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(b) Use model “budworm5” to estimate the odds of a female budworm dying if “ldose” equals 3. From (a) , d odds i = exp ⇣ ˆ ↵ F I F ( i ) + ˆ ↵ M I M ( i ) + ˆ β ldose i ⌘ So, for a container of 20 female budworms, at ldose = 3, an estimate of the odds of a randomly selected female budworm dying equals d odds ( Female, ldose = 3) = exp ( - 3 . 4732 + 1 . 0642(3)) = 0 . 7553 (after rounding) (c) Use model “budworm5” to estimate the expected number of budworms that would die in a container containing 20 female budworms, if ldose equals 3. If container i is the container with 20 females at ldose = 3, then b E (number of females dead in Female container at ldose = 3) = 20ˆ ⇡ i , where ˆ ⇡ i is the estimated probability of death corresponding to the estimated odds in part (2b). If we use notation ˆ γ i = d odds ( Female, ldose = 3), then ✓ ˆ ⇡ i 1 - ˆ ⇡ i ◆ = ˆ γ i () ˆ ⇡ i = ˆ γ i (1 - ˆ ⇡ i ) () ˆ ⇡ i (1 + ˆ γ i ) = ˆ γ i () ˆ ⇡ i = ˆ γ i (1 + ˆ γ i ) So ˆ ⇡ i = 0 . 7553 (1 + 0 . 7553) = 0 . 4303 , and b E (number of females dead in Female container at ldose = 3) = 20ˆ ⇡ i ⇡ 20(0 . 4303) ⇡ 8 . 6059 . Note: the expected value does not need to be an integer. Several students took the additional incorrect step of concluding that the expected number equals 9. Please ask if you have questions about expected values. (d) What is the change in the odds of a female budworm dying if ldose increases from 3 to 4? Show working. odds ( Female, ldose = 3) = exp ( ↵ F + β (3)) odds ( Female, ldose = 4) = exp ( ↵ F + β (4)) = odds ( Female, ldose = 3) ⇥ exp( β ) So the odds of a Female dying at ldose = 4 is exp( β ) times the odds for a Female dying at ldose = 3. In terms of the estimates based on these data: exp( ˆ β ) = exp(1 . 0642) = 2 . 8958 (rounded). Alternatively, d odds ( Female, ldose = 4) - d odds ( Female, ldose = 3) = exp ⇣ ˆ ↵ F + ˆ β (3) ⌘ ⇥ (exp( ˆ β ) - 1) = 1 . 4340 . (e) What is the change in the probability of a female budworm dying if ldose increases from 3 to 4? Show working. If γ F,j = odds ( Female, ldose = j ), and ⇡ F,j = the probability a randomly selected Female dies at ldose = j , then ✓ ⇡ F,j 1 - ⇡ F,j ◆ = γ F,j () ⇡ F,j = γ F,j (1 + γ F,j ) . So ⇡ F, 4 = γ F, 4 (1 + γ F, 4 ) = ( γ F, 3 ) e β (1 + ( γ F, 3 ) e β ) From (2c), ˆ ⇡ F, 3 = 0 . 4303. Similarly, ˆ ⇡ F, 4 = 2 . 1893 (1+2 . 1893) ⇡ 0 . 6865 . So ˆ ⇡ F, 4 - ˆ ⇡ F, 3 = 0 . 6865 - 0 . 4303 = 0 . 2562. (f) Would the di ↵ erence between predicted probabilities of death for males and females be the same at ldose=1, and at ldose=3? Why or why not? Draw a sketch/sketches with clear labels to justify your answer. The di ↵ erences between the log(odds) between males and females would be the same at ldose=1 and ldose=3. However, the di ↵ erences between predicted probabilities of death would not be equal, since logit is a non-linear transformation. In your sketch, the Male and Female prediction lines should be parallel on the logit scale, but not on the probability scale. You must clearly label all axes, not only sketch the pattern. You didn’t need to include the following, but for completeness: the predictions based on the current data show that ˆ ⇡ M, 1 - ˆ ⇡ M, 3 6 = ˆ ⇡ F, 1 - ˆ ⇡ F, 3 : d odds M, 1 = exp ⇣ ˆ ↵ M + ˆ β (1) ⌘ = exp( - 2 . 3724 + 1 . 0642(1)) ⇡ 0 . 2703 , so ˆ ⇡ M, 1 ⇡ 0 . 2703 (1 + 0 . 2703) = 0 . 2128 d odds M, 3 = exp ⇣ ˆ ↵ M + ˆ β (3) ⌘ = exp( - 2 . 3724 + 1 . 0642(3)) ⇡ 2 . 2710 , so ˆ ⇡ M, 3 ⇡ 2 . 2710 (1 + 2 . 2710) = 0 . 6943 d odds F, 1 = exp ⇣ ˆ ↵ F + ˆ β (1) ⌘ = exp( - 3 . 4732 + 1 . 0642(1)) ⇡ 0 . 0899 , so ˆ ⇡ F, 1 ⇡ 0 . 0899 (1 + 0 . 0899) = 0 . 0825 d odds F, 3 = exp ⇣ ˆ ↵ F + ˆ β (1) ⌘ = exp( - 3 . 4732 + 1 . 0642(3)) ⇡ 0 . 7553 , so ˆ ⇡ F, 3 ⇡ 0 . 7533 (1 + 0 . 7533) = 0 . 4303 (rounded) 3 This study source was downloaded by 100000875045266 from CourseHero.com on 10-29-2023 17:08:00 GMT -05:00 https://www.coursehero.com/file/81313248/Solutions-AdditionalPracticeProblemsForMidtermPreppdf/

(c) Based on the parameter estimates in model fit1, explain how the predicted odds of buying a car would di ↵ er for 2 households that have identical incomes but cars that di ↵ er in ages by one year. Consider household A which has income x 1 = a and oldest automobile x 2 = c years old, then using the parameter estimates ˆ β 0 , ˆ β 1 and ˆ β 2 , the estimated odds of household A buying a new car in the next year is d odds(household A buys a new car) = exp[ ˆ β 0 + ˆ β 1 a + ˆ β 2 c ] . Consider household B which has income x 1 = a and oldest automobile x 2 = ( c +1) years old, then the estimated odds of household B buying a new car is d odds(household B buys a new car) = exp[ ˆ β 0 + ˆ β 1 a + ˆ β 2 ( c + 1)] = d odds(household A buys a new car) exp[ ˆ β 2 ] . That is, the estimated odds that household B buys a new car in the next year, is exp[ ˆ β 2 ] = exp[0 . 59863] ⇡ 1 . 820 times the odds that household A buys a new car in the next year. It is good if you also note that age is not significant in this model (in addition to giving the answer above). > fit2 <- update(fit1, .~.-age) > summary(fit2) Call: glm(formula = purchase ~ income, family = binomial(link = logit)) Deviance Residuals: Min 1Q Median 3Q Max -1.5883 -0.8430 -0.7121 0.9262 1.7688 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.98079 0.85720 -2.311 0.0208 * income 0.04342 0.02011 2.159 0.0308 * --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 44.987 on 32 degrees of freedom Residual deviance: 39.305 on 31 degrees of freedom AIC: 43.305 Number of Fisher Scoring iterations: 4 # ------------------------------------------------------------------------------------------------------ > round( vcov(fit2), digits=5 ) # vcov provides the variance-covariance matrix of the (Intercept) income # main parameters in model fit2 (Intercept) 0.73478 -0.0154 income -0.01540 0.0004 (d) Using the results of model ”fit2” (i.e., assuming model ”fit2” is the correct model for the answer to this part of the question), find a 99% confidence interval estimate for β income . (Note: I would provide you with tables if asking this question in the midterm.) Using the approximate normality of ˆ β , the form of a 99% confidence interval is h ˆ β income ± z 0 . 005 se ( ˆ β income ) i , where z 0 . 005 ⇡ 2 . 575, ˆ β income = 0 . 04342, and se ( ˆ β income ) = 0 . 02011. Thus a 99% approximate confidence interval for β income is [0 . 04342 - 2 . 575(0 . 02011) , 0 . 04342 + 2 . 575(0 . 02011)] ⇡ [ - 0 . 00836 , 0 . 09520] . 6 This study source was downloaded by 100000875045266 from CourseHero.com on 10-29-2023 17:08:00 GMT -05:00 https://www.coursehero.com/file/81313248/Solutions-AdditionalPracticeProblemsForMidtermPreppdf/