8.2, 8.4, 8.8, 8.16, 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.

\[ \hat{y} = 120.07 - 1.93 * parity \]

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

    The slope indicates that for each increase in parity, baby birth weight decreases by 1.93 ounces.

Predicted birth weight:

#First born
120.07 - 1.93*-0
## [1] 120.07
#Otherwise
120.07 - 1.93*1
## [1] 118.14

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

    The p-value of the slope is 0.1052, which is greater than 0.05. Thus, there is not a statistically significant relationship between the average birth weight and parity.

8.4 Absenteeism. 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.

\[ \hat{y} = 18.93 - 9.11 * eth + 3.10 * sex + 2.15 * lrn \]

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

    The slope of eth indicates that, all else being equal, predicted absenteeism decreases by 9.11 days when the student is not aboriginal.

    The slope of sex indicates that, all else being equal, predicted absenteeism increases by 3.10 days when the student is male.

    The slope of lrn indicates that, all else being equal, predicted absenteeism increases by 2.15 day when the student is a slow learner.

  2. 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.

    18.93 - 9.11 * 0 + 3.10 * 1 + 2.15 * 1
    ## [1] 24.18
    #residual= actual missed days - predicted 

2-24.18
    ## [1] -22.18

  3. 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  
varResidual <- 240.57 
varAllStudents <- 264.17

R2 <- 1 - (varResidual / varAllStudents)

adjustedR2 <- 1 - (varResidual / varAllStudents) * ((n-1) / (n-k-1))

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?

The lrn variable should be removed from the model first.

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.

    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)

plot(temperature, damaged)

    plot(temperature, undamaged)

    #From the scatter plots, we see that there were more damaged O-rings at lower temperatures

  2. 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.

    This models uses temperature as a predictor of damaged and undamaged O-rings. The intercept is 11.6630 and the slope is -0.2162. Both estimates have p-vales less than 0.05, so they are significant.

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

\[ log (\frac {\hat{p}}{1-\hat{p}}) = 11.6630 = 0.2162 * Temperature \]

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

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 \[ log (\frac {\hat{p}}{1-\hat{p}}) = 11.6630 = 0.2162 * Temperature \] 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:

pˆ57 = 0.341 pˆ59 = 0.251 pˆ61 = 0.179 pˆ63 = 0.124 pˆ65 = 0.084 pˆ67 = 0.056 pˆ69 = 0.037 pˆ71 = 0.024

Solve logistic equation for p:

\[ \hat{p} = \frac {e^(11.6630 - 0.2162 * Temperature)} {1 + e ^ (11.6630 - 0.2162 * Temperature)} \]

    p <- function(temp)
    {
     damagedOring <- 11.6630 - 0.2162 * temp
  
     phat <- exp(damagedOring) / (1 + exp(damagedOring))
  
     return (round(phat*100,2))
    }

    #Probabilities for temperature of 51, 53, and 55
    p(51)
## [1] 65.4
    p(53)
## [1] 55.09
    p(55)
## [1] 44.32

  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.

    ShuttleMission <- data.frame(temperature, damaged, undamaged)

library(ggplot2)
ggplot(ShuttleMission,aes(x=temperature,y=damaged)) + geom_point() + 
stat_smooth(method = 'glm', family = 'binomial')
    ## Warning: Ignoring unknown parameters: family

  2. 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.

    We must assume that observations are independent to accept the model’s validity. However, the independence assumption may not be met if the O-rings being used are all manufactured by the same company and there are probabilities that an O-ring will be defective or not.