CHAPTER 8 HOMEWORK - 8.2, 8.4, 8.8, 8.16, 8.18
Baby weights, Part II. 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.
\(\widehat{weight}\) = 120.07 + -1.93 * parity
The predicted birth weight of first-borns is 1.93 oz lower than first born baby
No because the p-value is greater than .05.
Absenteeism. Researchers interested in the relationship between absenteeism from school and certain demographic characteristics of children collected data from 146 randomly sampled stu- dents 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).
\(\widehat{daysabsent}\) = 18.93 + -9.11 * eth + 3.10 * sex + 2.15 * lrn
If the student is aboriginal, they are likely to have 9.11 more days absent.
If the student is male, they are likely to have 3.10 more days absent.
If the student is categorized as a slow learner, they are likely to have 2.15 more days absent.
pred <- 18.93 + -9.11 * 0 + 3.10 * 1 + 2.15 * 1
resid <- 2 - pred
resid
## [1] -22.18
R2 <- 1 - 240.57/264.17
adjR2 <- 1 - 240.57 / 264.17 * (146 - 1) / (146 - 3 - 1)
The R2 is 0.08934 and the Adjusted R2 is 0.0701
Absenteeism, Part II. Exercise 8.4 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?
The learner status should be removed first as per the adjusted R squared method.
Challenger disaster, Part I. 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.
The lower the temperature, the higher the probability of a damaged O-ring.
The estimate coefficients and the p-values are the key components. 11.6630 is the starting place for the model and -0.2162 is added to the starting place for every degree of temperature. The p-values help us better determine the variable fit in the model.
\(\widehat{pFailure}\) = e^(11.6630-0.2162 * temp) / (1 + e^(11.6630-0.2162 * temp))
Yes concerns regarding O-rings are justified since the P-Value is less than 0.05 - we must reject the null which indicates there is no relationship between damaged rings and temperature.
Challenger disaster, Part II. Exercise 8.16 introduced us to O-rings that were identified as a plausible explanation for the breakup of the Challenger space shuttle 73 seconds into takeo↵ 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.
e <- exp(1)
temperature <- c(51, 53, 55)
round(e^(11.6630-(0.2162 * temperature)) / (1 + e^(11.6630-(0.2162 * temperature))), 3)
## [1] 0.654 0.551 0.443
temperature <- c(50:85)
probability <- round(e^(11.6630-(0.2162 * temperature)) / (1 + e^(11.6630-(0.2162 * temperature))), 3)
plot(temperature, probability, type = "l", main = "probability of O-ring failure")
The residuals must have a normal distribution and the variance constant.