Baby weights, Part I. (9.1, p. 350) The Child Health and Development Studies investigate a range of topics. One study considered all pregnancies between 1960 and 1967 among women in the Kaiser Foundation Health Plan in the San Francisco East Bay area. Here, we study the relationship between smoking and weight of the baby. The variable smoke is coded 1 if the mother is a smoker, and 0 if not. The summary table below shows the results of a linear regression model for predicting the average birth weight of babies, measured in ounces, based on the smoking status of the mother.
The variability within the smokers and non-smokers are about equal and the distributions are symmetric. With these conditions satisfied, it is reasonable to apply the model. (Note that we don’t need to check linearity since the predictor has only two levels.)
weight = -8.94 * smoke + 123.05
# The slope is saying that for each increaese in unit of smoke,
# the weight in oz of the baby decreases by 8.94.
# smoke is only binary however and is restricted to the set {0, 1}
# Non smoking
-8.94 * 0 + 123.05
## [1] 123.05
# smoking
-8.94 * 1+ 123.05
## [1] 114.11
It is statistically significant because the p-value for smoke is nearly 0.
That means we would reject the null hypothesis that there is no relationship
between birth weight and smoking.
Absenteeism, Part I. (9.4, p. 352) Researchers interested in the relationship between absenteeism from school and certain demographic characteristics of children collected data from 146 randomly sampled students in rural New South Wales, Australia, in a particular school year. Below are three observations from this data set.
The summary table below shows the results of a linear regression model for predicting the average number of days absent based on ethnic background (eth: 0 - aboriginal, 1 - not aboriginal), sex (sex: 0 - female, 1 - male), and learner status (lrn: 0 - average learner, 1 - slow learner).
avg_n_days_absent = 18.93 - 9.11(eth) + 3.10(sex) + 2.15(lrn)
Holding everything else constant:
if not aboriginal, will miss 9.11 less days
if male, will miss 3.10 more days
if slow learner, will miss 2.15 more days
eth = 0
sex = 1
lrn = 1
estimate = 18.93 - (9.11 * eth) + (3.10 * sex) + (2.15 * lrn)
resid = 2 - estimate
resid
## [1] -22.18
var_resid = 240.57
var = 264.17
n = 146
R2 = 1 - var_resid/var
R2
## [1] 0.08933641
k = 3
adjR2 = 1 - (var_resid/var)*((n-1)/(n-k-1))
adjR2
## [1] 0.07009704
Absenteeism, Part II. (9.8, p. 357) Exercise above considers a model that predicts the number of days absent using three predictors: ethnic background (eth), gender (sex), and learner status (lrn). The table below shows the adjusted R-squared for the model as well as adjusted R-squared values for all models we evaluate in the first step of the backwards elimination process.
Which, if any, variable should be removed from the model first?
Eliminate learner status, because when you exclude it you get a higher
adjusted R2
Challenger disaster, Part I. (9.16, p. 380) On January 28, 1986, a routine launch was anticipated for the Challenger space shuttle. Seventy-three seconds into the flight, disaster happened: the shuttle broke apart, killing all seven crew members on board. An investigation into the cause of the disaster focused on a critical seal called an O-ring, and it is believed that damage to these O-rings during a shuttle launch may be related to the ambient temperature during the launch. The table below summarizes observational data on O-rings for 23 shuttle missions, where the mission order is based on the temperature at the time of the launch. Temp gives the temperature in Fahrenheit, Damaged represents the number of damaged O-rings, and Undamaged represents the number of O-rings that were not damaged.
When temperatures are low, you have more damaged O rings.
The intercept of 11.6630 is the log odds ratio that an O-ring will
be damaged against an O-ring not being damaged at 0 degrees Farenheit.
The temperature coefficient of -0.2162 is the log odds ratio that
an O-ring will be damaged vs not be damaged will change with a
1 degree increase in Farenheit.
log( (p-hat) / (1 - p-hat)) = 11.6630 - (0.2162 * Temperature)
#the probability of becoming damaged is:
#p = (e^(11.6630 - 0.2162*Temp))/(1 + (e^(11.6630 - 0.2162*Temp))).
Temp = 65
(exp(11.6630 - 0.2162*Temp))/(1 + (exp(11.6630 - 0.2162*Temp)))
## [1] 0.08393843
# We can see that a fairly normal temperature of 65, there is
# already an 8% probability of failure of O-ring.
# The concerns are justified
Challenger disaster, Part II. (9.18, p. 381) Exercise above introduced us to O-rings that were identified as a plausible explanation for the breakup of the Challenger space shuttle 73 seconds into takeoff in 1986. The investigation found that the ambient temperature at the time of the shuttle launch was closely related to the damage of O-rings, which are a critical component of the shuttle. See this earlier exercise if you would like to browse the original data.
\begin{center} \end{center}
where \(\hat{p}\) is the model-estimated probability that an O-ring will become damaged. Use the model to calculate the probability that an O-ring will become damaged at each of the following ambient temperatures: 51, 53, and 55 degrees Fahrenheit. The model-estimated probabilities for several additional ambient temperatures are provided below, where subscripts indicate the temperature:
\[\begin{align*} &\hat{p}_{57} = 0.341 && \hat{p}_{59} = 0.251 && \hat{p}_{61} = 0.179 && \hat{p}_{63} = 0.124 \\ &\hat{p}_{65} = 0.084 && \hat{p}_{67} = 0.056 && \hat{p}_{69} = 0.037 && \hat{p}_{71} = 0.024 \end{align*}\]
Temp = 51
(exp(11.6630 - 0.2162*Temp))/(1 + (exp(11.6630 - 0.2162*Temp)))
## [1] 0.6540297
Temp = 53
(exp(11.6630 - 0.2162*Temp))/(1 + (exp(11.6630 - 0.2162*Temp)))
## [1] 0.5509228
Temp = 55
(exp(11.6630 - 0.2162*Temp))/(1 + (exp(11.6630 - 0.2162*Temp)))
## [1] 0.4432456
temps <- c(51,53,55,57,58,63,66,67,68,69,70,72,73,75,76,78,79,81)
probabilities = function(temperature){
mod = 11.6630 - 0.2162 * temperature
probs = (exp(mod) / (1+exp(mod)))
return(probs)
}
probs <- sapply(temps, probabilities)
plot(temps, probs)
lines(temps, probs)
The assumptions for the model validity is each outcome is independent of the other outcomes
and that each predictor is linearly related to logit(pi) if all other
predictors are held constant.
My concerns are that the outcomes are not necessarily independent. If there are 6 O-rings
on each shuttle, and each monitored for 23 shuttle missions, there may not be independence,
espeically if they are counting each O-ring as a sample from the same shuttle mission.
Also, there are currently no other predictors besides temperature, so another predictor
could be used to check the second assumption of linear related logit.