not they supported capital punishment, the following information was obtained.
| Do you support capital punishment? | Number of individuals |
| Yes | 40 |
| No | 60 |
| No Opinion | 50 |
We are interested in determining whether or not the opinions of the individuals (as to Yes, No, and No Opinion) are uniformly distributed. The conclusion of the test at the 5% level of significance is that the
| a. distribution is uniform. |
| b. distribution might have been normal. |
| c. null hypothesis cannot be rejected. |
| d. Marascuilo procedure is more applicable. |
10. The number of degrees of freedom for the appropriate chi-square distribution in a test of independence is
| a. A chi-square distribution is not used. |
| b. k – 1. |
| c. number of rows minus 1 times number of columns minus 1. |
| d. n – 1. |
11. A population where each of its element is assigned to one and only one of several classes or categories is a
| a. Poisson population. |
| b. multinomial population. |
| c. binomial population. |
| d. normal population. |
12. The table below gives beverage preferences for random samples of teens and adults.
| Teens | Adults | Total | |
| Coffee | 50 | 200 | 250 |
| Tea | 100 | 150 | 250 |
| Soft Drink | 200 | 200 | 400 |
| Other | 50 | 50 | 100 |
| 400 | 600 | 1000 |
We are asked to test for independence between age (i.e., adult and teen) and drink preferences. The expected number of adults who prefer coffee is
| a. 200. |
| b. .25. |
| c. 150. |
| d. .33. |
13. The following table shows the number of individuals in a sample of 300 who indicated they support the new tax proposal.
| Political Party | Support |
| Democrats | 100 |
| Republicans | 120 |
| Independents | 80 |
We are interested in determining whether or not the opinions of the individuals of the three groups are uniformly distributed. The expected frequency for each group is
| a. .333. |
| b. .50. |
| c. 50. |
| d. 100. |
14. Marascuilo procedure is used to test for a significant difference between pairs of population
| a. proportions. |
| b. variances. |
| c. means. |
| d. standard deviations. |
15. Last school year, the student body of a local university consisted of 30% freshmen, 24% sophomores, 26% juniors, and 20% seniors. A sample of 300 students taken from this year’s student body showed the following number of students in each classification.
| Freshmen | 83 |
| Sophomores | 68 |
| Juniors | 85 |
| Seniors | 64 |
We are interested in determining whether or not there has been a significant change in the classifications between the last school year and this school year. At a .05 level of significance, the null hypothesis
| a. cannot be tested. |
| b. should not be rejected. |
| c. was designed wrong. |
| d. should be rejected. |
16. You want to test whether or not the following sample of 30 observations follows a normal distribution. The mean of the sample equals 11.83 and the standard deviation equals 4.53.
| 2 | 3 | 5 | 5 | 7 | 8 | 8 | 9 | 9 | 10 |
| 11 | 11 | 12 | 12 | 12 | 12 | 13 | 13 | 13 | 14 |
| 15 | 15 | 15 | 16 | 16 | 17 | 17 | 18 | 18 | 19 |
The calculated value for the test statistic equals
| a. 1.67. |
| b. 0. |
| c. 6. |
| d. 2. |
17. Two hundred randomly selected people were asked to state their primary source for news about current events: (1) Television, (2) Radio, (3) Internet, or (4) Other. We wish to determine whether the percent distribution of responses is 10%, 30%, 50%, and 10% for news sources 1 through 4, respectively. What is the null hypothesis?
| a. H0: p = .50 |
| b. H0: the proportions are not all equal. |
| c. H0: p1= .10, p2 = .30, p3 = .50, p4 = .10 |
| d. H0: p1 = .25, p2 = .25, p3 = .25, p4 = .25 |
18. The test for goodness of fit
| a. is always an upper tail test. |
| b. is always a lower tail test. |
| c. can be a lower or an upper tail test. |
| d. is always a two-tailed test. |
19. The table below gives beverage preferences for random samples of teens and adults.
| Teens | Adults | Total | |
| Coffee | 50 | 200 | 250 |
| Tea | 100 | 150 | 250 |
| Soft Drink | 200 | 200 | 400 |
| Other | 50 | 50 | 100 |
| 400 | 600 | 1000 |
We are asked to test for independence between age (i.e., adult and teen) and drink preferences. The test statistic for this test of independence is
| a. 62.5. |
| b. 0. |
| c. 82.5. |
| d. 8.4. |
20. In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College. To see whether or not the proportions have changed, a sample of 300 students from the university was taken. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. The expected frequency for the Business College is
| a. 90. |
| b. .35. |
| c. 105. |
| d. .3. |
21. An important application of the chi-square distribution is
| a. testing for goodness of fit. |
| b. testing for the independence of two categorical variables. |
| c. making inferences about a single population variance. |
| d. All of these alternatives are correct. |
22. A population where each of its element is assigned to one and only one of several classes or categories is a
| a. multinomial population. |
| b. Poisson population. |
| c. normal population. |
| d. binomial population |
23. When individuals in a sample of 150 were asked whether or not they supported capital punishment, the following information was obtained.
| Do you support capital punishment? | Number of individuals |
| Yes | 40 |
| No | 60 |
| No Opinion | 50 |
We are interested in determining whether or not the opinions of the individuals (as to Yes, No, and No Opinion) are uniformly distributed. The conclusion of the test at the 5% level of significance is that the
| a. distribution is uniform. |
| b. null hypothesis cannot be rejected. |
| c. Marascuilo procedure is more applicable. |
| d. distribution might have been normal |
24. In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College. To see whether or not the proportions have changed, a sample of 300 students from the university was taken. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. This problem is an example of a
| a. z test for proportions. |
| b. test for independence. |
| c. multinomial population. |
| d. Marascuilo procedure. |
25. Last school year, the student body of a local university consisted of 30% freshmen, 24% sophomores, 26% juniors, and 20% seniors. A sample of 300 students taken from this year’s student body showed the following number of students in each classification.
| Freshmen | 83 |
| Sophomores | 68 |
| Juniors | 85 |
| Seniors | 64 |
We are interested in determining whether or not there has been a significant change in the classifications between the last school year and this school year. The expected frequency of seniors is
| a. 68. |
| b. 60. |
| c. 64. |
| d. 20%. |
26. The owner of a car wash wants to see if the arrival rate of cars follows a Poisson distribution. In order to test the assumption of a Poisson distribution, a random sample of 150 ten-minute intervals was taken. You are given the following observed frequencies:
| Number of Cars Arriving in a 10-Minute Interval | Frequency |
| 0 | 3 |
| 1 | 10 |
| 2 | 15 |
| 3 | 23 |
| 4 | 30 |
| 5 | 24 |
| 6 | 20 |
| 7 | 13 |
| 8 | 8 |
| 9 or more | 4 |
| 150 |
The calculated value for the test statistic equals
| a. 1.72. |
| b. .18. |
| c. 3.11. |
| d. 2.89 |
27. Which of the following Chi-Square tests can be used to determine if a data set follows a Normal distribution?
| a. Chi Square Test of Association |
| b. Chi-Square Test of Homogeneity |
| c. Chi-Square Goodness of Fit |
| d. Chi-Square Test of Equality of Three or More Population Proportions |
28. When individuals in a sample of 150 were asked whether or not they supported capital punishment, the following information was obtained.
| Do you support capital punishment? | Number of individuals |
| Yes | 40 |
| No | 60 |
| No Opinion | 50 |
We are interested in determining whether or not the opinions of the individuals (as to Yes, No, and No Opinion) are uniformly distributed. The conclusion of the test at the 5% level of significance is that the
| a. Marascuilo procedure is more applicable. |
| b. distribution might have been normal. |
| c. distribution is uniform. |
| d. null hypothesis cannot be rejected |
29. The table below gives beverage preferences for random samples of teens and adults.
| Teens | Adults | Total | |
| Coffee | 50 | 200 | 250 |
| Tea | 100 | 150 | 250 |
| Soft Drink | 200 | 200 | 400 |
| Other | 50 | 50 | 100 |
| 400 | 600 | 1000 |
We are asked to test for independence between age (i.e., adult and teen) and drink preferences. The expected number of adults who prefer coffee is
| a. .33. |
| b. 200. |
| c. .25. |
| d. 150 |
30. Marascuilo procedure is used to test for a significant difference between pairs of population
| a. variances. |
| b. standard deviations. |
| c. means. |
| d. proportions. |
31. With respect to the number of categories, k, when would a multinomial experiment be identical to a binomial experiment?
| a. k = 1 |
| b. k = 2 |
| c. k = 3 |
| d. k = 4 |
32. An important application of the chi-square distribution is
| a. testing for goodness of fit. |
| b. testing for the independence of two categorical variables. |
| c. making inferences about a single population variance. |
| d. All of these alternatives are correct. |
33. In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College. To see whether or not the proportions have changed, a sample of 300 students from the university was taken. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. The calculated value for the test statistic equals
| a. .01. |
| b. 4.38. |
| c. .75. |
| d. 4.29. |
34. The sampling distribution for a goodness of fit test is the
| a. t distribution. |
| b. chi-square distribution. |
| c. normal distribution. |
| d. Poisson distribution |
35. The following table shows the number of individuals in a sample of 300 who indicated they support the new tax proposal.
| Political Party | Support |
| Democrats | 100 |
| Republicans | 120 |
| Independents | 80 |
We are interested in determining whether or not the opinions of the individuals of the three groups are uniformly distributed. The calculated value for the test statistic equals
| a. 2. |
| b. 0. |
| c. 8. |
| d. 4. |
36. In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College. To see whether or not the proportions have changed, a sample of 300 students from the university was taken. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. The hypothesis is to be tested at the 5% level of significance. The critical value from the table equals
| a. 9.348. |
| b. 5.991. |
| c. 7.378. |
| d. 7.815 |
37. If there are three or more populations, then it is
| a. impossible to test for equality of the three population proportions, because chi-square tests deal with only two populations. |
| b. possible to test for equality of three or more population proportions. |
| c. reasonable to test for equality of multiple population proportions using chi-square lower tail tests. |
| d. customary to use a t distribution to test for equality of the three population proportions. |
38. The table below gives beverage preferences for random samples of teens and adults.
| Teens | Adults | Total | |
| Coffee | 50 | 200 | 250 |
| Tea | 100 | 150 | 250 |
| Soft Drink | 200 | 200 | 400 |
| Other | 50 | 50 | 100 |
| 400 | 600 | 1000 |
We are asked to test for independence between age (i.e., adult and teen) and drink preferences. The test statistic for this test of independence is
| a. 0. |
| b. 62.5. |
| c. 82.5. |
| d. 8.4. |
39. The following table shows the number of individuals in a sample of 300 who indicated they support the new tax proposal.
| Political Party | Support |
| Democrats | 100 |
| Republicans | 120 |
| Independents | 80 |
We are interested in determining whether or not the opinions of the individuals of the three groups are uniformly distributed. The number of degrees of freedom associated with this problem is
| a. 299. |
| b. 300. |
| c. 2. |
| d. 3. |
40. The test for goodness of fit
| a. is always a two-tailed test. |
| b. is always an upper tail test. |
| c. is always a lower tail test. |
| d. can be a lower or an upper tail test. |
41. Which function in Excel is used to perform a test of independence?
| a. CHISQ.TEST |
| b. T.TEST |
| c. NORM.S.DIST |
| d. Z.TEST |
42. A manufacturer of carpentry screws claims that its production machine is very accurate and that the standard deviation of the length of all screws produced is .08 mm or less. A sample of 30 screws showed a standard deviation of .10. The test statistic to test the manufacturer’s claim is _____.
| a. 18.56 |
| b. 19.2 |
| c. 45.31 |
| d. 23.2 |
43. We are interested in testing to see if the variance of a population is less than 7. The correct null hypothesis is
| a. σ2 ≥ 49. |
| b. σ < 49. |
| c. σ < 7. |
| d. σ2 ≥ 7. |
44. The symbol used for the variance of the sample is
| a. s2. |
| b. σ. |
| c. σ2. |
| d. s. |
45. The value of F.05 with 8 numerator and 19 denominator degrees of freedom is
| a. 2.48. |
| b. 2.58. |
| c. 2.96. |
| d. 3.63 |
46. Which of the following has an F distribution?
a.  |
b.  |
c.  |
d.  |
47. A researcher would like to test the hypothesis that population B has a smaller variance than population A, using a 5% level of significance for the hypothesis test. What is the critical value from the F distribution table?
| a. 2.39 |
| b. 2.84 |
| c. 2.22 |
| d. 1.96 |
48. The 95% confidence interval estimate of a population variance when a sample variance of 324 is obtained from a sample of 81 items is
| a. 16.42 to 194.35. |
| b. 243.086 to 453.520. |
| c. 254.419 to 429.203. |
| d. 14.14 to 174.94. |
49. We are interested in testing whether the variance of a population is significantly more than 625. The null hypothesis for this test is
| a. H0: σ2 ≤ 625. |
| b. H0: σ2 > 625. |
| c. H0: σ2 ≤ 25. |
| d. H0: σ2 ≥ 625. |
50. For a sample size of 21 at 95% confidence, the chi-square values needed for interval estimation are
| a. 10.283 and 35.479. |
| b. 2.700 and 19.023. |
| c. 8.260 and 37.566. |
| d. 9.591 and 34.170. |
51. The producer of a certain medicine claims that their bottling equipment is very accurate and that the standard deviation of all their filled bottles is .1 ounces or less. A sample of 20 bottles showed a standard deviation of .11 ounces. The test statistic to test the claim is
| a. 2.3. |
| b. 22.99. |
| c. 4.85. |
| d. 24.2. |
52. The critical value of F using α = .05 when there is a sample size of 21 for the sample with the smaller variance, and there is a sample size of 9 for the sample with the larger sample variance is
| a. 2.10. |
| b. 2.45. |
| c. 2.94. |
| d. 2.37. |
53. The chi-square value for a one-tailed (lower tail) test when the level of significance is .1 and the sample size is 15 is
| a. 23.685. |
| b. 7.790. |
| c. 21.064. |
| d. 6.571 |
54. Consider the following hypothesis problem.
| n = 30 | H0: σ2 = 500 |
| s2 = 625 | Ha: σ2 ≠ 500 |
The test statistic equals
| a. 37.50. |
| b. 23.20. |
| c. 24.00. |
| d. 36.25 |
55. A sample of 41 observations yielded a sample standard deviation of 5. If we want to test H0: σ2 = 20, the test statistic is
| a. 100.75. |
| b. 51.25. |
| c. 50. |
| d. 10. |
56. The value of F.05 with 8 numerator and 19 denominator degrees of freedom is
| a. 2.48. |
| b. 2.96. |
| c. 3.63. |
| d. 2.58 |
57. χ2.975 = 8.231 indicates that
| a. 2.5% of the chi-square values are greater than 8.231. |
| b. 97.5% of the chi-square values are greater than 8.231. |
| c. 5% of the chi-square values are equal to 8.231. |
| d. 97.5% of the chi-square values are less than 8.231. |
58. Consider the following sample information from Population A and Population B.
| Sample A | Sample B | |
| n | 24 | 16 |
| s2 | 32 | 38 |
We want to test the hypothesis that the population variances are equal. The test statistic for this problem equals
| a. 1.19. |
| b. 1.50. |
| c. .84. |
| d. .67. |
59. Given the sample information below, what test statistic should be used to determine whether the standard deviations are equal for two populations, prior to performing a hypothesis test for the equality of the population means?
| a. t = 18.75, with degrees of freedom = 19 |
| b. F = 1.58, with numerator degrees of freedom = 19 and denominator degrees of freedom = 29 |
| c. F = 2.49, with numerator degrees of freedom = 29 and denominator degrees of freedom = 19 |
| d. F = 0.40, with degrees of freedom = 19 |
60. Consider the following hypothesis problem.
| n = 23 | s2 = 60 | H0: σ2 ≤ 66 |
| Ha: σ2 > 66 |
The p-value is
| a. between .05 and .10. |
| b. less than .025. |
| c. between .025 and .05. |
| d. greater than .10. |
61. The sampling distribution used when making inferences about a single population’s variance is
| a. a normal distribution. |
| b. a t distribution. |
| c. a chi-square distribution. |
| d. an F distribution. |
62. The value of F.01 with 9 numerator and 20 denominator degrees of freedom is
| a. 2.91. |
| b. 2.94. |
| c. 2.39. |
| d. 3.46. |
63. Which of the following is not a property of a chi-square distribution?
| a. χ2 is skewed to the right. |
| b. The number of degrees of freedom defines the shape of the distribution of χ2. |
| c. χ2 can have both positive and negative values. |
| d. All of these choices are properties of the χ2 distribution. |
64. A sample of 41 observations yielded a sample standard deviation of 5. If we want to test H0: σ2 = 20, the test statistic is
| a. 10. |
| b. 51.25. |
| c. 50. |
| d. 100.75. |
65. Which of the following rejection rules is proper?
| a. Reject H0 if p-value ≥ α/2. |
| b. Reject H0 if F ≥ Fα. |
| c. Reject H0 if F ≤ Fα/2. |
| d. Reject H0 if p-value ≤ Fα. |
66. In practice, the most frequently encountered hypothesis test about a population variance is a
| a. one-tailed test, with rejection region in the lower tail. |
| b. two-tailed test, with equal-size rejection regions. |
| c. two-tailed test, with unequal-size rejection regions. |
| d. one-tailed test, with rejection region in the upper tail. |
67. What is the null hypothesis for testing whether the variance of a population differs from 2.5?
a.  |
b.  |
c.  |
d.  |
68. The critical value of F using α = .05 when there is a sample size of 21 for the sample with the smaller variance, and there is a sample size of 9 for the sample with the larger sample variance is
| a. 2.45. |
| b. 2.10. |
| c. 2.94. |
| d. 2.37. |
69. Consider the following hypothesis problem.
| n = 30 | H0: σ2 = 500 |
| s2 = 625 | Ha: σ2 ≠ 500 |
The test statistic equals
| a. 24.00. |
| b. 23.20. |
| c. 37.50. |
| d. 36.25. |
70. What is the critical value of F for a one-tailed test at the α=.10 level, when there is a sample size of 8 for the sample with the smaller variance and a sample size of 11 for the sample with the larger sample variance?
| a. 2.70 |
| b. 2.41 |
| c. 1.65 |
| d. 3.64 |
71. We are interested in testing to see if the variance of a population is less than 7. The correct null hypothesis is
| a. σ2 ≥ 49. |
| b. σ2 ≥ 7. |
| c. σ < 7. |
| d. σ < 49. |
72. Which of the following has a chi-square distribution?
| a. (n – 1)s/σ. |
| b. (n – 1)s2/σ2. |
| c. (n – 1)σ2/s2. |
| d. (n – 1)σ/s. |
73. Consider the following hypothesis problem.
| n = 23 | s2 = 60 | H0: σ2 ≤ 66 |
| Ha: σ2 > 66 |
The p-value is
| a. greater than .10. |
| b. between .05 and .10. |
| c. between .025 and .05. |
| d. less than .025 |
74. The 95% confidence interval estimate of a population variance when a sample variance of 324 is obtained from a sample of 81 items is
| a. 16.42 to 194.35. |
| b. 254.419 to 429.203. |
| c. 14.14 to 174.94. |
| d. 243.086 to 453.520 |
75. The value of F.01 with 9 numerator and 20 denominator degrees of freedom is
| a. 2.91. |
| b. 3.46. |
| c. 2.39. |
| d. 2.94 |
76. To avoid the problem of not having access to tables of the F distribution when a one-tailed test is required and with F values given for the lower tail, let the
| a. sample variance from the population with the smaller hypothesized variance be the numerator of the test statistic. |
| b. smaller sample variance be the numerator of the test statistic. |
| c. larger sample variance be the numerator of the test statistic. |
| d. sample variance from the population with the larger hypothesized variance be the numerator of the test statistic. |
77. The value of F.05 with 8 numerator and 19 denominator degrees of freedom is
| a. 2.58. |
| b. 2.48. |
| c. 2.96. |
| d. 3.63. |
78. The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. At the .05 level of significance, the null hypothesis
| a. should not be tested. |
| b. should not be rejected. |
| c. should be rejected. |
| d. should be revised |
79. A manufacturer of carpentry screws claims that its production machine is very accurate and that the standard deviation of the length of all screws produced is .08 mm or less. A sample of 30 screws showed a standard deviation of .10. The test statistic to test the manufacturer’s claim is _____.
| a. 19.2 |
| b. 45.31 |
| c. 23.2 |
| d. 18.56 |
80. A sample of 21 observations yielded a sample variance of 16. If we want to test H0: σ2 = 16, the test statistic is
| a. 50. |
| b. 5. |
| c. 21. |
| d. 20. |
81. A sample of 28 elements is selected to estimate a 95% confidence interval for the variance of the population. The chi-square values to be used for this interval estimation are
| a. 15.308 and 44.461. |
| b. 11.808 and 49.645. |
| c. 16.151 and 40.113. |
| d. 14.573 and 43.195. |
82. The producer of a certain medicine claims that their bottling equipment is very accurate and that the standard deviation of all their filled bottles is .1 ounces or less. A sample of 20 bottles showed a standard deviation of .11 ounces. The test statistic to test the claim is
| a. 2.3. |
| b. 4.85. |
| c. 22.99. |
| d. 24.2 |
83. A sample of 21 observations yielded a sample variance of 16. If we want to test H0: σ2 = 16, the test statistic is
| a. 20. |
| b. 50. |
| c. 21. |
| d. 5. |
84. To avoid the problem of not having access to Tables of F distribution when F values are needed for the lower tail, the numerator of the test statistic for a two-tailed test should be the one with
| a. the larger sample size. |
| b. the larger sample variance. |
| c. the smaller sample variance. |
| d. the smaller sample size. |
85. Consider the following sample information from Population A and Population B.
| Sample A | Sample B | |
| n | 24 | 16 |
| s2 | 32 | 38 |
We want to test the hypothesis that the population variances are equal. The test statistic for this problem equals
| a. 1.19. |
| b. 1.50. |
| c. .67. |
| d. .84. |
86. In practice, the most frequently encountered hypothesis test about a population variance is a
| a. one-tailed test, with rejection region in the upper tail. |
| b. one-tailed test, with rejection region in the lower tail. |
| c. two-tailed test, with unequal-size rejection regions. |
| d. two-tailed test, with equal-size rejection regions. |
87. A sample of 21 elements is selected to estimate a 90% confidence interval for the variance of the population. The chi-square value(s) to be used for this interval estimation is(are)
| a. 12.443 and 28.412. |
| b. 10.851 and 31.410. |
| c. 31.410. |
| d. 12.443. |
88. Consider the following hypothesis problem.
| n = 14 | H0: σ2 ≤ 410 |
| s = 20 | Ha: σ2 > 410 |
The null hypothesis is to be tested at the 5% level of significance. The critical value(s) from the chi-square distribution table is(are)
| a. 5.009 and 24.736. |
| b. 5.629 and 26.119. |
| c. 23.685. |
| d. 22.362. |
89. A sample of 60 items from population 1 has a sample variance of 8 while a sample of 40 items from population 2 has a sample variance of 10. If we want to test whether the variances of the two populations are equal, the test statistic will have a value of
| a. 1.56. |
| b. .8. |
| c. 1.25. |
| d. 1.5. |
89. The manager of the service department of a local car dealership has noted that the service times of a sample of 30 new automobiles has a standard deviation of 6 minutes. A 95% confidence interval estimate for the standard deviation of the service times (in minutes) for all their new automobiles is
| a. 4.778 to 8.066. |
| b. 22.833 to 65.059. |
| c. 16.047 to 45.722. |
| d. 2.93 to 6.31 |
90. Last school year, the student body of a local university consisted of 30% freshmen, 24% sophomores, 26% juniors, and 20% seniors. A sample of 300 students taken from this year’s student body showed the following number of students in each classification.
| Freshmen | 83 |
| Sophomores | 68 |
| Juniors | 85 |
| Seniors | 64 |
We are interested in determining whether or not there has been a significant change in the classifications between the last school year and this school year. The calculated value for the test statistic equals
| a. 1.66. |
| b. .65. |
| c. .54. |
| d. 6.66 |
91. The owner of a car wash wants to see if the arrival rate of cars follows a Poisson distribution. In order to test the assumption of a Poisson distribution, a random sample of 150 ten-minute intervals was taken. You are given the following observed frequencies:
| Number of Cars Arriving in a 10-Minute Interval | Frequency |
| 0 | 3 |
| 1 | 10 |
| 2 | 15 |
| 3 | 23 |
| 4 | 30 |
| 5 | 24 |
| 6 | 20 |
| 7 | 13 |
| 8 | 8 |
| 9 or more | 4 |
| 150 |
The p-value is
| a. between .05 and .1. |
| b. greater than .1. |
| c. between .025 and .05. |
| d. less than .01. |
92. Which of the following Chi-Square tests can be used to determine if a data set follows a Normal distribution?
| a. Chi-Square Goodness of Fit |
| b. Chi Square Test of Association |
| c. Chi-Square Test of Equality of Three or More Population Proportions |
| d. Chi-Square Test of Homogeneity |
93. Last school year, the student body of a local university consisted of 30% freshmen, 24% sophomores, 26% juniors, and 20% seniors. A sample of 300 students taken from this year’s student body showed the following number of students in each classification.
| Freshmen | 83 |
| Sophomores | 68 |
| Juniors | 85 |
| Seniors | 64 |
We are interested in determining whether or not there has been a significant change in the classifications between the last school year and this school year. The expected number of freshmen is
| a. 83. |
| b. 90. |
| c. 30. |
| d. 10 |
94. You want to test whether or not the following sample of 30 observations follows a normal distribution. The mean of the sample equals 11.83 and the standard deviation equals 4.53.
| 2 | 3 | 5 | 5 | 7 | 8 | 8 | 9 | 9 | 10 |
| 11 | 11 | 12 | 12 | 12 | 12 | 13 | 13 | 13 | 14 |
| 15 | 15 | 15 | 16 | 16 | 17 | 17 | 18 | 18 | 19 |
The calculated value for the test statistic equals
| a. 6. |
| b. 2. |
| c. 1.67. |
| d. 0. |
95. The test for goodness of fit
| a. is always an upper tail test. |
| b. is always a two-tailed test. |
| c. is always a lower tail test. |
| d. can be a lower or an upper tail test. |
96. The degrees of freedom for a table with 6 rows and 3 columns is
| a. 10. |
| b. 18. |
| c. 15. |
| d. 6. |
97. In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College. To see whether or not the proportions have changed, a sample of 300 students from the university was taken. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. This problem is an example of a
| a. z test for proportions. |
| b. multinomial population. |
| c. Marascuilo procedure. |
| d. test for independence. |
98. The number of degrees of freedom associated with the chi-square distribution in a test of independence is
| a. number of sample items minus 1. |
| b. number of populations minus 1. |
| c. number of populations minus number of estimated parameters minus 1. |
| d. number of rows minus 1 times number of columns minus 1. |
99. The following table shows the number of individuals in a sample of 300 who indicated they support the new tax proposal.
| Political Party | Support |
| Democrats | 100 |
| Republicans | 120 |
| Independents | 80 |
We are interested in determining whether or not the opinions of the individuals of the three groups are uniformly distributed. The expected frequency for each group is
| a. .333. |
| b. 50. |
| c. .50. |
| d. 100. |
100. The table below gives beverage preferences for random samples of teens and adults.
| Teens | Adults | Total | |
| Coffee | 50 | 200 | 250 |
| Tea | 100 | 150 | 250 |
| Soft Drink | 200 | 200 | 400 |
| Other | 50 | 50 | 100 |
| 400 | 600 | 1000 |
We are asked to test for independence between age (i.e., adult and teen) and drink preferences. With a .05 level of significance, the critical value for the test is
| a. 15.507. |
| b. 7.815. |
| c. 14.067. |
| d. 5.991. |
101.
Last school year, the student body of a local university consisted of 30% freshmen, 24% sophomores, 26% juniors, and 20% seniors. A sample of 300 students taken from this year’s student body showed the following number of students in each classification.
| Freshmen | 83 |
| Sophomores | 68 |
| Juniors | 85 |
| Seniors | 64 |
We are interested in determining whether or not there has been a significant change in the classifications between the last school year and this school year. The expected number of freshmen is
| a. 30. |
| b. 83. |
| c. 90. |
| d. 10. |
102. You want to test whether or not the following sample of 30 observations follows a normal distribution. The mean of the sample equals 11.83 and the standard deviation equals 4.53.
| 2 | 3 | 5 | 5 | 7 | 8 | 8 | 9 | 9 | 10 |
| 11 | 11 | 12 | 12 | 12 | 12 | 13 | 13 | 13 | 14 |
| 15 | 15 | 15 | 16 | 16 | 17 | 17 | 18 | 18 | 19 |
The calculated value for the test statistic equals
| a. 1.67. |
| b. 2. |
| c. 6. |
| d. 0. |
103. In order not to violate the requirements necessary to use the chi-square distribution for testing equality of three or more population proportions, each expected frequency in a goodness of fit test must be
| a. no more than 5. |
| b. at least 5. |
| c. less than 2. |
| d. at least 10 |
104. When individuals in a sample of 150 were asked whether or not they supported capital punishment, the following information was obtained.
| Do you support capital punishment? | Number of individuals |
| Yes | 40 |
| No | 60 |
| No Opinion | 50 |
We are interested in determining whether or not the opinions of the individuals (as to Yes, No, and No Opinion) are uniformly distributed. The expected frequency for each group is
| a. .333. |
| b. .50. |
| c. 1/3. |
| d. 50. |
105. If the sample size is n = 75, what are the degrees of freedom for the appropriate chi-square distribution when testing for independence of two variables each with three categories?
| a. 74 |
| b. 4 |
| c. 2 |
| d. 69 |
106. In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College. To see whether or not the proportions have changed, a sample of 300 students from the university was taken. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. The hypothesis is to be tested at the 5% level of significance. The critical value from the table equals
| a. 7.815. |
| b. 5.991. |
| c. 7.378. |
| d. 9.348. |
107. Last school year, the student body of a local university consisted of 30% freshmen, 24% sophomores, 26% juniors, and 20% seniors. A sample of 300 students taken from this year’s student body showed the following number of students in each classification.
| Freshmen | 83 |
| Sophomores | 68 |
| Juniors | 85 |
| Seniors | 64 |
We are interested in determining whether or not there has been a significant change in the classifications between the last school year and this school year. At a .05 level of significance, the null hypothesis
| a. was designed wrong. |
| b. should not be rejected. |
| c. cannot be tested. |
| d. should be rejected. |
108. The following table shows the number of individuals in a sample of 300 who indicated they support the new tax proposal.
| Political Party | Support |
| Democrats | 100 |
| Republicans | 120 |
| Independents | 80 |
We are interested in determining whether or not the opinions of the individuals of the three groups are uniformly distributed. The expected frequency for each group is
| a. .333. |
| b. 50. |
| c. 100. |
| d. .50 |
109. A random sample of 550 physicians was selected and classified according to the type of practice they operate and whether or not their practice was full (meaning, not accepting new patients). The counts are given in the table below.
The test statistic for a test of independence is χ2 = 34.893. What is the p-value for this test?
| a. p = .004 |
| b. p = .0025 |
| c. p = .05 |
| d. p < .0001 |
110. An important application of the chi-square distribution is
| a. making inferences about a single population variance. |
| b. testing for the independence of two categorical variables. |
| c. testing for goodness of fit. |
| d. All of these alternatives are correct. |
111. How many degrees of freedom does the chi-square test statistic for a goodness of fit have when there are 10 categories?
| a. 7 |
| b. 62 |
| c. 9 |
| d. 74 |
112. The following table shows the number of individuals in a sample of 300 who indicated they support the new tax proposal.
| Political Party | Support |
| Democrats | 100 |
| Republicans | 120 |
| Independents | 80 |
We are interested in determining whether or not the opinions of the individuals of the three groups are uniformly distributed. The expected frequency for each group is
| a. .50. |
| b. .333. |
| c. 100. |
| d. 50. |
113. When individuals in a sample of 150 were asked whether or not they supported capital punishment, the following information was obtained.
| Do you support capital punishment? | Number of individuals |
| Yes | 40 |
| No | 60 |
| No Opinion | 50 |
We are interested in determining whether or not the opinions of the individuals (as to Yes, No, and No Opinion) are uniformly distributed. The conclusion of the test at the 5% level of significance is that the
| a. distribution might have been normal. |
| b. null hypothesis cannot be rejected. |
| c. Marascuilo procedure is more applicable. |
| d. distribution is uniform. |
114. When individuals in a sample of 150 were asked whether or not they supported capital punishment, the following information was obtained.
| Do you support capital punishment? | Number of individuals |
| Yes | 40 |
| No | 60 |
| No Opinion | 50 |
We are interested in determining whether or not the opinions of the individuals (as to Yes, No, and No Opinion) are uniformly distributed. The p-value is
| a. less than .01. |
| b. between .05 and .1. |
| c. between .01 and .05. |
| d. larger than .1 |
115. A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a standard deviation of 2 years. We want to determine if the average age of all the students at the university is significantly more than 24. Assume the distribution of the population of ages is normal. The p-value is between
| a. .025 to .05. |
| b. .01 to .025. |
| c. .05 to .10. |
| d. .005 to .01. |
116. A Type I error is committed when
| a. the critical value is greater than the value of the test statistic. |
| b. sample data contradict the null hypothesis. |
| c. a true alternative hypothesis is not accepted. |
| d. a true null hypothesis is rejected. |
117. When the following hypotheses are being tested at a level of significance of α
H0: μ ≥ 500
Ha: μ < 500
the null hypothesis will be rejected, if the p-value is
| a. = α/2. |
| b. > α. |
| c. ≤ 1 – α/2. |
| d. ≤ α. |
118. If the null hypothesis is rejected at the .05 level of significance, it will
| a. sometimes not be rejected at the .10 level of significance. |
| b. always not be rejected at the .10 level of significance. |
| c. always be rejected at the .10 level of significance. |
| d. sometimes be rejected at the .10 level of significance. |
119. Read the z statistic from the normal distribution table and circle the correct answer. For a one-tailed test (lower tail) using α = .1020, z =
| a. -1.96. |
| b. -1.27. |
| c. -1.53. |
| d. -1.64 |
120. In a one-tailed hypothesis test (lower tail), the test statistic is determined to be -2. The p-value for this test is
| a. .0056. |
| b. .4772. |
| c. .0228. |
| d. .5228. |
121. A grocery store has an average sales of $8000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8300 per day. From past information, it is known that the standard deviation of the population is $1200. The correct null hypothesis for this problem is
| a. μ = 8000. |
| b. μ > 8300. |
| c. μ ≤ 8000. |
| d. μ ≤ 8300. |
122. Given the following information,
n = 49, x̄ = 50, s = 7
H0: μ ≥ 52
Ha: μ < 52
the test statistic is
| a. -2. |
| b. -1. |
| c. 1. |
| d. 2. |
123. For a one-tailed (upper tail) hypothesis test with a sample size of 18 and a .05 level of significance, the critical value of the test statistic t is
| a. 1.740. |
| b. 1.645. |
| c. 2.110. |
| d. 1.734. |
124. For a one-tailed hypothesis test (upper tail), the p-value is computed to be .034. If the test is being conducted at the 5% level of significance, the null hypothesis
| a. could be rejected or not rejected depending on the sample mean. |
| b. is rejected. |
| c. is not rejected. |
| d. could be rejected or not rejected depending on the sample size. |
125. A school’s newspaper reported that the proportion of students majoring in business is at least 30%. You plan on taking a sample to test the newspaper’s claim. The correct set of hypotheses is
| a. H0: p ≤ .30 Ha: p > .30. |
| b. H0: p ≥ .30 Ha: p < .30. |
| c. H0: p < .30 Ha: p ≥ .30. |
| d. H0: p > .30 Ha: p ≤ .30 |
126. then the probability of a Type II error (β) must be
| a. .95. |
| b. .05. |
| c. .025. |
| d. Cannot be computed. |
127. If the sample size increases for a given level of significance, the probability of a Type II error will
| a. increase. |
| b. decrease. |
| c. remain the same. |
| d. be equal to 1.0 regardless of α. |
128. The probability of committing a Type I error when the null hypothesis is true as an equality is
| a. greater than 1. |
| b. the level of significance. |
| c. the confidence level. |
| d. β. |
129. For the following hypothesis test,
H0: μ ≥ 150
Ha: μ < 150
the test statistic
| a. can be either negative or positive. |
| b. must be positive. |
| c. must be negative. |
| d. must be a number between zero and one. |
130. The average manufacturing work week in metropolitan Chattanooga was 40.1 hours last year. It is believed that the recession has led to a reduction in the average work week. To test the validity of this belief, the hypotheses are
| a. H0: μ ≥ 40.1 Ha: μ < 40.1. |
| b. H0: μ < 40.1 Ha: μ ≥ 40.1. |
| c. H0: μ = 40.1 Ha: μ ≠ 40.1. |
| d. H0: μ > 40.1 Ha: μ ≤ 40.1. |
131. In hypothesis testing, the critical value is
| a. a number that establishes the boundary of the rejection region. |
| b. the probability of a Type II error. |
| c. the probability of a Type I error. |
| d. the same as the p-value |
132. A weatherman stated that the average temperature during July in Chattanooga is 80 degrees or less. A sample of 32 Julys is taken to test the weatherman’s statement. The correct set of hypotheses is
| a. H0: μ ≤ 80 Ha: μ > 80. |
| b. H0: μ < 80 Ha: μ > 80. |
| c. H0: μ ≠ 80 Ha: μ = 80. |
| d. H0: μ ≥ 80 Ha: μ < 80. |
133. For a one-tailed hypothesis test (upper tail), the p-value is computed to be .034. If the test is being conducted at the 5% level of significance, the null hypothesis
| a. is rejected. |
| b. could be rejected or not rejected depending on the sample size. |
| c. could be rejected or not rejected depending on the sample mean. |
| d. is not rejected. |
134. Read the t statistic from the t distribution table and circle the correct answer. For a one-tailed test (upper tail), using a sample size of 18, and at the 5% level of significance, t =
| a. 2.12. |
| b. -1.740. |
| c. 1.740. |
| d. -2.12. |
135. A school’s newspaper reported that the proportion of students majoring in business is at least 30%. You plan on taking a sample to test the newspaper’s claim. The correct set of hypotheses is
| a. H0: p ≥ .30 Ha: p < .30. |
| b. H0: p ≤ .30 Ha: p > .30. |
| c. H0: p < .30 Ha: p ≥ .30. |
| d. H0: p > .30 Ha: p ≤ .30. |
136. For a two-tailed hypothesis test with a sample size of 20 and a .05 level of significance, the critical values of the test statistic t are
| a. -1.729 and 1.729. |
| b. -2.093 and 2.093. |
| c. -2.086 and 2.086. |
| d. -1.725 and 1.725. |
137. The average hourly wage of computer programmers with 2 years of experience has been $21.80. Because of high demand for computer programmers, it is believed there has been a significant increase in the average hourly wage of computer programmers. To test whether or not there has been an increase, the correct hypotheses to be tested are
| a. H0: μ < 21.80 Ha: μ ≥ 21.80. |
| b. H0: μ > 21.80 Ha: μ ≤ 21.80. |
| c. H0: μ = 21.80 Ha: μ ≠ 21.80. |
| d. H0: μ ≤ 21.80 Ha: μ > 21.80. |
138. A computer manufacturer claims its computers will perform effectively for more than 5 years. Which pair of hypotheses should be used to test this claim?
| a. H0: μ > 5; Ha: μ ≤ 5 |
| b. H0: μ < 5; Ha: μ ≥ 5 |
| c. H0: μ ≤ 5; Ha: μ > 5 |
| d. H0: μ ≥ 5; Ha: μ < 5 |
139. If a hypothesis test leads to the rejection of the null hypothesis,
| a. a Type I error may have been committed. |
| b. a Type I error must have been committed. |
| c. a Type II error may have been committed. |
| d. a Type II error must have been committed. |
140. When the following hypotheses are being tested at a level of significance of α
H0: μ ≥ 500
Ha: μ < 500
the null hypothesis will be rejected, if the p-value is
| a. = α/2. |
| b. ≤ α. |
| c. ≤ 1 – α/2. |
| d. > α. |
141. A grocery store has an average sales of $8000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8300 per day. From past information, it is known that the standard deviation of the population is $1200. The p-value is
| a. .9772. |
| b. .5475. |
| c. .0228. |
| d. 2.000 |
142. If the probability of a Type I error (α) is .05, then the probability of a Type II error (β) must be
| a. .025. |
| b. Cannot be computed. |
| c. .95. |
| d. .05 |
143. If the null hypothesis is rejected at the .05 level of significance, it will
| a. always not be rejected at the .10 level of significance. |
| b. sometimes not be rejected at the .10 level of significance. |
| c. sometimes be rejected at the .10 level of significance. |
| d. always be rejected at the .10 level of significance. |
144. Read the t statistic from the t distribution table and circle the correct answer. For a one-tailed test (upper tail), using a sample size of 18, and at the 5% level of significance, t =
| a. 1.740. |
| b. 2.12. |
| c. -2.12. |
| d. -1.740. |
145. The power curve provides the probability of
| a. correctly rejecting the null hypothesis. |
| b. incorrectly rejecting the null hypothesis. |
| c. correctly accepting the null hypothesis. |
| d. incorrectly accepting the null hypothesis. |
146. In hypothesis testing, the tentative assumption about the population parameter is
| a. the alternative hypothesis. |
| b. the null hypothesis. |
| c. either the null or the alternative. |
| d. neither the null nor the alternative. |
147. A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 80%. The test statistic is
| a. .05. |
| b. 2.00. |
| c. 1.25. |
| d. .80 |
148. For a given sample size in hypothesis testing,
| a. the sum of Type I and Type II errors must equal to 1. |
| b. the smaller the Type I error, the larger the Type II error will be. |
| c. the smaller the Type I error, the smaller the Type II error will be. |
| d. Type II error will not be effected by Type I error |
149. For a two-tailed test, the p-value is the probability of obtaining a value for the test statistic as
| a. unlikely as that provided by the sample. |
| b. likely as that provided by the population. |
| c. unlikely as that provided by the population. |
| d. likely as that provided by the sample. |
150. Read the z statistic from the normal distribution table and circle the correct answer. For a one-tailed test (lower tail) using α = .1020, z =
| a. -1.27. |
| b. -1.96. |
| c. -1.53. |
| d. -1.64. |
151. Read the z statistic from the normal distribution table and circle the correct answer. For a one-tailed test (upper tail) using α = .1230, z =
| a. 1.645. |
| b. 1.54. |
| c. 1.16. |
| d. 1.96 |
152. When the following hypotheses are being tested at a level of significance of α
H0: μ ≥ 500
Ha: μ < 500
the null hypothesis will be rejected, if the p-value is
| a. > α. |
| b. ≤ α. |
| c. ≤ 1 – α/2. |
| d. = α/2. |
153. The average monthly rent for one-bedroom apartments in Chattanooga has been $700. Because of the downturn in the real estate market, it is believed that there has been a decrease in the average rental. The correct hypotheses to be tested are
| a. H0: μ = 700 Ha: μ ≠ 700. |
| b. H0: μ < 700 Ha: μ ≥ 700. |
| c. H0: μ ≥ 700 Ha: μ < 700. |
| d. H0: μ > 700 Ha: μ ≤ 700 |
154. For a lower tail test, the p-value is the probability of obtaining a value for the test statistic
| a. at least as large as that provided by the sample. |
| b. at least as small as that provided by the sample. |
| c. at least as large as that provided by the population. |
| d. at least as small as that provided by the population. |
155. The following information was obtained from independent random samples taken of two populations.
Assume normally distributed populations with equal variances.
| Sample 1 | Sample 2 | |
| Sample Mean | 45 | 42 |
| Sample Variance | 85 | 90 |
| Sample Size | 10 | 12 |
The 95% confidence interval for the difference between the two population means is (use rounded standard error)
| a. -2.65 to 8.65. |
| b. -4.86 to 10.86. |
| c. -5.344 to 11.344. |
| d. -5 to 3. |
156. The following information was obtained from matched samples taken from two populations. Assume the population of differences is normally distributed.
| Individual | Method 1 | Method 2 |
| 1 | 7 | 5 |
| 2 | 5 | 9 |
| 3 | 6 | 8 |
| 4 | 7 | 7 |
| 5 | 5 | 6 |
The point estimate for the difference between the means of the two populations (Method 1 – Method 2) is
| a. 2. |
| b. 0. |
| c. -1. |
| d. -4. |
157. The following information was obtained from matched samples:
If the null hypothesis tested is H0: µd = 0, what is the test statistic for the difference between the two population means?
| a. –.48 |
| b. –.50 |
| c. –.24 |
| d. .51 |
158. The results of a recent poll on the preference of shoppers regarding two products are shown below.
| Product | Shoppers Surveyed | Shoppers Favoring This Product |
| A | 800 | 560 |
| B | 900 | 612 |
The 95% confidence interval estimate for the difference between the populations favoring the products is
| a. .6 to .7. |
| b. .046 to .066. |
| c. -.024 to .064. |
| d. -.024 to .7. |
159. Salary information regarding male and female employees of a large company is shown below.
| Male | Female | |
| Sample Size | 64 | 36 |
| Sample Mean Salary (in $1000) | 44 | 41 |
| Population Variance (σ2) | 128 | 72 |
The standard error of the difference between the two sample means is
| a. 2.0. |
| b. 4. |
| c. 4.24. |
| d. 7.46 |
160. A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information.
| Today | Five Years Ago | |
| x̄ | 82 | 88 |
| σ2 | 112.5 | 54 |
| n | 45 | 36 |
The 95% confidence interval for the difference between the two population means is
| a. -3.08 to 3.92. |
| b. -24.77 to 12.23. |
| c. -9.92 to -2.08. |
| d. -13.84 to -1.16. |
161. In a study of whether an exercise routine is effective, the weights of a random sample of individuals before they began the exercise plan and the weights of the same individuals after two months on the exercise plan are recorded. A hypothesis test is conducted to determine if the exercise plan is effective. What is the 95% confidence interval estimate of the mean of the population of differences if n = 30, d̄ = 10.5, and sd = 2.75?
| a. The means of the before and after weights must be known to compute the confidence interval. |
| b. 7.75 to 13.25 |
| c. 9.47 to 11.53 |
| d. 9.65 to 11.35 |
162. If we are interested in testing whether the proportion of items in population 1 is larger than the proportion of items in population 2, the
| a. null hypothesis should state p1 – p2 > 0. |
| b. alternative hypothesis should state p1 – p2 < 0. |
| c. null hypothesis should state p1 – p2 < 0. |
| d. alternative hypothesis should state p1 – p2 > 0. |
163. The management of a department store is interested in estimating the difference between the mean credit purchases of customers using the store’s credit card versus those customers using a national major credit card. You are given the following information.
| Store’s Card | Major Credit Card | |
| Sample size | 64 | 49 |
| Sample mean | $140 | $125 |
| Population standard deviation | $10 | $8 |
A 95% confidence interval estimate for the difference between the average purchases of all customers using the two different credit cards is
| a. 12.22 to 17.78. |
| b. 13.04 to 16.96. |
| c. 13.31 to 16.69. |
| d. 11.68 to 18.32. |
164. Production output (i.e., number of parts) for a random sample of days from two different plants is shown below.
What is the estimate of the standard deviation for the difference between the two means?
| a. 5.12 |
| b. 75 |
| c. 130.34 |
| d. 14.66 |
165. A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information.
| Today | Five Years Ago | |
| x̄ | 82 | 88 |
| σ2 | 112.5 | 54 |
| n | 45 | 36 |
The standard error of x̄1 – x̄2 is
| a. 4. |
| b. 2. |
| c. 12.9. |
| d. 9.3. |
166. The following information was obtained from matched samples taken from two populations. Assume the population of differences is normally distributed.
| Individual | Method 1 | Method 2 |
| 1 | 7 | 5 |
| 2 | 5 | 9 |
| 3 | 6 | 8 |
| 4 | 7 | 7 |
| 5 | 5 | 6 |
The point estimate for the difference between the means of the two populations (Method 1 – Method 2) is
| a. 2. |
| b. -1. |
| c. -4. |
| d. 0. |
167. The following information was obtained from matched samples taken from two populations. Assume the population of differences is normally distributed.
| Individual | Method 1 | Method 2 |
| 1 | 7 | 5 |
| 2 | 5 | 9 |
| 3 | 6 | 8 |
| 4 | 7 | 7 |
| 5 | 5 | 6 |
The 95% confidence interval for the difference between the two population means is
| a. -2.776 to 2.776. |
| b. -1.776 to 1.776. |
| c. -3.776 to 1.776. |
| d. -1.776 to 2.776 |
168. An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had in the previous year. The results are shown below.
| Under Age of 18 | Over Age of 18 |
| n1 = 500 | n2 = 600 |
| Number of accidents = 180 | Number of accidents = 150 |
We are interested in determining if the accident proportions differ between the two age groups. The pooled estimator of the population proportion is
| a. .450. |
| b. .305. |
| c. .300. |
| d. .027 |
169. The following information was obtained from independent random samples taken of two populations.
Assume normally distributed populations with equal variances.
| Sample 1 | Sample 2 | |
| Sample Mean | 45 | 42 |
| Sample Variance | 85 | 90 |
| Sample Size | 10 | 12 |
The 95% confidence interval for the difference between the two population means is (use rounded standard error)
| a. -5 to 3. |
| b. -5.344 to 11.344. |
| c. -2.65 to 8.65. |
| d. -4.86 to 10.86 |
170. A school administrator is interested in determining whether the proportion of students in the elementary school who are girls (pE) is significantly more than the proportion of students in the high school who are girls (pH). Which set of hypotheses would be most appropriate for answering the administrator’s question?
| a. H0 : pE – pH = 0 Ha : pE – pH ≠ 0 |
| b. H0 : pE – pH ≤ 0 Ha : pE – pH > 0 |
| c. H0 : pE – pH ≥ 0 Ha : pE – pH < 0 |
| d. H0: pE – pH < 0 Ha: pE – pH ≥ 0 |
171. Generally, the ________ sample procedure for inferences about two population means provides better precision than the _______ sample approach.
| a. matched, pooled |
| b. independent, pooled |
| c. matched, independent |
| d. single, independent |
172. The following information was obtained from independent random samples taken of two populations.
Assume normally distributed populations with equal variances.
| Sample 1 | Sample 2 | |
| Sample Mean | 45 | 42 |
| Sample Variance | 85 | 90 |
| Sample Size | 10 | 12 |
The degrees of freedom for the t distribution are
| a. 22. |
| b. 21. |
| c. 24. |
| d. 20 |
173. How many degrees of freedom will the t distribution have when constructing an interval estimate for the difference between the means of two populations if the two population standard deviations are unknown and assumed unequal and the samples sizes of groups 1 and 2 are n1 = 15 and n2 = 18?
| a. 33 degrees of freedom |
| b. 31 degrees of freedom |
| c. 17 degrees of freedom |
| d. Not enough information is given to compute the degrees of freedom |
174. A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information.
| Today | Five Years Ago | |
| x̄ | 82 | 88 |
| σ2 | 112.5 | 54 |
| n | 45 | 36 |
The test statistic for the difference between the two population means is
| a. -.65. |
| b. -3.0. |
| c. -1.5. |
| d. -.47 |
175. Two major automobile manufacturers have produced compact cars with engines of the same size. We are interested in determining whether or not there is a significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles. A random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive each automobile for a specified distance. The following data (in miles per gallon) show the results of the test. Assume the population of differences is normally distributed.
| Driver | Manufacturer A | Manufacturer B |
| 1 | 32 | 28 |
| 2 | 27 | 22 |
| 3 | 26 | 27 |
| 4 | 26 | 24 |
| 5 | 25 | 24 |
| 6 | 29 | 25 |
| 7 | 31 | 28 |
| 8 | 25 | 27 |
The test statistic is
| a. 1.616. |
| b. 2.096. |
| c. 1.906. |
| d. 2.256. |
176. In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the following data have been gathered.
| Downtown Store | North Mall Store | |
| Sample size | 25 | 20 |
| Sample mean | $9 | $8 |
| Sample standard deviation | $2 | $1 |
A 95% interval estimate for the difference between the two population means is
| a. .226 to 1.774. |
| b. .071 to 1.929. |
| c. 1.078 to 2.922. |
| d. 1.09 to 4.078. |
177. The management of a department store is interested in estimating the difference between the mean credit purchases of customers using the store’s credit card versus those customers using a national major credit card. You are given the following information.
| Store’s Card | Major Credit Card | |
| Sample size | 64 | 49 |
| Sample mean | $140 | $125 |
| Population standard deviation | $10 | $8 |
A point estimate for the difference between the mean purchases of all users of the two credit cards is
| a. 265. |
| b. 2. |
| c. 18. |
| d. 15. |
178. The results of a recent poll on the preference of shoppers regarding two products are shown below.
| Product | Shoppers Surveyed | Shoppers Favoring This Product |
| A | 800 | 560 |
| B | 900 | 612 |
The 95% confidence interval estimate for the difference between the populations favoring the products is
| a. .6 to .7. |
| b. -.024 to .064. |
| c. -.024 to .7. |
| d. .046 to .066. |
179. The following information was obtained from independent random samples. Suppose we are interested in testing H0: µ1 – µ2 = 15 and Ha: µ1 – µ2 ≠ 15.
What is the test statistic used in the hypothesis test for the difference between the two population means?
| a. –5.49 |
| b. .829 |
| c. 14.07 |
| d. –1.37 |
180. When each data value in one sample is paired with a corresponding data value in another sample for a sample of 35 individuals or objects and the corresponding differences are computed, what type of distribution will the difference data have?
| a. Matched pairs distribution |
| b. Exponential distribution |
| c. Uniform distribution |
| d. t distribution |
181. In testing the null hypothesis H0: μ1 – μ2 = 0, the computed test statistic is z = -1.66. The corresponding p-value is
| a. .0970. |
| b. .9515. |
| c. .0485. |
| d. .9030. |
182. The following table shows the predicted sales (in $1000s) and the actual sales (in $1000s) for six stores over a six-month period.
What is the mean of the matched samples data in the above table?
| a. 79.5 |
| b. –2.50 |
| c. 80.75 |
| d. 82 |
183. The management of a department store is interested in estimating the difference between the mean credit purchases of customers using the store’s credit card versus those customers using a national major credit card. You are given the following information.
| Store’s Card | Major Credit Card | |
| Sample size | 64 | 49 |
| Sample mean | $140 | $125 |
| Population standard deviation | $10 | $8 |
A 95% confidence interval estimate for the difference between the average purchases of all customers using the two different credit cards is
| a. 13.04 to 16.96. |
| b. 12.22 to 17.78. |
| c. 13.31 to 16.69. |
| d. 11.68 to 18.32. |
184. How many degrees of freedom will the t distribution have when constructing an interval estimate for the difference between the means of two populations if the two population standard deviations are unknown and assumed unequal and the samples sizes of groups 1 and 2 are n1 = 15 and n2 = 18?
| a. 33 degrees of freedom |
| b. 31 degrees of freedom |
| c. 17 degrees of freedom |
| d. Not enough information is given to compute the degrees of freedom. |
185. The following information was obtained from matched samples taken from two populations.
The daily production rates for a sample of workers before and after a training program are shown below. Assume the population of differences is normally distributed.
| Worker | Before | After |
| 1 | 20 | 22 |
| 2 | 25 | 23 |
| 3 | 27 | 27 |
| 4 | 23 | 20 |
| 5 | 22 | 25 |
| 6 | 20 | 19 |
| 7 | 17 | 18 |
The point estimate for the difference between the means of the two populations is
| a. -2. |
| b. -1. |
| c. 0. |
| d. 1. |
186. An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had in the previous year. The results are shown below.
| Under Age of 18 | Over Age of 18 |
| n1 = 500 | n2 = 600 |
| Number of accidents = 180 | Number of accidents = 150 |
We are interested in determining if the accident proportions differ between the two age groups. The p-value is
| a. .3. |
| b. less than .001. |
| c. .0228. |
| d. more than .10 |
187. Salary information regarding male and female employees of a large company is shown below.
| Male | Female | |
| Sample Size | 64 | 36 |
| Sample Mean Salary (in $1000) | 44 | 41 |
| Population Variance (σ2) | 128 | 72 |
At 95% confidence, the margin of error is
| a. 1.960. |
| b. 3.920. |
| c. 2.000. |
| d. 1.645. |
188. Two major automobile manufacturers have produced compact cars with engines of the same size. We are interested in determining whether or not there is a significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles. A random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive each automobile for a specified distance. The following data (in miles per gallon) show the results of the test. Assume the population of differences is normally distributed.
| Driver | Manufacturer A | Manufacturer B |
| 1 | 32 | 28 |
| 2 | 27 | 22 |
| 3 | 26 | 27 |
| 4 | 26 | 24 |
| 5 | 25 | 24 |
| 6 | 29 | 25 |
| 7 | 31 | 28 |
| 8 | 25 | 27 |
The mean of the differences is
| a. 2.0. |
| b. 2.5. |
| c. .5. |
| d. 1.5. |
189. When each data value in one sample is matched with a corresponding data value in another sample, the samples are known as
| a. independent samples. |
| b. matched samples. |
| c. pooled samples. |
| d. corresponding samples. |
190. Two independent simple random samples are taken to test the difference between the means of two populations whose standard deviations are not known, but are assumed to be equal. The sample sizes are n1 = 25 and n2 = 35. The correct distribution to use is the
| a. t distribution with 60 degrees of freedom. |
| b. t distribution with 58 degrees of freedom. |
| c. t distribution with 61 degrees of freedom. |
| d. t distribution with 59 degrees of freedom |
191. The following information was obtained from matched samples taken from two populations.
The daily production rates for a sample of workers before and after a training program are shown below. Assume the population of differences is normally distributed.
| Worker | Before | After |
| 1 | 20 | 22 |
| 2 | 25 | 23 |
| 3 | 27 | 27 |
| 4 | 23 | 20 |
| 5 | 22 | 25 |
| 6 | 20 | 19 |
| 7 | 17 | 18 |
The null hypothesis to be tested is H0: μd = 0. The test statistic is
| a. 0. |
| b. -1.96. |
| c. 1.00. |
| d. 1.77. |
192. Random samples of 100 parts from production line A had 12 parts that were defective and 100 parts from production line B had 5 that were defective. What is the test statistic for the hypothesis test of a difference between the two proportions?
| a. z = 1.96 |
| b. z = 2.16 |
| c. z = .07 |
| d. z = 1.77 |
193. In the past, 75% of the tourists who visited Chattanooga went to see Rock City. The management of Rock City recently undertook an extensive promotional campaign. They are interested in determining whether the promotional campaign actually increased the proportion of tourists visiting Rock City. The correct set of hypotheses is
| a. H0: p > .75 Ha: p ≤ .75. |
| b. H0: p ≤ .75 Ha: p > .75. |
| c. H0: p < .75 Ha: p ≥ .75. |
| d. H0: p ≥ .75 Ha: p < .75. |
194. Batteries produced by a manufacturing company have had a life expectancy of 135 hours. Because of an improved production process, the company believes that there has been an increase in the life expectancy of its batteries. A sample of 42 batteries showed an average life of 140 hours. From past information, the standard deviation of the population is known to be 24 hours. Test to determine whether there has been an increase in the life expectancy of batteries. What is the test statistic, and what conclusion can be made at the .10 level of significance?
| a. z = 2.33; Reject the null hypothesis |
| b. z = 1.35; Reject the null hypothesis |
| c. z = .088; Do not reject the null hypothesis |
| d. z = .208; Do not reject the null hypothesis |
195. Read the t statistic from the t distribution table and circle the correct answer. For a one-tailed test (upper tail), using a sample size of 18, and at the 5% level of significance, t =
| a. 1.740. |
| b. -2.12. |
| c. 2.12. |
| d. -1.740 |
196. The p-value is a probability that measures the support (or lack of support) for
| a. the null hypothesis. |
| b. either the null or the alternative hypothesis. |
| c. neither the null nor the alternative hypothesis. |
| d. the alternative hypothesis. |
197. A weatherman stated that the average temperature during July in Chattanooga is 80 degrees or less. A sample of 32 Julys is taken to test the weatherman’s statement. The correct set of hypotheses is
| a. H0: μ ≤ 80 Ha: μ > 80. |
| b. H0: μ ≠ 80 Ha: μ = 80. |
| c. H0: μ ≥ 80 Ha: μ < 80. |
| d. H0: μ < 80 Ha: μ > 80. |
198. The average monthly rent for one-bedroom apartments in Chattanooga has been $700. Because of the downturn in the real estate market, it is believed that there has been a decrease in the average rental. The correct hypotheses to be tested are
| a. H0: μ > 700 Ha: μ ≤ 700. |
| b. H0: μ < 700 Ha: μ ≥ 700. |
| c. H0: μ = 700 Ha: μ ≠ 700. |
| d. H0: μ ≥ 700 Ha: μ < 700. |
199. For a sample size of 30, changing from using the standard normal distribution to using the t distribution in a hypothesis test,
| a. Not enough information is given to answer this question. |
| b. will result in the area corresponding to the critical value being larger. |
| c. will result in the area corresponding to the critical value being smaller. |
| d. will have no effect on the area corresponding to the critical value. |
200. Read the t statistic from the t distribution table and circle the correct answer. For a one-tailed test (lower tail), using a sample size of 10, and at the 10% level of significance, t =
| a. -1.383. |
| b. -2.821. |
| c. 1.383. |
| d. 2.821. |
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