1. the Correct Hypotheses for the Test Are

Problem A The manager of a grocery store claims that the average time that customers spend in checkout lines is 20 minutes or less. A sample of 36 customers is taken. The average time spent on checkout lines for the sample is 24.6 minutes; and the sample standard deviation is 12 minutes. Conduct a hypothesis test (at 0.05 level of significance) to determine if the mean waiting time for the customer population is significantly more than 20 minutes. ANSWER QUESTIONS 1 - 4.

1. The correct hypotheses for the test are: a. H0 is:   20; Ha is:  < 20 b. H0 is:   20; Ha is:  > 20 c. H0 is:  > 20; Ha is:   20 d. H0 is:  < 20; Ha is:   20

2. The observed value of the test statistic is: a. 2.3 b. 0.38 c. -2.3 d. -0.38

3. The p-value is: a. 0.5107 b. 0.0214 c. 0.0137 d. 0.4893

4. The null hypothesis should: a. not be rejected b. be rejected c. the test is inconclusive d. none of the above answers are correct

Problem B A random sample of 16 students selected from the student body of a large university had an average age of 25 years with a standard deviation of 2 years. Conduct a hypothesis test (at 0.05 level of significance) to determine if the average age of the population of students at the university is significantly different from 24 years. Assume that the distribution of the population of ages is normal. BASED ON THIS PROBLEM - ANSWER QUESTIONS 5 THROUGH 7 BELOW.

5. The correct hypotheses for the test are: a. H0 is:   24; Ha is:  < 24 b. H0 is:   24; Ha is:  > 24 c. H0 is: is not equal to 24; Ha is:  = 24 d. H0 is:  = 24; Ha is: is not equal to 24. 6. The observed value of the test statistic is: a. 1.96 b. 2.00 c. 1.645 d. 0.05

7. It can be concluded that the mean age of the student population is: a. not significantly different from 24 b. significantly different from 24 c. significantly less than 24 d. significantly less than 25

Problem C: Use the following to answer questions 8-9: A major airline company is concerned that its proportion of late arrivals has substantially increased in the past month. Historical data shows that on the average 18% of the company airplanes have arrived late. In a random sample of 1,240 airplanes, 310 airplanes have arrived late. If we are conducting a hypothesis test of a single proportion to determine if the proportion of late arrivals has increased:

8. What is the correct statement of null and alternative hypothesis? A) H0: p < .18 and HA: p  .18 B) H0: p  .18 and HA: p > .18 C) H0: p = .18 and HA: p ≠ .18 D) H0: p  .18 and HA: p < .18 E) H0: p  .20 and HA: p > .20

9. What is the value of the calculated test statistic? A) Z = 6.416 B) Z = 3.208 C) Z = -3.208 D) Z = -6.416 E) Z = 1.833

10. A research and development specialist for a company that produces a certain soft drink wants to determine if a 12-ounce can of a certain brand of soft drink contains 120 calories as the labeling indicates. Using a random sample of 9 cans, the specialist determined that the average calories per can is 124 with a standard deviation of 6 calories. At the .05 level of significance, is there sufficient evidence that the average calorie content of a 12-ounce can is different from 120 calories? Assume that the number of calories per can is normally distributed.

A) No, t = .667 and .667 is less than the critical value of 2.306. B) No, t = 2 and 2 is less than the critical value of 2.306. C) Yes, t = 2.5 and 2.5 is greater than the critical value of 2.262. D) Yes, t = 2 and 2 is greater than the critical value of 1.833. E) No, t = .667 and .667 is less than the critical value of 2.262. ------Problem D: A statistical test using ANOVA gives the following results: SSTR = 6,750; SSE = 8000; nT = 20

H0:  1 =  2 =  3 =  4 Ha: Not all population means are equal ANSWER QUESTIONS 11 THROUGH 15 BELOW.

11. The treatment mean square value (MSTR) is: a. 400 b. 500 c. 1687.5 d. 2250

12. The error mean square (MSE) is: a. 400 b. 500 c. 1687.5 d. 2250

13. The computed (i.e., observed) F-value is: a. 0.22 b. 0.84 c. 4.22 d. 4.5

14. The null hypothesis is to be tested at the 5% level of significance. The critical F-value is: a. 2.87 b. 3.24 c. 4.08 d. 8.7

15. The null hypothesis should: a. not be rejected b. be rejected c. the test is inconclusive d. none of the above answers are correct ------Problem E: A company wants to measure the relationship between its employee productivity (measured in output/employee) and the number of employees. Sample data for the last four months are shown below. Use simple linear regression to estimate this relationship.

Independent Variable Dependent Variable Number of Employees Employee Productivity 15 5 12 7 10 9 7 11

ANSWER QUESTIONS 16 THROUGH 19 BELOW.

16. The least squares estimate of the slope b1 is: a. -0.7647 b. -0.13 c. 21.4 d. 16.41

17. The least squares estimate of the intercept b0 is: a. -7.647 b. -0.13 c. 21.4 d. 16.41

18. The estimated employee productivity when the number of employees is 5 is: a. 78 b. 12.59 c. 5.8 d. 32.6 19. If the sample covariance is -8.67; estimate the coefficient of correlation between the number of employees and employee productivity: a. –0.997 b. 0.997 c. 1.23 d. 1.02 ------Problem F: Consumer Research is an independent agency that is collecting data on annual income (INCOME) and household size (SIZE), to predict annual credit card charges. It runs a regression analysis on the data and an incomplete MS Excel output is shown below. ANSWER QUESTIONS 20 THROUGH 30 BELOW.

Regression Statistics Multiple R 0.88038239 R Square Adjusted R Square Standard Error 510.495493 Observations

ANOVA df SS MS F Significance F Regression 2 17960368.3 3.31446E-07 Residual 20 260605.648 Total 22

Coefficients Standard t Stat P-value Lower 95% Error Intercept 352.694714 4.15578994 0.00048872 730.0172039 INCOME 25.062956 8.47147285 2.95851223 0.00776734 7.391781505 SIZE 408.400776 71.808401 1.447E-05 258.6111461

20. The sample size is: a. 23 b. 22 c. 20 d. 21

21. The coefficient of determination is: a. 0.88 b. 0.775 c. 0.92 d. -0.38

22. The Sum of Squares for Error (i.e., Residual) is: a. 17960368.3 b. 5212112.97 c. 23172481.3 d. 260605.648

23. The Sum of Squares for Total (SST) is: a. 17960368.3 b. 5212112.97 c. 23172481.3 d. 260605.648

24. The Mean Square for Regression is a. 17960368.3 b. 5212112.97 c. 260605.648 d. 8980184.17

25. The observed or computed F-value is: a. 34.459 b. 0.029 c. 3.445 d. 0.29 26. The hypothesis to be tested is: H0: B1 = B2 = 0 Ha: At least one of the B is not equal to 0. The hypothesis is to be tested at the 5% level of significance. The null hypothesis is: a. not rejected b. rejected c. the test is inconclusive d. none of the above answers are correct

27. The hypothesis to be tested is: H0: B1 = 0 Ha: B1 ╪ 0

The hypothesis is to be tested at the 1% level of significance. The null hypothesis is: a. not rejected b. rejected c. the test is inconclusive d. none of the above answers are correct

28. The estimate of the intercept b0 is: a. 10010.2 b. 2810.3 c. 1465.5 d. 2641.5

29. The observed or computed t-stat (i.e., t-value) for the independent variable SIZE is: a. 2.96 b. 3.445 c. 4.16 d. 5.687

30. What is the estimated annual credit charges if INCOME = 20, and SIZE = 3? a. 9700 b. 12600 c. 3189 d. 5300 ------Problem G: Last 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 number of students in each classification. Freshmen 83 Sophomores 68 Juniors 85 Seniors 64

We want to know if there has been a significant change in the proportions of student classifications between the two years. ANSWER QUESTIONS 31 THROUGH 34 BELOW.

31. The expected number of freshmen in this year is: a. 83 b. 90 c. 30 d. 10

32. The number of degrees of freedom is: a. 4 b. 2 c. 3 d. 1 33. The hypothesis is to be tested at the 5% level of significance. The critical chi-square value from the table equals: a. 1.645 b. 1.96 c. 2.75 d. 7.815

34. If the chi-square value that is calculated equals 1.6615, then the null hypothesis is: a. not rejected b. rejected c. the test is inconclusive d. none of the above answers are correct

Problem H: Use the following Excel Output to answer questions 35-39:

Source Sum of Squares d.f. Between Groups 213.88125 3 Within Groups 11.208333 20 Total 225.0895 23

35. Consider the above one-way ANOVA table. What is the treatment mean square?

A) 71.297 B) 0.5604 C) 1.297 D) 213.881 E) 9.7

36. Consider the above one-way ANOVA table. What is the mean square error?

A) 71.297 B) 0.5604 C) 1.297 D) 213.8810 E) 9.7

37. Consider the above one-way ANOVA table. How many groups (treatment levels) are included in the study? A) 3 B) 4 C) 6 D) 20 E) 24

38. Consider the above one-way ANOVA table. If there are equal number of observations in each group, then each group (treatment level) consists of ______observations.

A) 3 B) 4 C) 6 D) 20 E) 24

39. What is the critical F-value at an alpha of 0.05?

A) 3.1 B) 3.86 C) 14.17 D) 4.94 E) 8.66 Problem I: Use the following to answer questions 40-42:

The following results were obtained from a simple regression analysis:

Yˆ = 37.2895 – (1.2024) * X r = – 0.6774

40. For each unit change in X (independent variable), the estimated change in Y (dependent variable) is equal to: A) –1.2024 B) 0.6774 C) 37.2895 D) 0.2934

41. When X (independent variable) is equal to zero, the estimated value of Y (dependent variable) is equal to: A) –1.2024 B) 0.6774 C) 37.2895 D) 0.2934

42. ______is the proportion of the variation explained by the simple linear regression model: A) 0.8230 B) 0.6774 C) 0.4589 D) 0.2934 E) 37.2895 ------43. Given the following information about a hypothesis test of the difference between two means based on independent random samples, which one of the following is the correct rejection region at a significance level of .05?

HA: A > B, X 1 = 12, X 2 = 9, s1 = 4, s2 = 2, n1 = 13, n2 = 10.

A) Reject H0 if Z > 1.96 B) Reject H0 if Z > 1.645 C) Reject H0 if t > 1.721 D) Reject H0 if t > 2.08 E) Reject H0 if t > 1.734

44. Given the following information about a hypothesis test of the difference between two means based on independent random samples, what is the calculated value of the test statistic? Assume that the samples are obtained from normally distributed populations having equal variances. (SEE Pages 389 & 390 IN YOUR TEXT BOOK) HA: A > B, X 1 = 12, X 2 = 9, s1 = 5, s2 = 3, n1 = 13, n2 = 10.

A) –1.674 B) 1.5 C) 2.823 D) 1.78 E) 1.063 ------Problem J: Use the following to answer questions 45-46: Consider the following partial computer output for a multiple regression model.

Predictor Coefficient (bi) Standard Dev (sb) Constant 99.3883 X1 –0.007207 0.0031 X2 0.0011336 0.00122 X3 0.9324 0.373 Analysis of Variance Source df SS Regression 4 32 Error (residual) 17 8

45. How many observations were taken? A) 21 B) 17 C) 22 D) 20 E) 4

46. What is the value of R2? A) 35% B) 76.95% C) 80% D) 77% E) 25% ------Problem K: Business travelers were asked to rate Miami Airport (on a scale of 1-10). Similarly business travelers were asked to rate Los Angeles airport. A hypothesis test (at alpha = 0.05) is conducted for any difference in the population means in the ratings. The Excel output is shown below. Use the following to answer questions 47- 48: t-Test: Two-Sample Assuming Unequal Variances Miami Los Angeles Mean 6.34 6.72 Variance 4.677959184 5.63428571 Observations 50 50 Hypothesized Mean Difference 0 df 97 t Stat –0.836742811 P(T<=t) one-tail 0.202396923 t Critical one-tail 1.660714588 P(T<=t) two-tail 0.404793846 t Critical two-tail 1.984722076

47. The Excel Output indicates that the null hypothesis is: a. not rejected b. rejected c. the test is inconclusive d. none of the above answers are correct

48. A 95% confidence interval of the difference between the mean ratings is: a. – 0.52 to 1.25 b. 1.67 to 2.43 c. –0.51 to 1.27 d. –1.28 to 0.52 e. –2.43 to 1.67 49. A simple random sample of 100 observations was taken from a large population. The sample mean was 320 and the population standard deviation is 200. The standard error of the mean is: a. 14.14 b. 22 c. 11 d. 20

50. Random samples of size 100 are taken from a population whose mean is 300 and standard deviation is 40. The expected value of x-bar and standard error of the mean are: a. 300 and 40 b. 100 and 40 c. 300 and 4 d. 300 and 3

51. A population has a mean of 80 and a standard deviation of 10. A sample of 100 observations is taken. The probability that the sample mean is larger than 78 is: a. 0.9772 b. 0.4772 c. 0.0228 d. 0.3413

52. A population has a mean of 180 and a variance of 1600. A sample of 100 observations is taken. The probability that the sample mean will be between 183 and 186 is: a. 0.75 b. 0.1598 c. 0.4332 d. 0.2734

53. A population has a variance of 22. What should be the sample size so that there is a 95.4% probability that a sample mean will be within 3 units of the population mean: a. 0.4772 b. 506 c. 10 d. 100

54. Random samples of size 40 are taken from an infinite population whose population proportion is 0.6. The standard error of the proportion is: a. 0. 24 b. 0.006 c. 0.0774 d. 0.02445

55. A sample of 100 observations will be taken from an infinite population. The population proportion equals 0.2. The probability that the sample proportion will be more than 0.23 is: a. 0.7734 b. 0.75 c. 0.2734 d. 0.2266

56. The t-value at a 95% confidence level and 25 degrees of freedom is: a. 1.711 b. 2.060 c. 1.708 d. 2.064

57. A random sample of 16 statistics exams was taken. The average score in the sample was 76.2 with a variance of 144. The 99% confidence interval for the population average examination score is: a. 49.67 to 102.72 b. 73.99 to 78.41 c. 67.36 to 85.04 d. 77.44 to 94.96

58. The sample size needed to provide a margin of error of 2 or less with a 95% probability when the population standard deviation equals 11 is: a. 10 b. 11 c. 125 d. 117 Problem I: A local university wants to estimate the average time/week spent on computer laboratory terminals by each student. Data was collected for a sample of 81 students. The sample mean is 9 hours and standard deviation is 1.2 hours.

ANSWER QUESTIONS 49 THROUGH 51 BELOW.

59. The standard error of the mean is: a. 7.5 b. 0.014 c. 0.16 d. 0.133

60. With 0.95 probability the margin of error is approximately: a. 0.26 b. 1.96 c. 1.21 d. 1.64

61. The 95% confidence interval is: a. 7.04 to 11.096 hours b. 7.36 to 10.64 hours c. 7.8 to 10.2 hours d. 8.74 to 9.26 hours ------62. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent (more than 90 days overdue). For this quarter, the auditing staff randomly selected 400 customer accounts and found that 80 of these accounts were delinquent. What is the 95% confidence interval for the proportion of all delinquent customer accounts at this manufacturing company? A) .1608 to .2392 B) .1992 to .2008 C) .1671 to .2329 D) .1485 to .2515 E) .1714 to .2286

63. In a manufacturing process, we are interested in measuring the average length of a certain type of bolt. Past data indicates that the standard deviation is .25 inches. How many bolts should be sampled in order to make us 95% confident that the sample mean bolt length is within .02 inches of the true mean bolt length? A) 25 B) 49 C) 423 D) 601 E) 1225

64. A plant quality control manager wishes to estimate the strength of a cable wire. A random sample of 64 wires is tested. The strength of the cable wire is measured in pounds per square inch. Based on the sample results, the sample mean is 238.4 pounds and the sample standard deviation is 40 pounds. What is the 99% confidence interval? A) 227.96 to 248.84 B) 237.03 to 239.77 C) 225.53 to 251.28 D) 236.4 to 240.4 E) 229.21 to 247.59 65. According to a recent major survey, 70% of the respondents indicated that there is too much violence on TV. For a random sample of 40 people, what is the 95% confidence interval of the proportion of people who feel that there is too much violence on TV? A) .70  .12 B) .70  .03 C) .70  .90 D) .70  .14 E) .70  .07

Problem J: Based on the above portion of an Excel output, answer questions 56-58.

Mean 6.525 Standard Error Median 6.45 Mode 6 Standard Deviation 0.5437 Sample Variance Count 20

Based on the above portion of an Excel output, answer questions 28-30.

66. The standard error is: a. 0.5437 b. 0.2236 c. 0.0272 d. 0.1215

67. The Margin of Error at a 95% confidence level is: a. 0.254 b. 1.138 c. 0.057 d. 0.468

68. The confidence interval is: A) 6.45  0.5437 B) 6.525  0.057 C) 6.525  0.254 D) 6.0  1.138 E) 6.525  0.468