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Solutions to Exercises in Statistical Analysis: Chapter 13, Assignments of Statistics

Solutions to exercises in a statistics course, specifically for chapter 13. It covers various statistical analyses, including regression analysis, and includes calculations and interpretations of results. Students can use this document to check their understanding of the concepts covered in the chapter.

Typology: Assignments

Pre 2010

Uploaded on 09/17/2009

koofers-user-57g
koofers-user-57g 🇺🇸

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Download Solutions to Exercises in Statistical Analysis: Chapter 13 and more Assignments Statistics in PDF only on Docsity! STA 6127: Solutions of Exercises for Chap. 13 1a. Use first equation (ignoring father’s education) to get 11 + 2(1) = 13 for whites and 11 for nonwhites, for which the difference = 2. b. See p. 102a of solutions manual. c. The coefficient of Z in the second equation, which is −.6. d. Substitute X = 12 in second equation, yielding 12.0 for whites (Z = 1) and 12.6 for nonwhites (Z = 0). 2a. The predicted proportion of pro-choice votes increased by .005 for each 1% increase in the per- centage nonwhite of the district, controlling for the other predictors in the model. b. The predicted proportion of pro-choice votes was .063 higher for women members, controlling for the other predictors. c. The predicted proportion of pro-choice votes was .167 lower for Democrats, controlling for the other predictors. No, the relationship could be very different when we ignore rather than control the other factors. d. Ideology seems to be, by far, the most important predictor of proportion of pro-choice votes. A standard deviation increase in ideology corresponds to a .83 standard deviation predicted increase in the response, controlling for the other variables in the model. 3a. Ŷ = 8.3 + 9.8F − 5.3G + 7.0M1 + 2.0M2 + 1.2M3 + .501X . b. Ŷ = 8.3 + 9.8(1) + 7.0(1) + .501(10) = 30.1. c. Predicted number of alcoholic drinks was 5.3 lower for females than for males, controlling for father’s death, marital status, and alcohol consumption three years ago. d. t = −5.3/1.6 = −3.3, P = .001; the mean alcohol consumption is lower for females than males, controlling for other variables. e. −5.3 ± 1.96(1.6) = (−8.4,−2.2); the mean alcohol consumption is between 2.2 and 8.4 drinks per month lower for females than males, controlling for other variables. f. It compares the mean for the divorced group to the baseline group (married), controlling for other predictors. To compare all four levels of marital status, we need to conduct the F test comparing this model to the simpler model without these three dummy variables. 4a. The regression model is E(Y ) = α + β(X) + β1Z. Here β denotes the partial effect of X on the mean of Y i.e it is the amount by which the price of a house would change for 1 unit change in its size controlling for the type (old or new). On the other hand, β1 denotes the differences in the means of Y between the two categories of houses (old and new), controlling for size of home. b. Ŷ = −26.1+72.6X+19.6Z. The predicted selling price is 19.6 thousand dollars higher for new homes, controlling for size. The separate equations are Ŷ = −26.1+72.6X for older homes and Ŷ = −6.5+72.6X for new homes. c. (i) Ŷ = −6.5 + 72.6(3) = 211.2, (ii) 191.6. d. Adjusted mean for older homes would be larger, since mean selling price would increase if the average size were larger. 5a. (i) Ŷ = −48.4 + 96.0X , (ii) Ŷ = −16.6 + 66.6X ; the coefficient 29.4 of the cross-product term is the difference between the two slopes. b. (i) 239.6 thousand dollars, (ii) 183.2 thousand dollars; the predicted difference is larger than with the no interaction model. c. (i) 95.6, (ii) 83.3; for the model allowing interaction, the difference in predicted prices increases as the size of home increases. d. t = 29.4/8.2 = 3.6, P = .0005, which provides very strong evidence of interaction. There seems to be a greater slope for new homes; that is, the effect of size of home on price is greater for new homes. 6a. Ŷ = −16.6 + 66.6X for older homes, and Ŷ = −10.6 + 71.6X for new homes; now the lines are quite similar. b. (i) 207.13 thousand dollars, (ii) 183.21 thousand dollars. On comparing the results with that of problem 13.4 (c) we conclude that the predicted difference is larger than with the no interaction model.