8.2

a) \(\overset { \wedge }{ y } =120.07-1.93*parity\)

b) All else equal, a child will have a predicted birth weight 1.93 ounces less when parity is true (not the first born).

first_borns <- 120.07
first_borns
## [1] 120.07
not_first_born <- 120.07 -1.93
not_first_born
## [1] 118.14

c) Given the t value of -1.62, there is not a significant relationship to the typical .05 alpha level.

8.4

a) \(\overset { \wedge }{ y } =18.93-9.11*eth+3.1*sex+2.15*lrn\)

b) All else equal, the child can be expected to be absent 9.11 less days when not aboriginal, 3.1 more days when male, and 2.15 more days when a slow learner.

c)

predicted <- 18.93 + 3.1 + 2.15
predicted
## [1] 24.18
residual <- predicted - 2
residual
## [1] 22.18

d)

\({ R }^{ 2 }=1-\frac { 240.57 }{ 264.17 }\) =
0.0893364

\({ R }_{ adj }^{ 2 }=1-\frac { 240.57 }{ 264.17 } *\frac { 145 }{ 145-3 }\) =
0.070097

8.8

The adjusted R squared method would have us remove learner status. Adjusted R squared increases without it. It would also be removed with the P value method.

8.16

a) It appears there is an inverse relationship between damaged rings and temperature. The four lowest temperatures occurr when the there is at least one broken ring.

b) The intercept serves mostly to center the line. The lowest recorded temperature is 53, so a temperature of 0 isn’t very reasonable. The log odds of a ring failure will decrease by .2162 for each additional degree in farenheight. This parameter is easily statiscially signficant.

c) \({ p }_{ fail }\quad =\quad \frac { { e }^{ 11.663-.2126*temp } }{ 1+{ e }^{ 11.663-.2126*temp } }\)

d) The temperature parameter is easily staistically signficant. It is also practically significant, as the high point of a 95% confidence interval still have a parameter of about -.1. With the point estimates, low temperatures indicate high probability of failure, which is what is observed in the data.

8.18

a,b)

p51<- exp(11.663-51*.2126)/(1+exp(11.663-51*.2126))
p51
## [1] 0.6943212
p53<- exp(11.663-53*.2126)/(1+exp(11.663-53*.2126))
p53
## [1] 0.5975339
p55<- exp(11.663-55*.2126)/(1+exp(11.663-55*.2126))
p55
## [1] 0.4925006
probs <- data.frame(pbs = c(p51, p53, p55, .341, .251, .179, .124, .084, .056, .037, .024), index = c(1:11))

ggplot(probs) + geom_smooth(aes(y = pbs, x = index))
## `geom_smooth()` using method = 'loess'

c) I believe the only assumption that a GLM retains from a linear model is the idependance of error terms. The number of failures from each set of rings is dependant. More domain knowledge would help, but you would think the rest of the rings would weaken if one fails. I think the model should have been constructed with 0 indicating no rings failed and 1 indicated at least one ring failed. There also isn’t a lot of data, but the model might not be appropriate. There could just be a temperature, below which rings will fail.