Chapter 8 - Multiple and Logistic Regression

Practice: Graded: 8.18

8.2 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.

  1. Write the equation of the regression line.
# (Weight in Ounces) = 120.07 - 1.93 * (Parity or Not)
  1. Interpret the slope in this context, and calculate the predicted birth weight of first borns and others.
# If the child is NOT first born, that 2nd, 3rd, etc. child would be predicted to be 1.93 ounces less than the first child's birth weight. With calculations, the first born weight is 120.07 ounches, whereas, every other child afterwards would be 129.07 - 1.93 * (1) = 127.14.
  1. Is there a statistically significant relationship between the average birth weight and parity?
# The t-value and P value is -.162 and 0.1052 respectively. Given that the P value is greater than 0.05, we state that there is NO statistically significant relationship between the average birth weight and parity.

8.4 Absenteeism, Part I. 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).

  1. Write the equation of the regression line.
# Absenteeism = 18.93 - 9.11 * (ethnic background) + 3.10 * (sex) + 2.15 * (learner status)
  1. Interpret each one of the slopes in this context.
# Ethnic background: The model predicts a 9.11 (absent) days decrease in non-aoriginal children, all else held constant.
# Sex: The model predicts a 3.10 (absent) days increase in males over females, all else held constant.
# Learner status: The model predicts a 2.15 (absent) days increase in slow learners over average learners, all else held constant.
  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.
# Model prediction: 
Absent <- 18.93 - 9.11 * (0) + 3.10 * (1) + 2.15 * (1)
Residual <- 2 - Absent
paste("Residual for this student: ", Residual)
## [1] "Residual for this student:  -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.
# R-squared = 1 - (variance of residuals)/(variance in outcome)
# R-squared adjusted. 1 - (variance of residuals)/(variance in outcome)*(n-1)/(n-k-1), where k is predictor in variables in model.

R2.Ab <- 1 - (240.57)/(264.17)
R2.Ab.adj <- 1 - (240.57/264.17)*((146-1)/(146-3-1))
paste("R-squared: ", round(R2.Ab,4))
## [1] "R-squared:  0.0893"
paste("R-squared adjusted: ", round(R2.Ab.adj,4))
## [1] "R-squared adjusted:  0.0701"

8.8 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?

# We should remove the learner status, given that the adjusted R-squared would improve from 0.0701 to 0.0723.

8.16 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.

  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.
# From visual inspection, it appears that colder temperatures had more damaged O-rings than at average or even higher temperatures. This effect starts to become noticable around the 50 degree range, particularly at 53 degrees temperature. Therefore, my preliminary conclusion is that the colder the ambient temperature, the more likely the O-rings were going to be damaged.
  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.
# Given that this is a logistic regression, it behaves fairly similarly to a multiple regression. Therefore, if the numerical variable temperature increases, in this case 1 degree, then the likelihood of being damaged decreases by .2162. Of course, this is a simplification, as the final result will need to be transformed to make sense of the information. (Please see the transformation formula below in part C.)
  1. Write out the logistic model using the point estimates of the model parameters.
# The logistic regression formula here is: logit(pi) = log((pi)/(1-pi)) = 11.6630 - 0.2162 * (temperature), where if the final result nears 1, this would indicate the more likelhiood of a damaged O-ring.
  1. Based on the model, do you think concerns regarding O-rings are justified? Explain.
# This can be evaluated by looking at the p values. The pvalue for temperature is 0.0000, thus making this very statistically significant, so yes, there is good justification for the concerns for temperature.

8.18 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 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.

  1. The data provided in the previous exercise are shown in the plot. The logistic model fit to these data may be written as: (Please refer to textbook for formula.)

where pˆ 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:

(Please refer to the texbook as well for the different model-estimated probabilities.)

# General formula
temperature <- NULL
logit <- 11.6630 - 0.2162 * temperature
O.ring.failure.prob <- exp(logit)/(1 + exp(logit))

# Temperature 51 degrees Fahrenheit
temperature51 <- 51
logit.51 <- 11.6630 - 0.2162 * temperature51
O.ring.failure.prob.51 <- exp(logit.51)/(1 + exp(logit.51))
paste("O Ring Failure Probability at Temp 51: ", round(O.ring.failure.prob.51, 4))
## [1] "O Ring Failure Probability at Temp 51:  0.654"
# Temperature 53 degrees Fahrenheit
temperature53 <- 53
logit.53 <- 11.6630 - 0.2162 * temperature53
O.ring.failure.prob.53 <- exp(logit.53)/(1 + exp(logit.53))
paste("O Ring Failure Probability at Temp 53: ", round(O.ring.failure.prob.53, 4))
## [1] "O Ring Failure Probability at Temp 53:  0.5509"
# Temperature 55 degrees Fahrenheit
temperature55 <- 55
logit.55 <- 11.6630 - 0.2162 * temperature55
O.ring.failure.prob.55 <- exp(logit.55)/(1 + exp(logit.55))
paste("O Ring Failure Probability at Temp 55: ", round(O.ring.failure.prob.55, 4))
## [1] "O Ring Failure Probability at Temp 55:  0.4432"
  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.
temp.x <- seq(from = 51, to = 71, by = 2)
y <- c(round(O.ring.failure.prob.51, 4), round(O.ring.failure.prob.53, 4), round(O.ring.failure.prob.55, 4), 0.341, 0.251, 0.179, 0.124, 0.084, 0.056, 0.037, 0.024)
plot(temp.x, y, type = "o", col = "blue")

  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.
# Logistic Regression requires that we fulfill their two key conditions for creating this model.
# 1. Each predictor x is linearly related to logit p if all other predictors are held constant.
# 2. Each outcome Y is independent of the other outcomes.
# Both conditions are difficult to verify. There has only been 23 shuttle missions, which may not be enough of a sample size to adequate see if the first criteria can be satisfied. And for number 2, shuttles are very complicated. It is unclear that the O ring is independent of other outcomes. This needs further investigation. Therefore, we may have difficulty assuming that this logistic regression can be used with the information that we have right now.