Chi-Square Goodness-of-Fit Test and Hypothesis Testing: Concepts and Applications, Exams of Nursing

Download Chi-Square Goodness-of-Fit Test and Hypothesis Testing: Concepts and Applications and more Exams Nursing in PDF only on Docsity! http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 1 of 72 1 CHAPTER 10 HYPOTHESIS TESTING MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, please circle the correct answer. 1. If a researcher takes a large enough sample, he/she will almost always obtain: a. virtually significant results b. practically significant results c. consequentially significant results d. statistically significant results ANSWER: d 2. The null and alternative hypotheses divide all possibilities into: a. two sets that overlap b. two non-overlapping sets c. two sets that may or may not overlap d. as many sets as necessary to cover all possibilities ANSWER: b 3. Which of the following is true of the null and alternative hypotheses? a. Exactly one hypothesis must be true b. both hypotheses must be true c. It is possible for both hypotheses to be true d. It is possible for neither hypothesis to be true ANSWER: a http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 2 of 72 2 4. One-tailed alternatives are phrased in terms of: a. ¹ b. < or > c. » or = d. ANSWER: b 5. The chi-square goodness-of-fit test can be used to test for: a. significance of sample statistics http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 5 of 72 5 a. the rejection level b. the acceptance level c. the significance level d. the error in the hypothesis test ANSWER: c 12. A study in which randomly selected groups are observed and the results are analyzed without explicitly controlling for other factors is called: a. an observational study b. a controlled study c. a field test d. a simple study ANSWER: a 13. The null hypothesis usually represents: a. the theory the researcher would like to prove. b. the preconceived ideas of the researcher c. the perceptions of the sample population d. the status quo ANSWER: d 14. The ANOVA test is based on which assumptions? I. the sample are randomly selected II. the population variances are all equal to some common variance III. the populations are normally distributed IV. the populations are statistically significant a. All of the above b. II and III only c. I, II, and III only d. I, and III only ANSWER: b 15. In statistical analysis, the burden of proof lies traditionally with: a. the alternative hypothesis b. the null hypothesis c. the analyst d. the facts http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 6 of 72 6 ANSWER: a 16. When one refers to “how significant” the sample evidence is, he/she is referring to the: a. value of b. the importance of the sample c. the p-value d. the F-ratio ANSWER: c http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 7 of 72 7 17. Which of the following values is not typically used for ? a. 0.01 b. 0.05 c. 0.10 d. 0.25 ANSWER: d 18. Smaller p-values indicate more evidence in support of: a. the null hypothesis b. the alternative hypothesis c. the quality of the researcher d. further testing ANSWER: b 19. The chi-square test can be too sensitive if the sample is: a. very small b. very large c. homogeneous d. predictable ANSWER: b 20. The hypothesis that an analyst is trying to prove is called the: a. elective hypothesis b. alternative hypothesis c. optional hypothesis d. null hypothesis ANSWER: b 21. A p-value is considered “convincing” if it is: http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 10 of 72 10 a. a Lilliefors test b. an empirical cdf c. a p-test d. a quantile-quantile plot ANSWER: d 28. Which of the following tests are used to test for normality? a. A t-test and an ANOVA test b. An Empirical CDF test and an F-test http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 11 of 72 11 c. A Chi-Square test and a Lilliefors test d. A Quantile-Quantile plot and a p-value test ANSWER: c 29. If a teacher is trying to prove that new method of teaching math is more effective than traditional one, he/she will conduct a: a. one-tailed test b. two-tailed test c. point estimate of the population parameter d. confidence interval ANSWER: a 30. A type I error occurs when: a. the null hypothesis is incorrectly accepted when it is false b. the null hypothesis is incorrectly rejected when it is true c. the sample mean differs from the population mean d. the test is biased ANSWER: b TEST QUESTIONS 31. A sport preference poll yielded the following data for men and women. Use the 5% significance level and test to determine is sport preference and gender are independent. Sport Preference Basketball Football Soccer Men 20 25 30 75 Gender http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 12 of 72 12 Women 18 12 15 45 38 37 45 120 ANSWER: http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 15 of 72 15 10 0.56 0.12 11 1.24 0.43 12 -1.16 -0.23 13 0.37 0.70 14 -0.52 -0.24 15 -0.09 -0.59 16 1.07 0.24 17 -0.88 0.66 18 0.44 -0.54 19 -0.21 0.55 20 0.84 0.08 http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 16 of 72 16 ANSWER: Test statistic: P-value=0.023 Since the P-values is less than 0.10, we reject the null hypothesis of equal variances and conclude that the variances of the stocks are not equal at the 10% level. QUESTIONS 34 THROUGH 37 ARE BASED ON THE FOLLOWING INFORMATION: BatCo (The Battery Company) produces your typical consumer battery. The company claims that their batteries last at least 100 hours, on average. Your experience with the http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 17 of 72 17 Hypothesized mean Sample mean Std error of mean Degrees of freedom t-test statistic 100.0 98.5 0.777 19 Test of 100 versus one-tailed alternative BatCo battery has been somewhat different, so you decide to conduct a test to see if the companies claim is true. You believe that the mean life is actually less than the 100 hours BatCo claims. You decide to collect data on the average battery life (in hours) of a random sample and the information related to the hypothesis test is presented below. 34. Can the sample size be determined from the information above? Yes or no? If yes, what is the sample size in this case? ANSWER: Yes. 19 + 1 = 20. 35. You believe that the mean life is actually less than 100 hours, should you conduct a one-tailed or a two-tailed hypothesis test? Explain your answer. ANSWER: One-tailed test. You are interested in the mean being less than 100. 36. What is the sample mean of this data? If you use a 5% significance level, would you conclude that the mean life of the batteries is typically more than 100 hours? Explain your answer. ANSWER: 98.5 hours. No. You would reject the null hypothesis in favor of the alternative, which is less than 100 hours (0.034 < 0.05). 37. If you were to use a 1% significance level in this case, would you conclude that the mean life of the batteries is typically more than 100 hours? Explain your answer. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 20 of 72 20 ANSWER: We fail to reject the null hypothesis of independence at the 5% significance level (since p-value = 0.083 > 0.05). We may conclude that demands for these two types of computers are independent 41. Suppose that you are asked to test versus at the = 0.05 significance level. Furthermore, suppose that you observe values of the sample mean and sample standard deviation when n = 50 that lead to the rejection of . Is it true that you might fail to reject if you were to observe the same values of the sample mean and standard deviation from a sample with n > 50? Why or why not? ANSWER: No. When n increases and the standard deviation of the sample mean stays the same, the standard error will decrease. Therefore, the test statistic will become more significant. If you rejected with n = 50, you will continue to reject with n > 50. QUESTIONS 42 THROUGH 44 ARE BASED ON THE FOLLOWING INFORMATION: Do graduates of undergraduate business programs with different majors tend to earn disparate starting salaries? Below you will find the StatPro output for 32 randomly selected graduate with majors in accounting (Acct), marketing (Mktg), finance (Fin), and information systems (IS). http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 21 of 72 21 Summary statistics for samples Acct. Mktg. Fin. IS Sample sizes 9 6 10 7 Sample means 32711.67 27837.5 30174 32869.3 Sample standard deviations 2957.438 754.982 1354.613 3143.906 Sample variances 8746437.5 569997.5 1834976.7 9884145.2 Weights for pooled variance 0.286 0.179 0.321 0.214 Number of samples 4 Total sample size 32 Grand mean 31039.22 Pooled variance 5308612.5 Pooled standard deviation 2304.043 One Way ANOVA table Source SS df MS F p-value Between variation 117609807 3 39203269 7.385 0.0009 Within variation 148641149 28 5308612 Total variation 266250955 31 Confidence Intervals for Differences Difference Mean diff Lower limit Upper limit Acct. - Mktg. 4874.167 1263.672 8484.661 Acct. – Fin. 2537.667 -609.890 5685.223 Acct. - IS -157.619 -3609.912 3294.674 Mktg. – Fin. -2336.500 -5874.048 1201.048 Mktg. - IS -5031.786 -8843.014 -1220.557 Fin. - IS -2695.286 -6071.216 680.644 http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 22 of 72 22 42. Assuming that the variances of the four underlying populations are equal, can you reject at a 5% significance level that the mean starting salary for each of the given business majors? Explain why or why not? ANSWER: Yes. Because of the F-test and the p-value is less than 0.05 (p-value = 0.0009) 43. Is there any reason to doubt the equal-variance assumption made in Question 42? Support your answer. ANSWER: Yes, there is some cause for concern. The F-test is rather robust, however, is this case, the sample sizes are rather small and of different sizes. 44. Use the information above related to the 95% confidence intervals for each pair of differences to explain which ones are statistically significant at a = 0.05. ANSWER: These confidence intervals show that the accounting majors stating salaries, on average, are larger than the marketing majors. There is not a significant difference for the other pairs using a 95% confidence interval. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 25 of 72 25 Sales (Female) Sales (Male) Sample sizes 25 22 ANSWER: Simultaneous confidence intervals for mean differences with confidence level of 90% Difference Mean difference Lower limit Upper limit Significant? Accounting - Marketing 5512.857 2510.523 8515.191 Yes Accounting - Finance 2900.357 246.644 5554.071 Yes Accounting - Management 6092.857 3090.523 9095.191 Yes Marketing - Finance -2612.500 -5535.603 310.603 No Marketing - Management 580.000 -2662.892 3822.892 No Finance - Management 3192.500 269.397 6115.603 Yes The a Accounting mean is significantly different (larger) than each of the others. Also, the Finance mean is significantly different (larger) than the Management mean. The other means are not significantly different from each other. QUESTIONS 48 THROUGH 52 ARE BASED ON THE FOLLOWING INFORMATION: Q-Mart is interested in comparing its male and female customers. Q-Mart would like to know if its female charge customers spend more money, on average, than its male charge customers. They have collected random samples of 25 female customers and 22 male customers. On average, women charge customers spend $102.23 and men charge customers spend $86.46. Additional information are shown below: Summary statistics for two samples http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 26 of 72 26 Sample means 102.23 86.460 Sample standard deviations 93.393 59.695 Test of difference=0 Sample mean difference 15.77 Pooled standard deviation 79.466 Std error of difference 23.23 t-test statistic 0.679 p-value 0.501 48. Given the information above, what is and for this comparison? Also, does this represent a one-tailed or a two-tailed test? Explain your answer. ANSWER: . This represents a one-tail test. 49. What are the degrees of freedom for the t-statistic in this calculation? Explain how you would calculate the degrees of freedom in this case. ANSWER: d.f = 25 + 22 – 2 = 45 50. What is the assumption in this case that allows you to use the pooled standard deviation for this confidence interval? ANSWER: The assumption is that the populations’ standard deviations are equal ( ). 51. Using a 10% level of significance, is there sufficient evidence for Q-Mart to conclude that women charge customers on average spend more than men charge customers? Explain your answer. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 27 of 72 27 ANSWER: No. There is not a statistical difference between women and men spending at Q- Mart, since p-value = 0.501 > 0.10. 52. Using a 1% level of significance, is there sufficient evidence for Q-Mart to conclude that women charge customers on average spend more than men charge customers? Explain your answer. ANSWER: No. There is not a statistical difference between women and men spending at Q- Mart, since p-value = 0.501 > 0.01. 53. The CEO of a software company is committed to expanding the proportion of highly qualified women in the organization’s staff of salespersons. He claims that the proportion of women in similar sales positions across the country in 1999 is less than 45%. Hoping to find support for his claim, he directs his assistant to http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 30 of 72 30 54. Are all three sandwich shops serving the same number of customers, on average, for lunch each weekday? Explain how you arrived at your answer. ANSWER: No. You should reject H o at a 5% significance level (p-value = 0.0022). Means are not all equal. 55. Explain why the weights for the pooled variance are the same for each of the samples. ANSWER: The weights for the pooled variance are the same for each of the samples, because sample sizes are equal (sample of 10 customers from each sandwich shop). 56. Use the information related to the 95% confidence interval to explain how the number of customers Joe has each weekday compares to his competition. ANSWER: These intervals show that there is not a significant difference between Joe’s and Bob’s. However, there is a significant difference between Joe’s and Ted’s using a 95% confidence interval. QUESTIONS 57 AND 58 ARE BASED ON THE FOLLOWING INFORMATION: The manager of a consulting firm in Lansing, Michigan, is trying to assess the effectiveness of computer skills training given to all new entry-level professionals. In an effort to make such an assessment, he administers a computer skills test immediately before and after the training program to each of 20 randomly chosen employees. The pre-training and post-training scores of these 20 individuals are shown in the table below. Employee Score before Score after 1 62 77 2 63 77 3 74 83 4 64 88 5 84 80 6 81 80 7 54 83 8 61 88 9 81 80 http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 31 of 72 31 10 86 88 11 75 93 12 71 78 13 86 82 14 74 84 15 65 86 16 90 89 17 72 81 18 71 90 19 85 86 20 66 92 http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 32 of 72 32 Hypothesized mean Sample mean Std error of mean Degrees of freedom t-test statistic 1500.0 1509.5 4.854 24 Test of 1500 versus one-tailed alternative 57. Using a 10% level of significance, do the given sample data support that the firm’s training programs is effective in increasing the new employee’s working knowledge of computing? ANSWER: Test statistic: t = - 4.471 (paired t-test) P-value = 0.00013 The test scores have improved by an average of 11 points. Since the P-value is virtually 0, there is enough evidence to conclude that the given sample data support that the firm’s training program is increasing the new employee’s knowledge of computing at the 10% significance level. 58. Re-do Question 57 using a 1% level of significance. ANSWER: Again, since the P-value is virtually zero, there is plenty of evidence to support the effectiveness of the program at the 1% level of significance. QUESTIONS 59 THROUGH 62 ARE BASED ON THE FOLLOWING INFORMATION: Suppose a firm that produces light bulbs wants to know whether it can claim that it light bulbs typically last more than 1500 hours. Hoping to find support for their claim, the firm collects a random sample and records the lifetime (in hours) of each bulb. The information related to the hypothesis test is presented below. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 35 of 72 35 The sample mean is not significantly different from 131 at even the 10% level because the p-value is greater than 0.10 64. For which values of the sample mean (i.e., average weekly grocery bill) would you decide to reject the null hypothesis at the significance level? For which values of the sample mean would you decide to reject the null hypothesis at http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 36 of 72 36 -value t-value Lower limit Upper limit the significance level? ANSWER: For either p-value (0.01 or 0.10), we find the t-value that would lead to the rejection of the null hypothesis, and then solve the equation on either side of 131. This leads to the following results: 0.01 2.680 126.758 135.242 0.10 1.677 128.346 133.654 For example, at the 10% level, if we would reject the null hypothesis. QUESTIONS 65 THROUGH 68 ARE BASED ON THE FOLLOWING INFORMATION: Do undergraduate business students who major in information systems (IS) earn, on average, higher annual starting salaries than their peers who major in marketing (Mktg)? Before addressing this question with a statistical hypothesis test, a comparison should be done to determine whether the variances of annual starting salaries of the two types of majors are equal. Below you will find the StatPro output for 20 randomly selected IS majors and 20 randomly selected Mktg majors. Summary statistics for two samples IS Salary Mktg Salary Sample sizes 20 20 Sample means 30401.35 27715.85 Sample standard deviations 1937.52 2983.39 Test of difference 0 Sample mean difference 2685.5 Pooled standard deviation 2515.41 NA Std error of difference 795.44 795.44 Degrees of freedom 38 33 t-test statistic 3.376 3.376 p-value 0.0009 0.0009 http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 37 of 72 37 Test of equality of variances Ratio of sample variances 2.371 p-value 0.034 65. Use the information above to perform the test of equal variance. Explain how the ratio of sample variances is calculated. What type of distribution is used to test for equal variances? Also, would you conclude that the variances are equal or not? Explain your answer. ANSWER: (2983.39)2 / (1937.52)2 = 2.371. Since the p-value is 0.034, you can conclude that there is a significant difference between the sample variance. They are not equal. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 40 of 72 40 ANSWER: (200)(0.55) = 110 72. Using a 5% significance level, can the marketing consultant conclude that the proportion of customers who prefer Coca-Cola exceeds 50%? Explain your answer. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 41 of 72 41 Sample proportion 0.53 Standard error of sample proportion 0.01578 Z test statistic 1.9008 p-value 0.0287 ANSWER: No. You cannot reject the null hypothesis at a 5% level of significance, since p- value = 0.07761 > 0.05. 73. If you were to use a 1% significance level, would the conclusion from part c change? Explain your answer. ANSWER: No. You still cannot reject the null hypothesis at a 1% level of significance, since p-value = 0.07761 > 0.01. QUESTIONS 74 THROUGH 77 ARE BASED ON THE FOLLOWING INFORMATION: The owner of a popular Internet-based auction site believes that more than half of the people who sell items on her site are women. To test this hypothesis, the owner sampled 1000 customers who sale items on her site and she found that 53% of the customers sampled were women. Some calculations are shown in the table below 74. If you were to conduct a hypothesis test to determine if greater than 50% of customers who use this Internet-based site are women, would you conduct a one- tail or a two-tail hypothesis test? Explain your answer. ANSWER: One-tailed, since the owner is interested in finding whether the proportion is actually greater than 50%. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 42 of 72 42 75. How many customers out of the 1000 sampled must have been women in this case? ANSWER: (1000)(0.53) = 530 76. Using a 5% significance level, can the owner of this site conclude that women make up more than 50% of her customers? Explain your answer. ANSWER: Yes. You can reject the null hypothesis at a 5% level of significance, since p- http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 45 of 72 45 ANSWER: The assumption is that the two populations standard deviations are equal; that is 81. Using a 5% level of significance, is there sufficient evidence for Q-Mart to conclude that customers who use the Q-Mart card charge, on average, more than those who use another charge card? Explain your answer. ANSWER: Yes. There is a statistical difference between those using the Q-Mart card and those who use other types of charge cards, since p-value = 0.006 < 0.05. 82. Using a 1% level of significance, is there sufficient evidence for Q-Mart to conclude that customers who use the Q-Mart card charge, on average, more than those who use another charge card? Explain your answer. ANSWER: Yes. There is still a statistical difference between those using the Q-Mart card and those who use other types of charge cards, since p-value = 0.006 < 0.01. 83. The number of cars sold by three salespersons over a 6-month period are shown in the table below. Use the 5% level of significance to test for independence of salespersons and type of car sold. Insurance Preference Chevrolet Ford Toyota Ali 15 9 5 29 Salesperson Bill 20 8 15 43 Chad 13 4 11 28 48 21 31 100 http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 46 of 72 46 ANSWER: http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 47 of 72 47 We fail to reject the null hypothesis of independence at the 5% significance level (since p-value = 0.305 > 0.05). We may conclude that salespersons and type of car sold are independent. QUESTIONS 84 AND 85 ARE BASED ON THE FOLLOWING INFORMATION: An automobile manufacturer needs to buy aluminum sheets with an average thickness of 0.05 inch. The manufacturer collects a random sample of 40 sheets from a potential supplier. The thickness of each sheet in this sample is measured (in inches) and recorded. The information below are pertaining to the Chi-square goodness-of-fit test. Upper limit Category Frequency Normal Distance measure 0.03 0.03 1 1.920 0.441 0.04 0.03 but 0.04 10 8.074 0.459 0.05 0.04 but 0.05 13 14.947 0.254 0.06 0.05 but 0.06 12 11.218 0.055 >0.06 4 3.842 0.007 Test of normal fit Chi-square statistic 1.214 p-value 0.545 84. Are these measurements normally distributed? Summarize your results. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 50 of 72 50 86. Is it appropriate to perform a paired-comparison t-test in this case? Explain why or why not. ANSWER: A two-sample, not paired-sample, procedure should be used because there is no evidence of pairing. 87. Perform an appropriate hypothesis test with a 1% significance level. Assume that the population variances are equal. ANSWER: , , Test statistic t = 6.22, P-value=0. Since P-value is virtually 0, we can conclude at the 1% level that the mean salary for CIS majors is indeed larger. 88. How large would the difference between the mean starting salaries of CIS and IB majors have to be before you could conclude that CIS majors earn more on average? Employ a 1% significance level in answering this question. ANSWER: P-value=0.01, t =2.41, and Standard error of difference = . Then A mean difference of 1312.20 is all that would be required to get the conclusion in Question 87 at the 1% level. 89. A statistics professor has just given a final examination in his linear models course. He is particularly interested in determining whether the distribution of 50 http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 51 of 72 51 exam scores is normally distributed. The data are shown in the table below. Perform the Lilliefors test. Report and interpret the results of the test. 1 77 71 78 83 84 71 81 82 79 7 2 73 89 74 75 93 74 88 83 90 8 4 79 62 73 88 76 76 76 80 84 8 91 70 76 74 68 80 87 92 84 7 http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 52 of 72 52 9 80 91 74 69 88 84 83 87 82 7 2 ANSWER: The maximum distance between the empirical and normal cumulative distributions is 0.0802. This is less than 0.1247, the maximum allowed with a sample size of 50. Therefore, the normal hypothesis cannot be rejected at the 5% level. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 55 of 72 55 We reject the null hypothesis of independence at the 5% significance level (since p-value = 0.019 < 0.05). We may conclude that there is no evidence that life insurance preference of male students is different than that of female students. QUESTIONS 91 THROUGH 93 ARE BASED ON THE FOLLOWING INFORMATION: The retailing manager of Meijer supermarket chain in Michigan wants to determine whether product location has any effect on the sale of children toys. Three different aisle locations are considered: front, middle, and rear. A random sample of 18 stores is selected, with 6 stores randomly assigned to each aisle location. The size of the display area and price of the product are constant for all the stores. At the end of one-month trial period, the sales volumes (in thousands of dollars) of the product in each store were as shown below: Front Aisle Middle Aisle Rear Aisle 10.0 4.6 6.0 8.6 3.8 7.4 6.8 3.4 5.4 7.6 2.8 4.2 6.4 3.2 3.6 5.4 3.0 4.2 http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 56 of 72 56 91. At the 0.05 level of significance, is there evidence of a significant difference in average sales among the various aisle locations? ASNWER: StatPro’s one-way ANOVA produces the following results: To test at the 0.05 level of significance whether the average sales volumes in thousands of dollars are different across the three store aisle locations, we conduct an F test: H0: H 1 : At least one mean is different. Since p-value = 0.0004 < = 0.05, we reject H0. There is enough evidence to conclude that the average sales volumes in thousands of dollars are different across the three store aisle locations. 92. If appropriate, which aisle locations appear to differ significantly in average sales? (Use = 0.05) http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 57 of 72 57 ANSWER: It appears that the front and middle aisles and also the front and rear aisles differ http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 60 of 72 60 95. Do you think any of the assumptions needed in Question 94 have been violated? Explain. ANSWER: The assumption of equal variances may be violated because the sample variance in Lansing is nearly four times the size of the sample variance in Grand Rapids and the two sample sizes are not small. Nevertheless, the results of the test for the differences in the two means were overwhelming (i.e., the p value is nearly 0). 96. Construct a 95% confidence interval estimate of the difference between the population means of Lansing and Grand Rapids. ANSWER: 97. Explain how to use the confidence interval in Question 96 to answer Question 94. ANSWER: Since the 95% confidence interval in Question 96 does not include 0, we reject the null hypothesis at the 5% level of significance that the average appraised values for single-family homes are the same in Lansing and Grand Rapids. QUESTIONS 98THROUGH 100 ARE BASED ON THE FOLLOWING INFORMATION: In a survey of 1,500 customers who did holiday shopping on line during the 2000 holiday season, 270 indicated that they were not satisfied with their experience. Of the customers that were not satisfied, 143 indicated that they did not receive the products in time for the holidays, while 1,197 of the customers that were satisfied with their experience indicated that they did receive the products in time for the holidays. The following complete summary of results were reported: http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 61 of 72 61 Received Products in Time for Holidays Satisfied with their Experience Yes No Total Yes 1,197 33 1,230 No 127 143 270 Total 1,324 176 1,500 http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 62 of 72 62 98. Is there a significant difference in satisfaction between those who received their products in time for the holidays, and those who did not receive their products in time for the holidays? Test at the 0.01 level of significance. ANSWER: Populations: 1 = received product in time, 2 = did not receive product in time Decision rule: If Z < -2.5758 or Z > 2.5758, reject H0. Test statistic: Decision: Since Z calc = 23.248 is well above the upper critical bound of Z = 2.5758, reject H0. There is sufficient evidence to conclude that a significant difference in satisfaction exists between those who received their products in time for the holidays and those who did not receive their products in time for the holidays. 99. Find the p-value in Question 98 and interpret its meaning. ANSWER: The p-value is virtually 0. The probability of obtaining a difference in two sample proportions as large as 0.7166 or more is virtually 0 when is true. 100. Based on the results of Questions 98 and 99, if you were the marketing director of a company selling products online, what would you do to improve the satisfaction of the customers? http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 65 of 72 65 109. The probability of making a Type I error and the level of significance are the same. ANSWER: T 110. The p-value of a test is the smallest level of significance at which the null hypothesis can be rejected. ANSWER: T 111. If a null hypothesis about a population mean is rejected at the 0.025 level of significance, it must be rejected at the 0.01 level. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 66 of 72 66 ANSWER: F 112. In order to determine the p-value, it is unnecessary to know the level of significance. ANSWER: T 113. If we reject a null hypothesis about a population proportion p at the 0.025 level of significance, then we must also reject it at the 0.05 level. ANSWER: T 114. Using the confidence interval when conducting a two-tailed test for the population mean , we do not reject the null hypothesis if the hypothesized value for falls between the lower and upper confidence limits. ANSWER: T 115. A professor of statistics refutes the claim that the proportion of independent voters in Minnesota is at most 40%. To test the claim, the hypotheses: , , should be used. ANSWER: F 116. Using the confidence interval when conducting a two-tailed test for the population proportion p, we reject the null hypothesis if the hypothesized value for p falls inside the confidence interval. ANSWER: F 117. When testing the equality of two population variances, the test statistic is the ratio of the population variances; namely . http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 67 of 72 67 ANSWER: F 118. Tests in which samples are not independent are referred to as matched pairs. ANSWER: T 119. The pooled-variances t-test requires that the two population variances are different. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 70 of 72 70 127. Statistics practitioners use the analysis of variance (ANOVA) technique to compare more than two population means. ANSWER: T 128. Given the significance level 0.01, the F-value for the degrees of freedom, d.f. = (6,9) is 7.98. ANSWER: F 129. The analysis of variance (ANOVA) technique analyzes the variance of the data to determine whether differences exist between the population means. ANSWER: T 130. The F-test of the analysis of variance requires that the populations be normally distributed with equal variances. ANSWER: T 131. One-way ANOVA is applied to four independent samples having means 13, 15, 18 and 20, respectively. If each observation in the forth sample were increased by 30, the value of the F-statistics would increase by 30. ANSWER: F 132. The degrees of freedom for the denominator of a one-way ANOVA test for 4 population means with 10 observations sampled from each population are 40. ANSWER: F 133. A test for independence is applied to a contingency table with 4 rows and 4 columns. The degrees of freedom for this chi-square test must equal 9. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 71 of 72 71 ANSWER: T 134. The number of degrees of freedom for a contingency table with r rows and c columns is rc - 1 , provided that both r and c are greater than or equal to 2. ANSWER: F 135. The Lilliefors test is used to test for normality. http://www2.gsu.edu/~dscaas/testbank/TB%20Ch%2010.doc 5/3/18, 12@47 AM Page 72 of 72 72 ANSWER: T 214