Exercise 8.1 introduces a data set on birth weight of babies. Another variable we consider is parity, which is 0 if the child is the first born, and 1 otherwise. The summary table below shows the results of a linear regression model for predicting the average birth weight of babies, measured in ounces, from parity.
and certain demographic characteristics of children collected data from 146 randomly sampled students in rural New SouthWales, Australia, in a particular school year. Below are three observations from this data set.
$ e_1 = y_1 ??? = 2 ??? 24.18 = ???22.18 $
$R^2 = 1 - = 0.0893364 $ \(R^2_adj = 1 - \frac{240.57}{264.17} * \frac{145}{145 - 3} = 0.070097\)
The no learner status has the highest \(R_{adj}^2 = 0.0723\), and it is higher than the \(R_{adj}^2 = 0.0701\) of the full model. The learner status variable should be eliminated from the model first.
a). Out of 23 shuttle missions, it appears that missions launched at higher ambient tempreatures produced the least or no damaged O-rings. Also, the launch with the most damaged O-rings (5) happens to be the mission launched on the coldest ambient temperature (53 F). All missions launched in temperature below 65 F had at least one damaged O-ring. Out of 13 missions launched in temperature 70 F or higher, only 3 showed damaged O-rings. These two variables seem to be inversely related.
b). From the summary table, there is an inverse relationship between temperature and O-ring failures. This implies that Increase in temperature decreases the probability of an O-ring failure by 0.216.
c). \(log(\frac{\hat{p}}{1-\hat{p}}) = 11.6630 - 0.2162 * Temperature\)
d). The p-value is practically 0 which shows that the failures are likely not due to chance. Moreover, there appears to be an inverse relationship between temperature and failures of the O-RINGS. So, I think it is justified to be concerned.
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' and formula 'y ~ x'