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solutions6

# solutions6 - Stat 665(Spring 2011 Kaizar Solutions 6...

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Stat 665 (Spring 2011) Kaizar Solutions 6 Exercises 40 points total 1. (8 points) Problem 4.8 (a) From the R output: Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -3.6947 0.8802 -4.198 2.70e-05 *** wtkg 1.8151 0.3767 4.819 1.45e-06 *** The prediction equation is: logit π ) = - 3 . 69 + 1 . 82 X where X is the weight of the crab in kg. (b) Substituting X=1.2, 2.44, and 5.2 into the above prediction equation, and then taking the inverse of the logit (using the predict function in R) gives probabilities of 0.180, 0.676, and 0.997, respectively. (c) When ˆ π = 0, logit π ) = 0, which occurs when 1 . 82 X = 3 . 69, or when the weight = 3.69/1.82 = 2.04kg. (d) The linear approximation of the slope in probability is ˆ β ˆ π (1 - ˆ π ) = 1 . 8151 * 0 . 25 = 0 . 454. That is, for every increase in one kg weight (near 2.04kg), the probability of having a satellite increases by about 0.45. To convert to an increase in 0.1kg, I need to convert the meaning of β . If β is the increase in log odds for every one kg increase in weight, then 0 . 1 β is the increase in log odds for every 0.1 kg increase in weight. Thus, the estimated increase in probability of having a satellite for every increase in 0.1 kg is 0.1*0.454 = 0.045. Similarly, the extimated increase in probability of having a satellite for every increase in one SD of weight is 0.58*0.454= 0.263. (e) A 95% CI for β is ˆ β ± 1 . 96 * 0 . 38: (1.076834, 2.553455). Exponentiating gives a 95% CI for the increase in odds of a satellite for every one kg increase in weight: (2.94 12.85). (f) From the output above, we can see that the Wald p-value is 1.45e-06. Thus, I reject the null hypothesis that the coefficient on weight is zero. That is, I conclude that weight does have a significant association with the odds of having a satellite. To find the LRT statistic, we subtract the residual and null deviances: Null deviance: 225.76 on 172 degrees of freedom Residual deviance: 195.74 on 171 degrees of freedom 1

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LRT = 225.76 - 195.74 = 30.02. I compare this to a χ 2 1 distribution to get the p-value = 4.273103e-08. Thus, I reject the null hypothesis that the coefficient on weight is zero. That is, I conclude that weight does have a significant association with the odds of having a satellite. 2. (6 points) Exercise 4.9, parts a-d (a) Output from R: Estimate Std. Error z value Pr(>|z|) (Intercept) 1.0986 0.6667 1.648 0.0994 . colorfactor2 -0.1226 0.7053 -0.174 0.8620 colorfactor3 -0.7309 0.7338 -0.996 0.3192 colorfactor4 -1.8608 0.8087 -2.301 0.0214 * Looking at Table 3.2, I can interpret the meanings of each of the colors: 1=light medium (LM), 2=medium (M), 3=dark medium (DM), 4=dark (D). The prediction equation is: logit π ) = 1 . 0986 - 0 . 1226 M - 0 . 7309 DM - 1 . 8608 D where for each color indicator, the covariate =1 if the crab is that color, and 0 otherwise. I interpret the first indicator variable “colorfactor2” = -0.1226 to mean that the log odds of a satellite are -0.1226 smaller for crabs with medium color than crabs with light medium color. That is, the log odds ratio of satellite for medium and light medium crabs is -0.1226. (b) To test whether color has any effect, we need to consider that all of the coefficients for all the colors = 0. Thus, we can not use a Wald test, but instead must rely on a LRT: Null deviance: 225.76 on 172 degrees of freedom Residual deviance: 212.06 on 169 degrees of freedom LRT = 225.76 - 212.06 = 13.70.
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solutions6 - Stat 665(Spring 2011 Kaizar Solutions 6...

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