3 26 03 252y0321 Page layout view ECO252 QBA2 SECOND HOUR EXAM March 27 28 2003 Name KEY Hour of Class Registered I 40 points Do all the following 2points each unless noted otherwise A table identifying methods for comparing 2 samples is at the end of the exam 1 A powerful women s group has claimed that men and women differ in attitudes about sexual discrimination A group of 50 men group 1 and 40 women group 2 were asked if they thought sexual discrimination is a problem in the United States Of those sampled 11 of the men and 19 of the women did believe that sexual discrimination is a problem If the p value turns out to be 0 035 which is not the real value in this data set then a at 0 05 we should fail to reject H0 b at 0 04 we should reject H0 c at 0 03 we should reject H0 d None of the above would be correct statements Explanation The rule on p value says if the p value is less than the significance level alpha reject the null hypothesis if the p value is greater than or equal to the significance level do not reject the null hypothesis Table 12 4 A few years ago Pepsi invited consumers to take the Pepsi Challenge Consumers were asked to decide which of two sodas Coke or Pepsi they preferred in a blind taste test Pepsi was interesting in determining what factors played a role in people s taste preferences One of the factors studied was the gender of the consumer Below are the results of analyses comparing the taste preferences of men and women with the proportions depicting preference for Pepsi Males n 109 pSM 0 422018 Females n 52 pSF 0 25 pSM pSF 0 172018 z 2 11825 Note that H 1 may differ in 2 and 3 2 Referring to Table 12 4 to determine if a difference exists in the taste preferences of men and women give the correct alternative hypothesis that Pepsi would test a H1 M F 0 b H1 M F 0 c H1 pM pF 0 d H1 pM pF 0 3 Referring to Table 12 4 suppose Pepsi wanted to test to determine if the males preferred Pepsi less than the females Using the test statistic given z 2 11825 compute the appropriate p value for the test a 0 0170 b 0 0340 c 0 9660 d 0 9830 Explanation Alternative hypothesis is now pM pF or pM pF 0 If z 2 12 we want P z 2 12 5 4830 9830 3 26 03 252y0321 4 The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 1 minutes The probability is that a product is assembled in more than 20 minutes Solution x N 15 2 1 20 15 P x 20 P z P z 2 38 5 4913 0087 2 1 Make a diagram 5 The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 3 minutes Find x 275 for this distribution Solution x N 15 2 We want a point x 275 so that P x x 275 2750 The corre sponding value of z has P z z 275 2750 and P z z 275 7250 If this is true and zero is the median we must have P 0 z z 275 2250 The closest we can come on the Normal table is P 0 z 0 60 2257 So z 275 0 60 and x 275 z 15 0 60 2 3 16 38 6 A local real estate appraiser analyzed the sales prices of homes in 2 neighborhoods to the corresponding appraised values of the homes The goal of the analysis was to compare the distribution of sale to appraised ratios from homes in the 2 neighborhoods Random and independent samples were selected from the 2 neighborhoods from last year s homes sales 8 from each of the 2 neighborhoods Identify the nonparametric method that would be used to analyze the data a the Wilcoxon Signed Ranks Test D5b using the test statistic Z b the Wilcoxon Signed Ranks Test D5b using the table test statistic W c the Wilcoxon Rank Sum Test D5a using the table test statistic T d the Wilcoxon Rank Sum Test using the test statistic Z 7 Turn in your computer output from computer problem 1 only tucked inside this exam paper 3 points 2 point penalty for not handing this in TABLE 10 5 To test the effects of a business school preparation course 8 students took a general business test before the course The same eight students took the test after the course The data are given below Student Before x1 1 2 3 4 5 6 7 8 530 690 910 700 450 820 820 630 After x2 670 770 1000 710 550 870 770 610 diff Note Last column and sums not in original problem d x 2 x1 140 80 90 10 100 50 50 20 400 d2 19600 6400 8100 100 10000 2500 2500 400 49600 2 Two tests were run using Minitab Only one should have been run The results follow 3 3 26 03 252y0321 Table 1 Results for 2x0321 1 MTW MTB TwoSample c2 c1 SUBC Alternative 1 Two Sample T Test and CI after before Two sample T for after vs before after before N 8 8 Mean 744 694 StDev 144 155 SE Mean 51 55 Difference mu after mu before Estimate for difference 50 0 95 lower bound for difference 82 6 T Test of difference 0 vs T Value 0 67 P Value 0 258 DF 13 MTB Paired c2 c1 SUBC Alternative 1 Paired T Test and CI after before Paired T for after before after before Difference N 8 8 8 Mean 743 8 693 8 50 0 StDev 143 9 155 4 65 0 SE Mean 50 9 54 9 23 0 95 lower bound for mean difference 6 4 T Test of mean difference 0 vs 0 T Value 2 17 8 P Value 0 033 Compute the standard deviation of the d column and use it to fill in the underlined blanks in the second of the two tests in Table 1 Show your work 4 points 2 point penalty for not computing the variance d 2 nd 2 49600 8 50 2 Solution s d2 4228 5714 n 027 1 7 4228 5714 65 sd sd 65 027 8 4228 5714 22 9907 8 9 What was the alternate hypothesis tested in Table 1 1 H 1 D 0 or H 1 1 2 0 or H 1 1 2 10 Using the correct test state the null hypothesis and conclusion explaining what numbers in the tests led you to your conclusion 2 Solution Since this is paired data we use the second test The null …
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