Refer to Problem 13.12. Fit the Poisson and negative binomial GLMs using identity link. Show that the
Question:
Data from Problem 13.12:
One question in the 1990 General Social Survey asked subjects how many times they had sexual intercourse in the preceding month. Table 13.9 shows responses, classified by gender.
a. The sample means were 5.9 for males and 4.3 for females; the sample variances were 54.8 and 34.4. The mode for each gender was 0. Does an ordinary Poisson GLM seem appropriate? Explain.
b. The Poisson GLM with log link and a dummy variable for gender (1 = males, 0 = females) has gender estimate 0.308 (SE = 0.038). Explain why this implies a ratio of 1.36 for the fitted means. (This is also the ratio of sample means, since this model has fitted means equal to sample means.) Show that the WaId 95% confidence interval for the ratio of means for males and females is (1.26, 1.47).
c. For the negative binomial model, the log likelihood increases by 248.7 (deviance decreases by 497.3). The estimated difference between the log means is also 0.308, but now SE = 0.127. Show that the 95% confidence interval for the ratio of means is (1.06, 1.75). Compare to the Poisson GLM, and interpret.
d. The mode for the Poisson distribution is the integer part of the mean, rather than 0. Argue that a possibly more realistic mixture model assumes for gender i a proportion Ïi that has a Poisson distribution with mean 0 and a proportion 1 €“ Ïi that has distribution that is a gamma mixture of Poissons. Explain why the corresponding marginal distribution for each gender is a mixture of a degenerate distribution at 0 and a negative binomial distribution.
Table 13.9:
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Step by Step Answer:
The estimated difference of means is 1552 for each model but SE ...View the full answer
