8.2

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.

  1. Write the equation of the regression line.

y = -1.93*parity + 120.07

  1. Interpret the slope in this context, and calculate the predicted birth weight of first borns and others.

If the child is not a first born, the weight is estimated to be 1.93 less than if it was a first born.

  1. Is there a statistically significant relationship between the average birth weight and parity?

No, the p-value is too high to be deemed satistically significant.

8.4

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.

  1. Write the equation of the regression line.

y = -9.11(eth) + 3.10(sex) + 2.15(lrn) + 18.93

  1. Interpret each one of the slopes in this context.

While holding all other constants equal, we estimate that an not aboriginal eth estimates absenteesim of 9.11 lower.

While holding all other constants equal, we estimate that being female estimates absenteesim of 3.10 lower.

While holding all other constants equal, we estimate that an average learner estimates absenteesim of 2.15 lower.

  1. Calculate the residual for the first observation in the data set: a student who is aboriginal, male, a slow learner, and missed 2 days of school.
daysMissed <- 2
predictedDays <- -9.11*(0) + 3.10*(1) + 2.15*(1) + 18.93
residuals <- daysMissed-predictedDays 
residuals
## [1] -22.18
  1. The variance of the residuals is 240.57, and the variance of the number of absent days for all students in the data set is 264.17. Calculate the R2 and the adjusted R2. Note that there are 146 observations in the data set.
n <- 146
k <-3
var_residuals <- 240.57
var_students <- 264.17

R2 <- round(1 - (var_residuals/var_students),6)
adjusted_R2 <- round(1-(var_residuals/var_students)*((n-1)/(n-k-1)),6)

print(paste0("R2: ",R2,", adjusted R2: ",adjusted_R2 ))
## [1] "R2: 0.089336, adjusted R2: 0.070097"

8.8

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 lrn variable has the highest adjusted R2, and should be removed.

8.16

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.

  1. Each column of the table above represents a different shuttle mission. Examine these data and describe what you observe with respect to the relationship between temperatures and damaged O-rings.
temperature <- c(53,57,58,63,66,67,67,67,68,69,70,70,70,70,72,73,75,75,76,76,78,79,81)
damaged <- c(5,1,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,0)
undamaged <- c(1,5,5,5,6,6,6,6,6,6,5,6,5,6,6,6,6,5,6,6,6,6,6)
Shuttle_Mission <- data.frame(temperature, damaged, undamaged)

plot(Shuttle_Mission)

It appears the as temperatures rise, there are less damages to the O-rings.

  1. Failures have been coded as 1 for a damaged O-ring and 0 for an undamaged O-ring, and a logistic regression model was fit to these data. A summary of this model is given below. Describe the key components of this summary table in words.

The components are the intercept (11.6630) and the Temperature variable (-0.2162). Both are significant. We estimate that for every 1 increase in temperature (Fahrenheit) there is a decrease of 0.2161 of probability in a damaged O-ring.

  1. Write out the logistic model using the point estimates of the model parameters.

log(p/(1-p)) = -0.2161(Temperature) + 11.6630

  1. Based on the model, do you think concerns regarding O-rings are justified? Explain.

Yes, we estimate O-ring damages through Temperature fairly well. This can prevent mission failures and even deaths.

8.18

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 takeoin 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.

  1. The data provided in the previous exercise are shown in the plot. The logistic model fit to these data may be written as

where ˆp 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:

f51 = 51
f53 = 53
f55 = 55
 
damage_prob_51 = round(exp(11.6630 - (0.2162*f51))/(1+exp(11.6630-(0.2151*f51))),6)
damage_prob_53 = round(exp(11.6630 - (0.2162*f53))/(1+exp(11.6630-(0.2151*f53))),6)
damage_prob_55 = round(exp(11.6630 - (0.2162*f55))/(1+exp(11.6630-(0.2151*f55))),6)

print(paste0("Damaged probabality at 51: ",damage_prob_51,", Damaged probabality at 53: ",damage_prob_53,", Damaged probabality at 55: ",damage_prob_55))
## [1] "Damaged probabality at 51: 0.630244, Damaged probabality at 53: 0.533285, Damaged probabality at 55: 0.431322"
  1. Add the model-estimated probabilities from part (a) on the plot, then connect these dots using a smooth curve to represent the model-estimated probabilities.
library(ggplot2)
predict_probability <- function(temp) 
{
return (round(exp(11.6630 - (0.2162*temp))/(1+exp(11.6630-(0.2151*temp))),4))
}

temperatures <- c(51:81)
probabilities <- predict_probability(temperatures)

temp_df <- as.data.frame(cbind(temperatures,probabilities))

ggplot(temp_df,aes(temperatures,probabilities)) + geom_point() + stat_smooth(method = 'glm', family = 'binomial')
## Warning: Ignoring unknown parameters: family

  1. Describe any concerns you may have regarding applying logistic regression in this application, and note any assumptions that are required to accept the model’s validity

The logistic regression appears to be a linear fit, while the data points do not show they are linear. We must assume that each point is independent of one another in order for the model to be valid.