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.

Answer:

babyWeight = 123.05 - 8.94 x smoke

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

Answer:

The slope indicates the estimated weight of babies born to mothers who smoke, vs. mothers who do not smoke.

smoke_b_weight <- 123.05 - 8.94 * 1

paste('Predicted baby weight of mothers that smoke:', smoke_b_weight, 'ounces.')
## [1] "Predicted baby weight of mothers that smoke: 114.11 ounces."
non_smoke_b_weight <- 123.05 - 8.94 * 0

paste('Predicted baby weight of mothers that don\'t smoke:', non_smoke_b_weight, 'ounces.')
## [1] "Predicted baby weight of mothers that don't smoke: 123.05 ounces."
  1. Is there a statistically significant relationship between the average birth weight and smoking?

Answer:

Due to the fact that the P value of “smoke” is at, or close to zero, we can reject the null hypotheseis (there is no difference between the body weight of babies born to smokers and non smokers). We can therefore conclude that there is a significant relationship between smoking and baby weights.

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.

Answer

y = 18.93 - 9.11 x eth + 3.10 x sex + 2.15 x lrn

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

Answer:

eth indicates a decrease of 9.11 days of absenteeism for non-aborignal students.

sex indicates that male students are absent 3.10 days more than female students.

lrn indicates that slow learners are absent 2.15 days more than average learners.

  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.

Answer:

absenteeism <- 18.93 - 9.11 * 0 + 3.10 * 1 + 2.15 * 1
residual <- 2 - absenteeism

paste('The residual for the given student is', residual)
## [1] "The residual for the given student is -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 \(R^2\) and the adjusted \(R^2\). Note that there are 146 observations in the data set.

Answer:

R2 <- 1 - 240.57 / 264.17
R2_adjusted <- 1 - (240.57 / 264.17) * ((146 - 1) / (146 - 3 - 1))

paste('R-squared is:', round(R2, 4))
## [1] "R-squared is: 0.0893"
paste('R-squared adjusted is:', round(R2_adjusted, 4))
## [1] "R-squared adjusted is: 0.0701"

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?

Answer:

Eliminating the “learner status” variable will give us the highest \(R^2\) value, so this should be the first variable to be removed.

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.

Answer:

On first inspection, it appears that there are more damaged O-rings at lower temperatures, than there are at higher temperatures. This could be summarized as follows:

As temperature increases, the amount and frequency of damaged O-rings decreases.

  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.

Answer:

The Intercept suggests that at 0 degrees Fahrenheit, there would be an estimate of 12 O-ring failures. The Temperature Estimate implies that for every increase in temperature, the likelihood of O-Ring damage decreases by 0.2162. This is statistically significant because the Z-value is -4.07, and the p-value is less than 0.05.

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

Answer:

log(p / (1 - p)) = 11.6630 - 0.2162 x Temperature

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

Answer:

Based on the model, I do think concerns regarding O-rings are justified. The low p-value for temperature implies that temperature is strongly associated with failing O-rings.

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.

  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:

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

Answer:

damage_probability <- function(temperature, sentence = FALSE) {
  for (temp in temperature) {
    damaged_o_ring <- 11.6630 - 0.2162 * temp
    phat <- exp(damaged_o_ring) / (1 + exp(damaged_o_ring))
    probability <- round(phat * 100, 4)
    if (sentence == TRUE) {
      probability <- round(phat * 100, 2)
      result <- paste('The probability of O-ring damage at', temp, 'degrees Farenheit is', probability, '%')
    }
    else {
      result <- probability
    }
  }

  if (sentence == TRUE) {
    return(cat(result, sep = '\n\n'))
  }
  else {
    return(result)
  }
}

damage_probability(list(c(51, 53, 55)), TRUE)
## The probability of O-ring damage at 51 degrees Farenheit is 65.4 %
## 
## The probability of O-ring damage at 53 degrees Farenheit is 55.09 %
## 
## The probability of O-ring damage at 55 degrees Farenheit is 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.
library(ggplot2)

Temperature <- 51:71
model_esitmated_probs <- data.frame(Temperature, damage_probability(list(Temperature)))

ggplot(model_esitmated_probs, aes(x = Temperature, y = damage_probability(list(Temperature)))) + geom_smooth() + geom_point()

  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.

Answer:

The 2 key conditions that most be met in order to apply a logistic regression model are as follows:

1. Each predictor xi should be linearly related to logit(pi) if all other predictors are held constant.

2. Each outcome Yi is independent of the other outcomes.

In terms of our application, it is hard to assess if these conditions have been met or not. Our dataset only contains data for 23 missions, so it would be hard to confidently claim that condition 1 above has been met. For condition 2, it is hard to prove the independence of outcomes. The observations may be related to engineers becoming more familiar with the O-ring issue overtime, and thus making improvements prior to each mission, rather than an increase in temperature resulting in less damaged O-rings.