HYPOTHESIS TESTING QUESTOINS WITH CORRECT ANSWERS, Exams of Nursing

Download HYPOTHESIS TESTING QUESTOINS WITH CORRECT ANSWERS and more Exams Nursing in PDF only on Docsity! 214 214 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 215 215 4. One-tailed alternatives are phrased in terms of: 𝛼. b. < or > c. or = d. or ANSWER: b 218 218 Hypothesis Testing 10. A two-tailed test is one where: a. results in only one direction can lead to rejection of the null hypothesis b. negative sample means lead to rejection of the null hypothesis c. results in either of two directions can lead to rejection of the null hypothesis d. no results lead to the rejection of the null hypothesis ANSWER: c 11. The value set for 𝛼 is known as: 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? 219 219 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 220 220 Chapter 10 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 ANSWER: a 16. When one refers to “how significant” the sample evidence is, he/she is referring to the: a. value of 𝛼 b. the importance of the sample c. the p-value d. the F-ratio ANSWER: c 17. Which of the following values is not typically used for 𝛼 ? 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: 223 223 24. Typically one-way ANOVA is used in which of the following situations? I. there are several distinct populations II. there are two sample populations over 4000 III. randomized experiments IV. randomly selected populations a. All of the above b. II and III only c. I, II, and III only d. I, and III only ANSWER: d 224 224 Chapter 10 25. The chi-square test is not very effective if the sample is: a. small b. large c. irregular d. heterogeneous ANSWER: a 26. The alternative hypothesis is also known as the: a. elective hypothesis b. optional hypothesis c. research hypothesis d. null hypothesis ANSWER: c 27. An informal test for normality that utilizes a scatterplot and looks for clustering around a 45 line is known as: 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 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: 225 225 a. one-tailed test b. two-tailed test c. point estimate of the population parameter d. confidence interval ANSWER: a 228 228 not independent; that is, there is evidence that sport preference of men is different from that of women. 32. Suppose that we observe a random sample of size n from a normally distributed population. If we are able to reject H0 : 0 in favor of Ha : 0 at the 5% significance level, is it true that we can definitely reject H0 in favor of the appropriate one-tailed alternative at the 2.5% significance level? Why or why not? 229 229 Hypothesis Testing ANSWER: This is not true for certain. Suppose 0 =50 and the sample mean we observe is X =55. If the alternative for the one-tailed test is Ha : ∈ 50, then we obviously can’t reject the null because the observed sample mean X is in the wrong direction. But if the alternative is H1 : > 50, we can reject the null at the 2.5% level. The reason is that we know the p-value for the two-tailed test was less than 0.05. The p-value for a one-tailed test is half of this, or less than 0.025, which implies rejection at the 2.5% level. 33. An investor wants to compare the risks associated with two different stocks. One way to measure the risk of a given stock is to measure the variation in the stock’s daily price changes. The investor obtains a random sample of 20 daily price changes for stock 1 and 20 daily price changes for stock 2. These data are shown in the table below. Show how this investor can compare the risks associated with the two stocks by testing the null hypothesis that the variances of the stocks are equal. Use 𝛼 = 0.10 and interpret the results of the statistical test. Day Price Change for stock 1 Price Cha for stock nge 2 1 1.86 0.87 2 1.80 1.33 3 1.03 -0.27 4 0.16 -0.20 5 -0.73 0.25 6 0.90 0.00 7 0.09 0.09 8 0.19 -0.71 9 -0.42 -0.33 10 0.56 0.12 11 1.24 0.43 230 230 0 1 2 1 2 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 ANSWER: n1 =20, s1 =0.8487, n2 =20, s2 =0.5291 H : 2 / 2 =1 H : 2 / 2 1 a 1 2 Test statistic: F =s2 / s2 =2.573 233 233 1 2 1 2 Hypothesis Testing ANSWER: Yes. You cannot reject the null hypothesis at a 1% level of significance (0.034 > 0.01). QUESTIONS 38 AND 39 ARE BASED ON THE FOLLOWING INFORMATION: Two teams of workers assemble automobile engines at a manufacturing plant in Michigan. A random sample of 145 assemblies from team 1 shows 15 unacceptable assemblies. A similar random sample of 125 assemblies from team 2 shows 8 unacceptable assemblies. 38. Construct a 90% confidence interval for the difference between the proportions of unacceptable assemblies generated by the two teams. ANSWER: n1 =145, P̂ =0.1034, n =125, P̂ =0.0640, z - multiple =1.645 1 2 2 SE(P̂ - P̂ ) = =0.0334 1 2 (P̂ - P̂ ) Z SE(P̂ - P̂ ) =0.0394 0.0549 Lower limit = -0.0155, and Upper limit = 0.0943 39. Based on the confidence interval constructed in Question 38, is there sufficient evidence o conclude, at the 10% significance level, that the two teams differ with respect to their proportions of unacceptable assemblies? P̂ (1- P̂ ) 1 1 n1 + P̂ (1- P̂ ) 2 2 n2 234 234 ANSWER: Because the 90% confidence interval includes the value 0, we cannot reject the null hypothesis of equal proportions. 40. Staples, a chain of large office supply stores, sells a line of desktop and laptop computers. Company executives want to know whether the demands for these two types of computers are related in any way. Each day's demand for each type of computers is categorized as Low, Medium-Low, Medium-High, or High. The data shown in the table below is based on 200 days of operation. Based on these data, can Staples conclude that demands for these two types of computers are independent? Test at the 5% level of significance. Desktops Low Med-Low Med-High High 235 235 Chapter 10 Low 3 14 14 4 35 Laptops Med-Low 6 18 17 22 63 Med-High 13 16 11 16 56 High 8 14 15 9 46 30 62 57 51 200 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 H0 : =100 versus Ha : ∈100 at the 𝛼 = 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 H0 . Is it true that you might fail to reject H 0 if you were to observe the same 238 238 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 -6071.216 680.644 239 239 Chapter 10 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 𝛼 = 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. QUESTIONS 45 THROUGH 47 ARE BASED ON THE FOLLOWING INFORMATION: Do graduates of undergraduate business programs with different majors tend to earn disparate average starting salaries? Consider the data given in the table below. Accounting Marketing Finance Management 240 240 $37,220 $30,950 $32,630 $31,350 $29,410 $37,330 $35,700 $28,620 $27,750 $27,650 $27,640 $28,340 $29,870 $31,700 $31,740 $32,750 $30,550 $29,250 $28,890 $30,150 $28,600 $27,450 $26,410 $27,340 $27,300 243 243 Chapter 10 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 Sales (Female) Sales (Male) Sample sizes 25 22 Sample means 102.23 86.460 Sample standard deviations 93.393 59.695 Test of difference=0 244 244 48. Given the information above, what is H0 and Ha for this comparison? Also, does this represent a one-tailed or a two-tailed test? Explain your answer. ANSWER: H0 : F M , Ha : F > M . This represents a one-tail test. Sample mean difference Pooled standard deviation Std error of difference t-test statistic p- value 15.77 79.466 23.23 0.679 0.501 245 245 Hypothesis Testing 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 ( F M ). 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. 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 collect a random sample of salespersons employed by his company, which is 248 248 Difference Mean diff Lower Upper Joe’s – Bob’s 4.500 -0.282 9.282 Joe’s – Ted’s 7.200 2.418 11.982 Bob’s – Ted’s 2.700 -2.082 7.482 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 Ho at a 5% significance level (p-value = 0.0022). Means are not all equal. 249 249 Hypothesis Testing 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 88 250 250 9 81 80 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 92 253 253 Hypothesis Testing 60. The firm believes that the mean life is actually greater than 1500 hours, should you conduct a one-tailed or a two-tailed hypothesis test? Explain your answer. ANSWER: One-tailed, since the firm is interested in finding whether the mean is actually greater than 1500. 61. What is the sample mean of this data? If you use a 5% significance level, would you conclude that the mean life of the light bulbs is typically more than 1500 hours? Explain your answer. ANSWER: 1509.5 hours. Yes, you would reject the null hypothesis in favor of the mean being greater than 1500 hours (0.031 < 0.05). 62. If you were to use a 1% significance level in this case, would you conclude that the mean life of the light bulbs is typically more than 1500 hours? Explain your answer. ANSWER: No. You cannot reject the null hypothesis at a 1% level of significance (0.031 > 0.01). QUESTIONS 63 AND 64 ARE BASED ON THE FOLLOWING INFORMATION: A study is performed in San Antonio to determine whether the average weekly grocery bill per five-person family in the town is significantly different from the national average. A random sample of 50 five-person families in San Antonio showed a mean of $133.474 and a standard deviation of $11.193. 63. Assume that the national average weekly grocery bill for a five-person family is 254 254 $131. Is the sample evidence statistically significant? If so, at what significance levels can you reject the null hypothesis? ANSWER: H0 : =131 H𝛼 : 131 Test statistic: t = 1.563 p-value: 0.124 The sample mean is not significantly different from 131 at even the 10% level because the p-value is greater than 0.10 255 255 Chapter 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 𝛼 =0.01 significance level? For which values of the sample mean would you decide to reject the null hypothesis at the 𝛼 =0.10 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 t =( X - 131) /1.583 for X on either side of 131. This leads to the following results: 𝛼 -value t-value Lower limit Upper limit 0.01 2.680 126.758 135.242 0.10 1.677 128.346 133.654 For example, at the 10% level, if the null hypothesis. X ∈ 128.346 or X >133.654, we would reject 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 2983.39 258 258 Use the 5% significance level and test to determine if party affiliation is independent of the educational level of the voters. Party Affiliation Democrat Republican Independent Didn't Complete High School 95 80 115 290 Educational Level Has High School Diploma 135 85 105 325 Has College Degree 160 105 120 385 390 270 340 1000 ANSWER: 259 259 Sample proportion Standard error of sample proportion Z test statistic p-value 0.55 0.03518 1.4213 0.07761 Chapter 10 We fail to reject the null hypothesis of independence at the 5% significance level (since p-value = 0.087 > .05). We may conclude that party affiliation is independent of the educational level of the voters. QUESTIONS 70 THROUGH 73 ARE BASED ON THE FOLLOWING INFORMATION: A marketing research consultant hired by Coca-Cola is interested in determining if the proportion of customers who prefer Coke to other brands is over 50%. A random sample of 200 consumers was selected from the market under investigation, 55% favored Coca- Cola over other brands. Additional information is presented below. 70. If you were to conduct a hypothesis test to determine if greater than 50% of customers prefer Coca-Cola to other brands, would you conduct a one-tail or a two-tail hypothesis test? Explain your answer. 260 260 ANSWER: One-tailed, since the consultant is interested in finding whether the proportion is actually greater than 50%. 263 263 Chapter 10 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- value = 0.0287 < 0.05. 77. If you were to use a 1% significance level, would the conclusion from Question 76 change? Explain your answer. ANSWER: Yes. Your answer would now change. You cannot reject the null hypothesis at a 1% level of significance, since p-value = 0.0287 > 0.01. QUESTIONS 78 THROUGH 82 ARE BASED ON THE FOLLOWING INFORMATION: Q-Mart is interested in comparing customer who used it own charge card with those who use other types of credit cards. Q-Mart would like to know if customers who use the Q- Mart card spend more money per visit, on average, than customers who use some other type of credit card. They have collected information on a random sample of 38 charge customers and the data is presented below. On average, the person using a Q-Mart card spends $192.81 per visit and customers using another type of card spend $104.47 per visit. Summary statistics for two samples Q-Mart Other Charges Sample sizes 13 25 Sample means 192.81 104.47 Sample standard deviations 115.243 71.139 264 264 Test of difference=0 78. Given the information above, what is H0 and Ha for this comparison? Also, does this represent a one-tailed or a two-tailed test? Explain your answer. ANSWER: H0 : Q- Mart Others , Ha : Q- Mart > Others . This represents a one-tail test. Sample mean difference Pooled standard deviation Std error of difference t-test statistic p- value 88.34 88.323 30.201 2.925 0.006 265 265 Hypothesis Testing 79. 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 = 13 + 25 – 2 = 36 80. 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 two populations standard deviations are equal; that is Q- Mart Others 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. 268 268 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 0.545 269 269 Hypothesis Testing 84. Are these measurements normally distributed? Summarize your results. ANSWER: Yes. Based on the Chi-square test, with a p-value of 0.545, you can conclude that the values are normally distributed. The frequency distribution also shows that the values are fairly close to the expected values. 85. Are there any weaknesses or concerns about your conclusions in Question 84? Explain your answer. ANSWER: Yes. There are a couple of concerns. The sample size is rather small (n = 40), you should use a larger sample size for this test to be more effective. Also, the test depends on which and how many categories are used for the histogram. A different choice could result in a different answer. QUESTIONS 86 THROUGH 88 ARE BASED ON THE FOLLOWING INFORMATION: Do undergraduate business students who major is computer information systems (CIS) earn, on average, higher annual starting salaries than their peers who major in international business (IB)?. To address this question through a statistical hypothesis test, the table shown below contains the starting salaries of 25 randomly selected CIS majors and 25 randomly selected IB majors. Graduate Finance Marketing 1 29,522 28,201 2 31,444 29,009 3 29,275 29,604 4 26,803 26,661 5 28,727 26,094 6 32,531 22,900 7 33,373 24,939 8 31,755 23,071 9 31,393 29,852 270 270 10 26,124 27,213 11 30,653 23,935 12 30,795 25,794 13 30,319 28,897 14 31,654 27,890 15 27,214 27,400 16 30,579 26,818 17 30,249 27,603 18 31,024 26,880 19 31,940 28,791 20 31,387 24,000 21 29,479 25,877 22 30,735 24,825 23 29,271 28,423 24 30,215 28,956 25 31,587 29,758 273 273 Hypothesis Testing Normal (smooth) and empirical cumulative distributions 1.0 0.8 0.6 0.4 0.2 0.0 Standardized values of Score 90. An insurance firm interviewed a random sample of 600 college students to find out the type of life insurance preferred, if any. The results are shown in the table below. Is there evidence that life insurance preference of male students is different than that of female students. Test at the 5% significance level. -2 .9 3 -2 .7 1 -2 .4 9 -2 .2 7 -2 .0 5 -1 .8 3 -1 .6 1 -1 .3 9 -1 .1 7 -0 .9 5 -0 .7 3 -0 .5 1 -0 .2 9 -0 .0 7 0 .1 5 0 .3 6 0 .5 8 0 .8 0 1 .0 2 1 .2 4 1 .4 6 1 .6 8 1 .9 0 2 .1 2 2 .3 4 2 .5 6 2 .7 8 3 .0 0 3 .2 2 274 274 Insurance Preference Term Whole Life No Insurance Gender Male Female 80 50 30 40 240 160 350 250 130 70 400 600 275 275 Chapter 10 ANSWER: 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 NSWER: Confidence intervals for mean differences Confidence level 95.0% Tukey method Difference Mean diff Front Ailse - Middle Aisle 4.000 Front Ailse - Rear Aisle 2.059 Middle Aisle - Rear Aisle -1.667 278 Lower 2.037 a.3a7o -3.630 Upper 5.963 Age 0.297 Signif? Yes Yes Na 279 279 Chapter 10 It appears that the front and middle aisles and also the front and rear aisles differ significantly in average sales at 𝛼 = 0.05. 93. What should the retailing manager conclude? Fully describe the retailing manager’s options with respect to aisle locations? ANSWER: The front aisle is best for the sale of this product. The manager should evaluate the tradeoff in switching the location of this product and the product that is currently intended for the front location. QUESTIONS 94THROUGH 97 ARE BASED ON THE FOLLOWING INFORMATION: A real estate agency wants to compare the appraised values of single-family homes in two cities in Michigan. A sample of 60 listings in Lansing and 99 listings in Grand Rapids yields the following results (in thousands of dollars): Lansing Big Rapids X 191.33 172.34 S 32.60 16.92 n 60 99 94. Is there evidence of a significant difference in the average appraised values for single-family homes in the two Michigan cities? Use 0.05 level of significance. ANSWER: Populations: 1 = Lansing, 2 = Grand Rapids H0: 1 = 2 (The average appraised values for single-family homes are the same in Lansing and Grand Rapids) H1: 1 2 (The average appraised values for single-family homes are not the same in Lansing and Grand Rapids) 280 280 S p ∣ 2 1 1 n1 n2 � + ∣ 578.0822 1 + 1 ∣ 60 99 ∣ � Decision rule: df = 157. If t < – 1.9752 or t > 1.9752, reject H0. (n – 1) S 2 + (n – 1) S 2 (59) 32.602 + (98) 16.922 S 2 = 1 1 2 2 = = 578.0822 p Test statistic: (n1 – 1) + (n2 – 1) 59 + 98 t = ( X 1 – X 2 ) – ( 1 – 2 ) = (191.33 – 172.34) – 0 = 4.8275 283 283 0.82 1- 0.82 1 1 ∣ 1324 176 ∣ + � Chapter 10 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 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 H0 : P1 =P2 H1 : P1 P2 Decision rule: If Z < -2.5758 or Z > 2.5758, reject H0. Test statistic: 0.9041 – 0.1875 = =23.248 Decision: Since Zcalc = 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. ( p̂1 – p̂2 ) p̂ (1 – p̂ ) 1 + 1 c c ∣ n1 n2 � ∣ Z = 284 284 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 H0 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? ANSWER: Ensuring that the customers receive their products in time for the holidays will be one effective way to improve the satisfaction of the customers. 285 285 Hypothesis Testing TRUE / FALSE QUESTIONS 101. The p-value of a test is the probability of observing a test statistic at least as extreme as the one computed given that the null hypothesis is true. ANSWER: T 102. The p-value is usually 0.01 0r 0.05. ANSWER: F 103. A null hypothesis is a statement about the value of a population parameter. It is usually the current thinking, or “status quo”. ANSWER: T 104. An alternative or research hypothesis is usually the hypothesis a researcher wants to prove. ANSWER: T 105. A two-tailed alternative is one that is supported by evidence in a single direction. ANSWER: F 106. A one-tailed alternative is one that is supported by evidence in either direction. ANSWER: F 107. A Type I error probability is represented by 𝛼 ; it is the probability of incorrectly rejecting a null hypothesis that is true. 288 288 1 2 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 2 / 2 . ANSWER: F 118. Tests in which samples are not independent are referred to as matched pairs. 289 289 0 1 2 1 2 Hypothesis Testing ANSWER: T 119. The pooled-variances t-test requires that the two population variances are different. ANSWER: F. 120. In testing the difference between two population means using two independent samples, we use the pooled variance in estimating the standard error of the sampling distribution of the sample mean difference are normal with equal variances. ANSWER: T x1 - x2 if the populations 121. In conducting hypothesis testing for difference between two means when samples are dependent, the variable under consideration is D ; the sample mean difference between n pairs. ANSWER: T 122. The number of degrees of freedom associated with the t test, when the data are gathered from a matched pairs experiment with 12 pairs, is 22. ANSWER: F 123. The test statistic employed to test H : 2 / 2 =1 is F =s2 / s2 , which is F distributed with n1 - 1 and n2 - 1 degrees of freedom. ANSWER: T 290 290 0 1 2 1 2 124. When the necessary conditions are met, a two-tail test is being conducted to test the difference between two population proportions. The two sample proportions are p1 =0.35 an d p2 =0.42 , and the standard error of the sampling distribution of p1 - p2 is 0.054. The calculated value of the test statistic will be 1.2963. ANSWER: F 125. The equal-variances test statistic of 1 - 2 is Student t distributed with n1 + n2 -2 degrees of freedom, provided that the two populations are normally distributed. ANSWER: T 126. When the necessary conditions are met, a two-tail test is being conducted at 𝛼 = 0.05 to test H : 2 / 2 =1 . The two sample variances are s2 =700 and s2 =875 ,