- 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. |
2. 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. |
3. 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. |
4. 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. |
5. 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. |
6. 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 |
7. 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. |
8. 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. |
9. 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. |
10. 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 |
11. 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. |
12. then the probability of a Type II error (β) must be
| a. .95. |
| b. .05. |
| c. .025. |
| d. Cannot be computed. |
13. 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. |
14. 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. |
15. 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. |
16. 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 |
17. 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. |
18. 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. |
19. 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. |
20. 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 |
21. 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. |
22. 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. |
23. 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. |
24. 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 |
25. 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. |
26. 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. |
27. 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. |
28. 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. |
29. 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. |
30. 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. |
31. 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. |
32. 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. |
33. 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. |
34. 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. |
35. 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. |
36. 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. |
37. 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 |
38. 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. |
39. 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. |
40. 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. |
41. 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. |
42. 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 |
43. 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. |
44. 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. |
45. 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 |
46. 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 |
47. 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. |
48. 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. |
49. 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 |
50. 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. |
51. 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. |
52. 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. |
53. 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. |
54. 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. |
55. 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 |
56. 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 |
57. 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. |
58. 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. |
59. 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. |
60. 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. |
61. 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 |
62. 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. |
63. 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. |
64. 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 |
65. 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 |
66. 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. |
67. 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 |
68. 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. |
69. 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. |
70. 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. |
71. 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. |
72. 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. |
73. 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 |
74. 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 |
75. 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. |
76. 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. |
77. 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 |
78. 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. |
79. 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 |
80. 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. |
81. 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. |
82. 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. |
83. 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. |
84. 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. |
85. 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. |
86. 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 not 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 rejected. |
87. If the cost of making a Type I error is high, a smaller value should be chosen for the
| a. confidence coefficient. |
| b. level of significance. |
| c. critical value. |
| d. test statistic |
88. Read the z statistic from the normal distribution table and circle the correct answer. For a one-tailed test (upper tail) at α = .0630, z =
| a. 1.645. |
| b. 1.50. |
| c. 1.53. |
| d. 1.96. |
89. If the cost of making a Type I error is high, a smaller value should be chosen for the
| a. confidence coefficient. |
| b. level of significance. |
| c. test statistic. |
| d. critical value. |
90. 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. |
91. Your investment executive claims that the average yearly rate of return on the stocks she recommends is at least 10.0%. You plan on taking a sample to test her claim. The correct set of hypotheses is
| a. H0: μ < 10.0% Ha: μ ≥ 10.0%. |
| b. H0: μ ≥ 10.0% Ha: μ < 10.0%. |
| c. H0: μ > 10.0% Ha: μ ≤ 10.0%. |
| d. H0: μ ≤ 10.0% Ha: μ > 10.0%. |
92. The sum of the values of α and β
| a. is always 1. |
| b. is always .5. |
| c. gives the probability of taking the correct decision. |
| d. is not needed in hypothesis testing |
93. 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 population. |
| b. unlikely as that provided by the sample. |
| c. likely as that provided by the population. |
| d. likely as that provided by the sample |
94. 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. Using α = .05, it can be concluded that the population mean age is
| a. not significantly different from 24. |
| b. significantly more than 24. |
| c. significantly less than 24. |
| d. significantly different from 24 |
95. If the level of significance of a hypothesis test is raised from .01 to .05, the probability of a Type II error
| a. will decrease. |
| b. will also increase from .01 to .05. |
| c. will not change. |
| d. will increase. |
96. 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. |
97. 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 |
98. The probability of making a Type I error is denoted by
| a. β. |
| b. α. |
| c. 1 – β. |
| d. 1 – α. |
99. 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 value of the test statistic is
| a. 2.00. |
| b. 80.00. |
| c. .25. |
| d. 8.25. |
100. 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 p-value is
| a. .2112. |
| b. .025. |
| c. .05. |
| d. .1056. |
101. 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. Using α = .05, it can be concluded that the population mean age is
| a. significantly less than 24. |
| b. significantly different from 24. |
| c. significantly more than 24. |
| d. not significantly different from 24. |
102. If the cost of making a Type I error is high, a smaller value should be chosen for the
| a. confidence coefficient. |
| b. level of significance. |
| c. critical value. |
| d. test statistic. |
103. An assumption made about the value of a population parameter is called a(n)
| a. error. |
| b. conclusion. |
| c. hypothesis. |
| d. probability |
104. Read the z statistic from the normal distribution table and circle the correct answer. For a two-tailed test using α = .0160, z =
| a. 2.41. |
| b. 1.96. |
| c. 1.14. |
| d. .86. |
105. The critical value of t for a two-tailed test with 6 degrees of freedom using α = .05 is
| a. 1.985. |
| b. 2.365. |
| c. 2.447. |
| d. 1.943 |
106. Which of the following approaches cannot be used to perform a two-tailed hypothesis test about μ?
| a. Compare the p-value to the value of α. |
| b. Compare the confidence interval estimate of μ to the hypothesized value of μ. |
| c. Compare the level of significance to the confidence coefficient. |
| d. Compare the value of the test statistic to the critical value |
107. When the null hypothesis is rejected, it is
| a. not possible a Type I error has occurred. |
| b. possible either a Type I or a Type II error has occurred. |
| c. possible a Type II error has occurred. |
| d. possible a Type I error has occurred. |
108. When the p-value is used for hypothesis testing, the null hypothesis is rejected if
| a. α < p-value. |
| b. p-value = x̄. |
| c. p-value < ζ. |
| d. p-value ≤ α. |
109. The manager of a laptop computer dealership is considering a new bonus plan in order to increase sales. Currently, the mean sales rate per salesperson is five laptops per week. The correct set of hypotheses for testing the effect of the bonus plan is
| 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. |
110. In hypothesis tests about a population proportion, p0 represents the
| a. observed sample proportion. |
| b. observed p-value. |
| c. probability that H0 is correct. |
| d. hypothesized population proportion. |
111. If the null hypothesis is rejected at the 5% level of significance, it
| a. will always be rejected at the 1% level. |
| b. will never be tested at the 1% level. |
| c. will always not be rejected at the 1% level. |
| d. may be rejected or not rejected at the 1% level. |
112. 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. -1.725 and 1.725. |
| b. -2.093 and 2.093. |
| c. -1.729 and 1.729. |
| d. -2.086 and 2.086. |
113. In a lower tail hypothesis test situation, the p-value is determined to be .2. If the sample size for this test is 51, the t statistic has a value of
| a. -.849. |
| b. -1.299. |
| c. .849. |
| d. 1.299. |
114. For a two-tailed hypothesis test about μ, we can use any of the following approaches except
| a. compare the p-value to the value of α. |
| b. compare the value of the test statistic to the critical value. |
| c. compare the level of significance to the confidence coefficient. |
| d. compare the confidence interval estimate of μ to the hypothesized value of μ. |
115. 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. -2.821. |
| b. 2.821. |
| c. -1.383. |
| d. 1.383 |
116. 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. |
117. In a one-tailed hypothesis test (lower tail), the test statistic is determined to be -2. The p-value for this test is
| a. .4772. |
| b. .0228. |
| c. .0056. |
| d. .5228. |
118. 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. |
119. Read the z statistic from the normal distribution table and circle the correct answer. For a two-tailed test using α = .1388, z =
| a. .86. |
| b. 1.09. |
| c. 1.96. |
| d. 1.48. |
120. When the null hypothesis is rejected, it is
| a. possible a Type II error has occurred. |
| b. possible a Type I error has occurred. |
| c. possible either a Type I or a Type II error has occurred. |
| d. not possible a Type I error has occurred. |
121. 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. -2.821. |
| d. -1.383. |
122. Given the following information,
n = 49, x̄ = 50, s = 7
H0: μ ≥ 52
Ha: μ < 52
the test statistic is
| a. 1. |
| b. -2. |
| c. 2. |
| d. -1 |
123. 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 = .208; Do not reject the null hypothesis |
| c. z = .088; Do not reject the null hypothesis |
| d. z = 1.35; Reject the null hypothesis |
124. The p-value is
| a. a probability. |
| b. the same as the z statistic. |
| c. a distance. |
| d. a sample statistic. |
125. In hypothesis testing, the critical value is
| a. the same as the p-value. |
| b. the probability of a Type I error. |
| c. the probability of a Type II error. |
| d. a number that establishes the boundary of the rejection region. |
126. 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 |
What is the conclusion that can be reached about the difference in the average final examination scores between the two classes? (Use a .05 level of significance.)
| a. There is no statistically significant difference in the average final examination scores between the two classes. |
| b. The students who enrolled in statistics today are the same students who enrolled five years ago. |
| c. It is impossible to make a decision on the basis of the information given. |
| d. There is a statistically significant difference in the average final examination scores between the two classes. |
127.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 point estimate for the difference between the two population proportions in favor of this product is
| a. .68. |
| b. .02. |
| c. .44. |
| d. .07. |
128. 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. .300. |
| b. .027. |
| c. .450. |
| d. .305. |
129. 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. 1. |
| b. 0. |
| c. -2. |
| d. -1. |
130. 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. 20. |
| b. 24. |
| c. 21. |
| d. 22. |
131. 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 |
At α = .10, the null hypothesis
| a. should not be tested. |
| b. should not be rejected. |
| c. should be revised. |
| d. should be rejected. |
132. 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. -.47. |
| b. -.65. |
| c. -3.0. |
| d. -1.5. |
133. 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. .0228. |
| c. more than .10. |
| d. less than .001. |
134. 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 p-value for the difference between the two population means is
| a. .0026. |
| b. .4987. |
| c. .0013. |
| d. .9987 |
135. 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. -13.84 to -1.16. |
| b. -3.08 to 3.92. |
| c. -24.77 to 12.23. |
| d. -9.92 to -2.08. |
136. 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. -1.776 to 2.776. |
| b. -2.776 to 2.776. |
| c. -1.776 to 1.776. |
| d. -3.776 to 1.776. |
137. 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 point estimate for the difference between the means of the two populations is
| a. -6. |
| b. 9. |
| c. -9. |
| d. 58.5. |
138. Generally, the ________ sample procedure for inferences about two population means provides better precision than the _______ sample approach.
| a. single, independent |
| b. matched, pooled |
| c. matched, independent |
| d. independent, pooled |
139. 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 point estimate for the difference between the two population proportions in favor of this product is
| a. .02. |
| b. .07. |
| c. .68. |
| d. .44. |
140. 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 |
At 95% confidence, the margin of error is
| a. 3.32. |
| b. 1.96. |
| c. 15. |
| d. 1.694. |
141. When each data value in one sample is matched with a corresponding data value in another sample, the samples are known as
| a. corresponding samples. |
| b. pooled samples. |
| c. independent samples. |
| d. matched samples. |
142. 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. alternative hypothesis should state p1 – p2 < 0. |
| d. null hypothesis should state p1 – p2 < 0. |
143. 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. 2. |
| b. 265. |
| c. 15. |
| d. 18. |
144. he 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 |
At 95% confidence, the margin of error is
| a. .064. |
| b. .044. |
| c. .025. |
| d. .0225 |
145. In hypothesis tests about p1 – p2, the pooled estimator of p is a(n)
| a. weighted average of p̄1 and p̄2. |
| b. geometric average of p̄1 and p̄2. |
| c. simple average of p̄1 and p̄2. |
| d. exponential average of p̄1 and p̄2. |
146. 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. 18. |
| b. 15. |
| c. 2. |
| d. 265 |
147. 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. |
148. A researcher is interested in determining whether the mean of group 1 is 10 units larger than the mean of group 2. Which of the following represents the alternative hypothesis for testing the researcher’s theory?
| a. Ha: µ1 – µ2 > 10 |
| b. Ha: µ 1 – µ2 ≤ 10 |
| c. Ha: µ1 = µ2 |
| d. Ha:µ1 – µ2 = 0 |
149. The results of a recent poll on the preference of teenagers regarding the types of music they listen to are shown below.
| Music Type | Teenagers Surveyed | Teenagers Favoring This Type |
| Pop | 800 | 384 |
| Rap | 900 | 450 |
The point estimate of the difference between the two population proportions is
| a. .048. |
| b. -.5. |
| c. -.02. |
| d. .52. |
150. Regarding inferences about the difference between two population means, the sampling design that uses a pooled sample variance in cases of equal population standard deviations is based on
| a. research samples. |
| b. conditional samples. |
| c. pooled samples. |
| d. independent samples. |
151. 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. -1.776 to 2.776. |
| b. -2.776 to 2.776. |
| c. -3.776 to 1.776. |
| d. -1.776 to 1.776. |
152. 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. .305. |
| b. .300. |
| c. .450. |
| d. .027. |
153. 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. less than .001. |
| b. .3. |
| c. more than .10. |
| d. .0228 |
154. The local cable company is interested in determining whether or not the proportion of subscribers has increased during the past year. A random sample of households selected last year is compared with a random sample of households selected this year. Results are summarized below.
What is the value of the pooled estimate of p?
| a. .708 |
| b. .657 |
| c. .085 |
| d. .23 |
155. Of the two production methods, a company wants to identify the method with the smaller population mean completion time. One sample of workers is selected and each worker first uses one method and then uses the other method. The sampling procedure being used to collect completion time data is based on
| a. independent samples. |
| b. matched samples. |
| c. pooled samples. |
| d. worker samples. |
156. 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 point estimate for the difference between the two population proportions in favor of this product is
| a. .68. |
| b. .44. |
| c. .07. |
| d. .02. |
157. In order to determine whether or not there is a significant difference between the mean hourly wages paid by two companies (of the same industry), the following data have been accumulated.
| Company A | Company B | |
| Sample size | 80 | 60 |
| Sample mean | $16.75 | $16.25 |
| Population standard deviation | $1.00 | $.95 |
The test statistic is
| a. 3.01. |
| b. 2.75. |
| c. 1.645. |
| d. .098. |
158. In hypothesis tests about p1 – p2, the pooled estimator of p is a(n)
| a. exponential average of p̄1 and p̄2. |
| b. weighted average of p̄1 and p̄2. |
| c. simple average of p̄1 and p̄2. |
| d. geometric average of p̄1 and p̄2 |
159. The local cable company is interested in determining whether or not the proportion of subscribers has increased during the past year. A random sample of households selected last year is compared with a random sample of households selected this year. Results are summarized below.
What is the value of the pooled estimate of p?
| a. .657 |
| b. .708 |
| c. .23 |
| d. .085 |
160. 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 |
If you are interested in testing whether or not the population average salary of males is significantly greater than that of females, the test statistic is
| a. 2.0. |
| b. 1.645. |
| c. 1.5. |
| d. 1.96 |
161. 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 |
162. 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 standard error of p̄1 – p̄2 is
| a. .025. |
| b. .0225. |
| c. .68. |
| d. .044 |
163. To compute an interval estimate for the difference between the means of two populations, the t distribution
| a. is not restricted to small sample situations. |
| b. can be applied when the populations have equal means. |
| c. can be applied only when the populations have equal standard deviations. |
| d. is restricted to small sample situations. |
164. The sampling distribution of p̄1 – p̄2 is approximated by a
| a. t distribution with n1 + n2 – 1 degrees of freedom. |
| b. p̄1 – p̄2 distribution. |
| c. normal distribution. |
| d. t distribution with n1 + n2 degrees of freedom. |
165. The standard error of x̄1 – x̄2 is the
| a. variance of the sampling distribution of x̄1 – x̄2. |
| b. standard deviation of the sampling distribution of x̄1 – x̄2. |
| c. margin of error of x̄1 – x̄2. |
| d. pooled estimator of x̄1 – x̄2 |
166. 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. pooled samples. |
| c. matched samples. |
| d. corresponding samples. |
167. To construct an interval estimate for the difference between the means of two populations when the standard deviations of the two populations are unknown and it can be assumed the two populations have equal variances, we must use a t distribution with (let n1 be the size of sample 1 and n2 the size of sample 2)
| a. (n1 + n2 – 1) degrees of freedom. |
| b. (n1 – n2 + 2) degrees of freedom. |
| c. (n1 + n2) degrees of freedom. |
| d. (n1 + n2 – 2) degrees of freedom |
168. 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. 1.09 to 4.078. |
| d. 1.078 to 2.922. |
169. 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 61 degrees of freedom. |
| c. t distribution with 58 degrees of freedom. |
| d. t distribution with 59 degrees of freedom. |
170. In hypothesis tests about p1 – p2, the pooled estimator of p is a(n)
| a. simple average of p̄1 and p̄2. |
| b. weighted average of p̄1 and p̄2. |
| c. exponential average of p̄1 and p̄2. |
| d. geometric average of p̄1 and p̄2. |
171. 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 point estimate for the difference between the means of the two populations is
| a. 2. |
| b. 15. |
| c. 3. |
| d. 0. |
172. 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 95% confidence interval for the difference between the means of the two populations is
| a. -2 to 2. |
| b. 0 to 6.92. |
| c. -1.96 to 1.96. |
| d. -.92 to 6.92. |
173. 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. |
174. 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. |
175. 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 point estimate for the difference between the two population means is
| a. 2. |
| b. 1. |
| c. 3. |
| d. 4. |
176. A sample of 1400 items had 280 defective items. For the following hypothesis test,
H0: p ≤ .20
Ha: p > .20
the test statistic is
| a. .20. |
| b. .28. |
| c. .14. |
| d. zero. |
177. Read the t statistic from the t distribution table and circle the correct answer. For a one-tailed test (upper tail) with a sample size of 26 and at the .10 level, t =
| a. -1.316. |
| b. 1.740. |
| c. -1.740. |
| d. 1.316. |
178. 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. |
179. Which of the following does not need to be known in order to compute the p-value?
| a. the probability distribution of the test statistic |
| b. knowledge of whether the test is one-tailed or two-tailed |
| c. the level of significance |
| d. the value of the test statistic |
180. A student believes that the average grade on the final examination in statistics is at least 85. She plans on taking a sample to test her belief. The correct set of hypotheses is
| a. H0: μ ≥ 85 Ha: μ < 85. |
| b. H0: μ < 85 Ha: μ ≥ 85. |
| c. H0: μ ≤ 85 Ha: μ > 85. |
| d. H0: μ > 85 Ha: μ ≤ 85 |
181. Read the t statistic from the t distribution table and circle the correct answer. For a two-tailed test with a sample size of 20 and using α = .20, t =
| a. 2.539. |
| b. 1.325. |
| c. 1.328. |
| d. 2.528. |
182. 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. |
183. The p-value is a probability that measures the support (or lack of support) for
| a. the alternative hypothesis. |
| b. the null hypothesis. |
| c. either the null or the alternative hypothesis. |
| d. neither the null nor the alternative hypothesis. |
184. Read the z statistic from the normal distribution table and circle the correct answer. For a one-tailed test (upper tail) at α = .0630, z =
| a. 1.96. |
| b. 1.50. |
| c. 1.53. |
| d. 1.645 |
185. Generally, the ________ sample procedure for inferences about two population means provides better precision than the _______ sample approach.
| a. matched, independent |
| b. single, independent |
| c. independent, pooled |
| d. matched, pooled |
186. The results of a recent poll on the preference of teenagers regarding the types of music they listen to are shown below.
| Music Type | Teenagers Surveyed | Teenagers Favoring This Type |
| Pop | 800 | 384 |
| Rap | 900 | 450 |
The 95% confidence interval for the difference between the two population proportions is
| a. .48 to .5. |
| b. .028 to .068. |
| c. .5 to .52. |
| d. -.068 to .028. |
187. 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. -1.96. |
| b. 1.00. |
| c. 0. |
| d. 1.77. |
188. 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. 14.66 |
| b. 5.12 |
| c. 75 |
| d. 130.34 |
189. In testing the null hypothesis H0: μ1 – μ2 = 0, the computed test statistic is z = -1.66. The corresponding p-value is
| a. .0485. |
| b. .9030. |
| c. .0970. |
| d. .9515. |
190. Two independent simple random samples are taken to test the difference between the means of two populations whose variances are not known, but are assumed to be equal. The sample sizes are n1 = 32 and n2 = 40. The correct distribution to use is the
| a. t distribution with 72 degrees of freedom. |
| b. t distribution with 71 degrees of freedom. |
| c. t distribution with 73 degrees of freedom. |
| d. t distribution with 70 degrees of freedom |
191. 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. –1.37 |
| b. 14.07 |
| c. .829 |
| d. –5.49 |
192. 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. 2. |
| b. 12.9. |
| c. 4. |
| d. 9.3. |
193. 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. t distribution |
| b. Matched pairs distribution |
| c. Uniform distribution |
| d. Exponential distribution |
194. 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. -3.776 to 1.776. |
| b. -1.776 to 1.776. |
| c. -2.776 to 2.776. |
| d. -1.776 to 2.776 |
195. Two approaches to drawing a conclusion in a hypothesis test are
| a. p-value and critical value. |
| b. one-tailed and two-tailed. |
| c. Type I and Type II. |
| d. null and alternative. |
196. 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 test statistic is
| a. .05. |
| b. 1.96. |
| c. 2.00. |
| d. 1.65 |
197. For a given sample size in hypothesis testing,
| a. the smaller the Type I error, the larger the Type II error will be. |
| b. the sum of Type I and Type II errors must equal to 1. |
| 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. |
198. 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 |
199. 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. |
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