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 = 123.05 - 8.94(smoke)

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

The slope represents the change in baby weight dependent on mother smoking status. In this instance, if a mother smokes (smoke variable = 1), then the baby is likely to be 8.94 oz less than if a mother does not smoke.

weight(smoke) = 123.05 - 8.94 (1) = 114.11 oz weight(non-smoke) = 123.05 - 8.94 (0) = 123.05 oz

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

Yes, there is a statistically significant relationship between average birth weight and smoking. We know this because the p-value is less than 0.05 (it is 0), which means we can reject the null (there is no relationship).

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.

num_days_absent = 18.93 - 9.11 (ethnicity) + 3.10 (sex) + 2.15 (lrn)

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

Ethnicity: -9.11 ; If the individual is not aboriginal, then the number of days absent decreases by 9.11. Sex: 3.10 ; If the individual is a male, then the number of days absent increases by 3.10 Lrn: 2.15 ; If the individual is a slow learner, the the number of days absent increases by 2.15

  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.
eth <- 0
sex <- 1
lrn <- 1

estimate <- 18.93 - (9.11 *eth) + (3.10 *sex) + (2.15* lrn)
resid <- 2 - estimate

resid
## [1] -22.18

The residual 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.
varResid <- 240.57
var <- 264.17
n <- 146
k <- 3

r2 <- 1 - (varResid/var)
rsAdj <- 1 - (varResid/var)*((n-1)/(n-k-1))

r2
## [1] 0.08933641
rsAdj
## [1] 0.07009704

\(R^2\) = 0.08933641

\(R^2\) adj = 0.07009704

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?

The learner status variable should be removed first because removing it produces the highest adjusted \(R^2\) value.

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 lower temperatures tend to have a higher number of damaged 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 intercept is 11.6630, which means that at base temperature, there tend to be 11.6630 damaged O-rings. For every unit of increase in temperature, the number of failures decreases by 0.2162. A p-value of 0 shows that there is a statistically significant relationship between temperature and the number of damaged O-rings.

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

log_e((p_i)/(1-p_i)) = 11.6630 − 0.2162(Temperature)

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

Yes, I think the concerns regarding o-rings are justified because there is a statistically significant relationship between temperature and number of damaged 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.

\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:

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

t1 <- 51
t2 <- 53
t3 <- 55

p1 <- 11.6630 - (0.2162 * t1)
p2 <- 11.6630 - (0.2162 * t2)
p3 <- 11.6630 - (0.2162 * t3)

exp(p1) / (1+exp(p1))
## [1] 0.6540297
exp(p2) / (1+exp(p2))
## [1] 0.5509228
exp(p3) / (1+exp(p3))
## [1] 0.4432456

Probability at 51 degrees: 65.40297%

Probability at 53 degrees: 55.09228%

Probability at 55 degrees: 44.32456%

  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.
oringsData <- c(51,53,55,57,58,63,66,67,68,69,70,72,73,75,76,78,79,81)

probabilities = function(temperature){
  mod  = 11.6630 - 0.2162 * temperature
  probs = (exp(mod) / (1+exp(mod)))
  return(probs)
}

finalProbs <- sapply(oringsData,probabilities)

plot(finalProbs,oringsData)
lines(finalProbs, oringsData)

  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 biggest thing I can think of is that the sample size is not large enough. However, we could just collect more data to account for this. There are two key conditions for fitting a logistic regression model: 1. Each outcome Yi is independent of the other outcomes. 2. Each predictor xi is linearly related to logit(pi) if all other predictors are held constant.