TABLE 10-5TABLE 10-7TABLE 10-13/26/03 252y0321 ECO252 QBA2 Name KEY (Page layout view!) SECOND HOUR EXAM Hour of Class Registered _______ March 27 - 28, 2003I. (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 H0b) *at = 0.04, we should reject H0c) at = 0.03, we should reject H0d) 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 12-4A 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 interestingin 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 = 0.25pSM – pSF = 0.172018 z = 2.11825 (Note that 1H 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: 0– FMb) H1: 0– FMc) *H1: pM – pF 0d) H1: pM – pF 03. 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 11825.2z, compute the appropriate p value for the test.a) 0.0170b) 0.0340c) 0.9660d) *0.9830Explanation: Alternative hypothesis is now : pM < pF or : pM – pF < 0. If 12.2z, we want .9830.4830.5.12.2 zP3/26/03 252y03214. 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: 1.2,15~ Nx. 0087.4913.5.38.21.2152020 zPzPxP 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 275.x for this distribution. Solution: ..2,15~ Nx We want a point, 275.x, so that .2750.275.xxP The corre-sponding value of z has .2750.275.zzP and .7250.275.zzPIf this is true, and zero is the median, we must have .2250.0275. zzPThe closest we can come on the Normal table is .2257.60.00 zP So ,60.0275.zand .38.163.260.015275.zx6. 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 Zb) the Wilcoxon Signed-Ranks Test, D5b, using the table test statistic Wc) *the Wilcoxon Rank Sum Test, D5a, using the table test statistic Td) the Wilcoxon Rank Sum Test, using the test statistic Z7. 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 10-5To 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 After diff Note: Last column and sums not in original problem. 1x 2x 12xxd 2d 1 530 670 140 19600 2 690 770 80 6400 3 910 1000 90 8100 4 700 710 10 100 5 450 550 100 10000 6 820 870 50 2500 7 820 770 -50 2500 8 630 610 -20 400 400 49600 2Two tests were run using Minitab. (Only one should have been run!). The results follow: 33/26/03 252y0321Table 1:Results for: 2x0321-1.MTWMTB > TwoSample c2 c1;SUBC> Alternative 1.Two-Sample T-Test and CI: after, beforeTwo-sample T for after vs before N Mean StDev SE Meanafter 8 744 144 51before 8 694 155 55Difference = mu after - mu beforeEstimate for difference: 50.095% lower bound for difference: -82.6T-Test of difference = 0 (vs >): T-Value = 0.67 P-Value = 0.258 DF = 13MTB > Paired c2 c1;SUBC> Alternative 1.Paired T-Test and CI: after, beforePaired T for after - before N Mean StDev SE Meanafter 8 743.8 143.9 50.9before 8 693.8 155.4 54.9Difference 8 50.0 65.0 23.0 95% lower bound for mean difference: 6.4T-Test of mean difference = 0 (vs > 0): T-Value = 2.17 P-Value = 0.0338. 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.)Solution: 5714.422875084960012222ndndsd027.655714.4228 ds9907.2285714.42288027.65ds9. What was the alternate hypothesis tested in Table 1? (1)0:1DH or 0:211Hor 211:H10. Using the correct test, state the null hypothesis and conclusion,
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