STAT 453/653 Homework 6 Solutions

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2 STAT 453/653 Homework 6 Solutions By Virajitha Karnatapu Ajay Kumar November 4, ) In the first nine decades of the twentieth century in baseball s National League, the percentage of times the starting pitcher pitched a complete game were: 72.7 ( ) 63.4, 50.0, 44.3, 41.6,32.8, 27.2, 22.5, 13.3 ( ) a) Treating the number of games as the same in each decade, the linear probability model has ML fit ˆπ = x, where x = decade (x=1,2,...,9). Interpret The negative slope says that ˆπ(x) decreases as x increases. Therefore the estimated probability that the starting pitcher pitches a complete game decreases per decade. b) Substituting x=12, predict the percentage of complete games for Is this prediction plausible? Why? ˆπ = (12) = -7.5%. Since the percentage is a negative value, this prediction is not plausible. c) The logistic regression ML fit is ˆπ = exp( x) [1+exp( x)]. Obtain ˆπ for x=12. Is this more plausible than the prediction in (b)? ˆπ(12) = exp( (12) [1+exp( x) = 6.7%. This prediction is more plausible than the one in part(b). 4.4) Consider the snoring and heart disease data of Table 3.1 in section With scores 0,2,4,5 for snoring levels, the logistic regression ML fit is logit(ˆπ) = x. a) Interpret the sign of the estimated effect of the x. We have β = The positive value of β says that the estimated probability of heart disease increases as snoring level increases. 2

3 b) Estimate the probabilities of heart disease at snoring levels 0 and 5. ˆπ(0) = ˆπ(5) = exp( (5 [1+exp( (5) = exp( (5) [1+exp( (5) = c) Describe the estimated effect of snoring on the odds of heart disease. The odds of a heart disease is given by π(x) 1 π(x) = exp(α + βx) = e(α+βx). Thus the probability of having a heart disease is e (α+βx) times the probability of not having a heart disease for all snoring levels 0,2,4,5. Since e (α+βx) increases as x = 0,2,4,5 increases, we conclude that the probability of having disease increases as snoring level increases. 4.10) An International poll quoted in an Associated Press story reported low approval ratings for President George W.Bush among traditional allies of the United States, such as 32% in Canada, 30% in Britain, 19% in Spain, and 17% in Germany. Let Y indicate approval of Bush s performance (1 = yes, 0 = no), π = P (Y = 1), c 1 = 1 for Canada and 0 otherwise, c 2 = 1 for Britain and 0 otherwise, and c 3 = 1 for Germany and 0 otherwise. a) Explain why these results suggest that for the identity link function, ˆπ = c c c 3. The probability of approval for Canada is P(Y=1) = ˆπ = (1) = 0.32 = 32%. The probability of approval for Britain is P(Y=1) = ˆπ = (1) = 0.30 = 30%. 3

4 The probability of approval for Spain is P(Y=1) = ˆπ = (1) = 0.19 = 19%. The probability of approval for Germany is P(Y=1) = ˆπ = 0.17 = 17%. b) Show that the prediction equation for the logit link function is logit(ˆπ) = c c c 3. logit(ˆπ) = ln ˆπ 1 ˆπ can be found for each country with the given probabilities. logit(ˆπ) for Germany is ln = logit(ˆπ) for Cananda is ln = 0.75 = (1). logit(ˆπ) for Britain is ln = 0.85 = (1). logit(ˆπ) for Spain is ln = 1.45 = (1). Hence the prediction equation for the logit link function is logit(ˆπ = c c c ) A sample of subjects were asked their opinion about current laws legalizing abortion(support,oppose). For the explanatory variables gender(female,male), religious affiliation(protestant,catholic,jewish),and political party affiliation(democrat, Republican, Independent), the model for the probability π of supporting legalized abortion, logit(π) = α + β G h + βr i + β P j has reported parameter estimates (setting the parameter for the last category of a variable equal to 0.0) ˆα = 0.11, ˆβ 1 G = 0.16, ˆβ 2 G = 0.0, ˆβ 1 R = 0.57, ˆβ 2 R = 0.66, ˆβ 3 R = 0.0, ˆβ 1 P = 0.84, ˆβ 2 P = 1.67, ˆβ 3 P = 0.0. a) Interpret how the odds of supporting legalized abortion depend on gender. The estimated difference in logit between females and males as below is = ˆβ 1 G ˆβ 2 G = = The estimate of odds of females supporting legalized abortion is exp(0.16) = 1.17 times the odds for males. b) Find the estimated probability of supporting legalized abortion for (i)male Catholic Republicans and (ii) female Jewish Democrats. The estimated probability π of supporting legalized abortion for male Catholic Republicans is π = π = exp(ˆα+ ˆβ G 2 + ˆβ R 2 + ˆβ P 2 ) 1+exp(ˆα+ ˆβ G 2 + ˆβ R 2 + ˆβ P 2 ) exp( ) 1+exp( ) =

5 The estimated probability π of supporting legalized abortion for female Jewish Democrats is π = π = exp(ˆα+ ˆβ G 1 + ˆβ R 3 + ˆβ P 1 ) 1+exp(ˆα+ ˆβ G 1 + ˆβ R 3 + ˆβ P 1 ) exp( ) 1+exp( ) = 0.71 c) If we defined parameters such that the first category of a variable has value 0, then what would ˆβ 2 G equal? Show then how to obtain the odds ratio that describes the conditional effect of gender. Male gender has value ˆβ 1 G = 0, then the difference in logit between females and males is 0.16, the value of ˆβ 2 G = The estimated odds ratio that describes effect of gender still remains the same which is exp(0.16)=1.17. d) If we defined parameters such that they sum to 0 across the categories of a variable, then what would ˆβ 1 Gand ˆβ 2 G equal? Show then how to obtain the odds ratio that describes the conditional effect of gender. If the sum of logit between females and males becomes 0, then to get the sum of 0 and and difference of 0.16, the value of ˆβ 1 Gand ˆβ 2 G should be 0.08 and respectively. The estimated odds ratio that describes effect of gender still remains to be exp(0.16)=1.17 5