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UW-Madison SOC 360 - Mock Exam 3 Answer Key

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Mock Exam 3 Answer Key1. T 9. H2. F 10. A3. T 11. B4. T 12. D5. T 13. A6. A 14, A7. A 15. B8. BPart II. 1a. df = 20-1 = 19. 99% CI, t*= 2.861b. P: (1-.98)/2 = 0.01, z* = 2.326c. df = 20-2 = 18, two-tailed P: 0.04/2 = 0.02, t*= 2.2142a. White protestants: 104/267 = 0.39White catholics: 75/230 = 0.33b. (1) Ho : P = 0.5Ha: P < 0.5(2) z = (0.39-0.5)/ [ (0.5*0.5)/267] = -0.11/0.306 = 3.59(3) For z = 3.59, P-value < 0.0002. At alpha = 0.01 level, we reject the null and thehistorian’s claim is supported.c. a 95% confidence interval for the difference between the two proportions:Large Sample Method:phat1 = 104/267 = 0.390phat2 = 75/230 = 0.326(0.390 - 0.326) +/- 1.96* [{(0.390*(1-0.390))/267} + {0.326*(1-0.326)/230}]=0.064 +/- 1.96* 0.043=0.064 +/- 0.084= (-0.020, 0.148)Plus-4 estimate:phat1 = (104+1)/(267+2)=0.390phat2 = (75+1)/(230+2)=0.328(0.390 - 0.328) +/- 1.96* [{(0.390*(1-0.390))/267+2} + {0.328*(1-0.328)/230}]=0.062 +/- 1.96* 0.043=0.062 +/-0 .084= (-0.022, 0.146)3. Ho: : = 64Ha: : > 64(a) df = 25-1 = 24(b) SE = (10/5) = 2(c) t = (80-64)/2 = 8(d) t* = 2.492, we will reject the null hypothesis at the alpha = 0.01 level4a. sample size = n1 + n2 = 50 > 40. Although the population of income is veryhighly skewed, we can still use a t-test to examine two means since the sample sizeis large enough. b. It’s a SRS; both counts are greater than 10 and the population is at least 10 timesas large as the sample size. Therefore, we can use the normal distribution to testthe null hypothesis. 4c1 No. Normality of residuals is required for prediction intervals, regardless ofsample size.4c2. Yes, sample size is sufficiently large.5a. Ho: :single = :marriedb. a 95% CI: (41.33138, 43.74638)c. t = -2.0016, df = 200-1 = 199, 0.025> P-value > 0.026a. 2b. two possible answersNo, spread of residuals is highly unequal over the range of X, so cannot trust thehypothesis tests for regression statistics.OR No, there is evidence of nonlinearity in the original relationship based on theresidual graph. For low and high values of X, residuals are mostly greater thanzero. For middle values of X, the residuals are mostly less than zero.7a. P-value < 0.0005. It is significant at the alpha = 0.01 level that there is arelationship between the number of hours of tv watched and education. b. A 90% CI for ß: -0.1895677 +/- 1.646*0.0189853 = (-0.2208, -0.1583)c. Predicted y = 5.433871 - 0.1895677*16 = 2.40d. Predicted y = 5.433871 - 0.1895677*12 = 3.16e. prediction of part c will probably be more accurate8Before After Score Difference =(after-before)deviation deviation21 400 550 150 95 90252 450 610 160 105 110253 520 490 -30 -85 72254 540 540 0 -55 30255 480 460 -20 -75 56256 550 620 70 15 225Mean = 55 SD = 85.0294Ho: : = 0 (: = mean of the score difference) Ha: : > 0 t = (55-0)/ (85.0294/(6^.5)) = 55/34.7131 = 1.58, df = 6-1= 5. Therefore, 0.10> P-value>0.05. We have some evidence against the null hypothesis, but not strong


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UW-Madison SOC 360 - Mock Exam 3 Answer Key

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