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.)

  1. Write the equation of the regression line.

\[Weight = -8.94 * Smoke + 123.05\]

  1. Interpret the slope in this context, and calculate the predicted birth weight of babies born to smoker and non-smoker mothers.

In this context (since the variable ‘Smoke’ is either a 1 or a 0) the slope indicates that the predicted weight of the baby of a smoking mother is 8.94oz lower than that of a non-smoking mother.

The predicted weight of babies born to a non-smoker is 123.05 oz and \(123.05 - 8.94 = 114.11 oz\) for a smoker.

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

Due to the very low p-value, we can conclude that there is a statistically significant relationship between smoking and average birth weight.

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).

  1. Write the equation of the regression line.

\[abs = -9.11eth + 3.1sex + 2.15lrn + 18.93\]

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

Each of the following statements about the three slopes assumes that the other variables not spoken about are being held equal:

  1. The ethnicity slope of -9.11 indicates that the average non-aboriginal student missed 9.11 fewer days of school.
  2. The sex slope of 3.1 indicates that females missed 3.1 fewer days of school on average.
  3. the 2.15 slope of the learner variable indicates that slow learners on average missed 2.15 more days of school.
  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.
missed1 <- 2
eth1 <- 0
sex1 <- 1
lrn1 <- 1

Estimate1 <- -9.11*eth1 + 3.1 * sex1 + 2.15 * lrn1 + 18.93
Residual1 <- Estimate1 - missed1
Residual1
## [1] 22.18

The residual for the first student is 22.18 days.

  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 \(R^2\) and the adjusted \(R^2\). Note that there are 146 observations in the data set.
n <- 146
k <- 3
ResVar <- 240.57
StuVar <- 264.17

Res_Rat <- (ResVar / StuVar)

R_2 <- 1 - Res_Rat
R_2_Adj <- 1 - Res_Rat * ((n-1)/(n-k-1))

The $R^2 = $0.0893364 and the $Adj R^2 = $0.070097

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?

Based on the \(R^2\) values in the table, I would remove the lrn variable first, then re-evaluate the model.

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.

  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.

The data appears to show a trend towards more damaged O-rings at lower temperatures, but this is strongly affected by the one extreme reading at 53 degrees.

  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 model consists of two components - the intercept and slope related to the temperature. The Estimate column gives us values for these two things, with the Std. Error column supplying an error evaluation. The *z-value** and *Pr(>|z|)** values also provide a measure of how significant each of the parameters is to the predictive accuracy of the model.

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

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

  1. Based on the model, do you think concerns regarding O-rings are justified? Explain.
Temp_at_0 <- (11.663)/ 0.2162

The model indicates that launches at temperatures below 53.9454209 have an expected O-ring failure above 0, with each drop of 4.6253469 degrees leading to an additional expected failure. Since the reading at 53 degrees shows a rapid increase in failure at only that reading, more data would be ideal to generate a more precise model that better represents how sudden the expected failure rate increases. Based on this model, I wouldn’t recommend running any launches below 54 degrees with the O-ring design used here.

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}

  1. The data provided in the previous exercise are shown in the plot. The logistic model fit to these data may be written as \[\begin{align*} \log\left( \frac{\hat{p}}{1 - \hat{p}} \right) = 11.6630 - 0.2162\times Temperature \end{align*}\]

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:

\[\begin{align*} &\hat{p}_{57} = 0.341 && \hat{p}_{59} = 0.251 && \hat{p}_{61} = 0.179 && \hat{p}_{63} = 0.124 \\ &\hat{p}_{65} = 0.084 && \hat{p}_{67} = 0.056 && \hat{p}_{69} = 0.037 && \hat{p}_{71} = 0.024 \end{align*}\]

Solving for \(\hat{p}\) we get \[\hat{p} = \frac{e^{11.663-0.2162*Temp}}{1+e^{11.663-0.2162*Temp}}\]

p_hat <- function(x){
  model <- 11.663 - 0.2162 * x
  num <- exp(model)
  den <- 1 + num
  return(round(num/den,3))
}

p_hat(51)
## [1] 0.654
p_hat(53)
## [1] 0.551
p_hat(55)
## [1] 0.443
  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.
plot(orings[,1] + 
       c(rep(0, 5), c(-0.1, 0, 0.1), 0, 0, -0.07, -0.07, 0.07, 0.07, 
         rep(0, 4), -0.07, 0.07, 0, 0, 0), 
     orings[,2]/6, 
     xlab = "", ylab = "Probability of damage", 
     xlim = c(50, 82), ylim = c(0,1), 
     col = COL[1,2], pch = 19)
temperature <- c(51,53,55,57,59,61,63,65,67,69,71)
p <- p_hat(temperature)
lines(smooth.spline(temperature, p,spar=0.35),col="green")
mtext("Temperature (Fahrenheit)", 1, 2)

  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.

My concern would be theat there is very little data on the tail end of the model (lower temperatures) which is where the prediction starts to rise significantly. Additional data points below 60F would be helpful, possibly run in a lab test setting.