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: weight of baby = 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 of regression line is -8.94 and the regression line intersecting y-axis at 123.05, where indepdent variable is smoke and dependent variable is weight of baby. Therefore, we can interpret from regression line equation that with increased smoking, the the average weights of babies decrease. But, we need to keep in mind that negative weight is not possible.

When smoke = 0 i.e. non-smoking mothers, weight of baby = -8.94 x 0 + 123.05 = 123.05
When smoke = 1 i.e. smoking mothers, weight of baby = -8.94 x 1 + 123.05 = 114.11

Therefore, according to the model, babies born to mothers who smoke, weigh 8.94 ounces less than those born to mothers who doesn’t smoke.

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

Answer:

Here p-value for smoke is 0. From the perspetive of hypothesis testing:

H0: B1 = 0 (no difference in variable)
HA: B1 != 0 (there is a difference)

Since, the p-value is negligible, we can reject null hypothesis and conclude that the slope (B1) is not zero. So, we can say that there is an association between smoking and birth 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: The equation of multi regression line is: y = - 9.11 x eth + 3.10 x sex + 2.15 x lrn + 18.93

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

Answer:

eth: When subject is NO Aboriginal, there’s a 9.11 day decrease in absenteeism.
sex: When subject is Male, there’s a 3.10 day increase in absenteeism.
lrn: When subject is Slow Learner, there’s a 2.15 day increase in absenteeism.

  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:

eth <- 0     # Aboriginal
sex <- 1     # Male
lrn <- 1     # Slow Learner

missed.days <- 2     # Missed 2 days of school 

predict.days <- 18.93 - 9.11 * eth + 3.1 * sex + 2.15 * lrn

residual <- missed.days - predict.days

residual
## [1] -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:

s <- 146  # Sample size
n <- 3    # number of predictor variables

res_Var = 240.57         # Variance of residual
all_var = 264.17         # Variance for all students

R2 <- 1 - (res_Var / all_var)
R2
## [1] 0.08933641
adjusted_R2 <- 1 - (res_Var / all_var) * ( (s - 1) / (s - n - 1) )
adjusted_R2
## [1] 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?

Answer:

Adjusted R-squared was 0.07009704 (or 0.701) from last exercise. In the above table, “No learner status” = 0.0723. So, R-squared improves, when learner status is removed. Therefore, lrn variable should be removed from the model first.

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: We see there were more damaged of O-rings at lower temperatres, than at higher.

  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: If the temperature increases by 1 degree, then the probability of damage decreases by 0.2162.

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

Answer: Let p_hat be the model estimated probability.

Then the logistical model is: log(p_hat / (1 - p_hat)) = 11.6630 - 0.2162 x Temperature.

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

Answer:
From the model, we can derive:
p_hat = e^(11.6630 - 0.2162 x Temperature) / (1 + e^(11.6630 - 0.2162 x Temperature)).

Yes, the concerns are justifiable as the temperature decreases, the probability increases. At 40 degree, the probability is higher than 95%.

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:

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

Answer: In previous exercise, we arrived at:
p_hat = e^(11.6630 - 0.2162 x Temperature) / (1 + e^(11.6630 - 0.2162 x Temperature))

Now, we’ll write this as a function and compute some of the given values:

p <- function(Temperature)
{
  oring <- 11.6630 - 0.2162 * Temperature
  p_hat <- exp(oring) / (1 + exp(oring))
  return (round(p_hat * 100, 2))
}

p(51)
## [1] 65.4
p(53)
## [1] 55.09
p(55)
## [1] 44.32
p(40)
## [1] 95.32
  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.

Answer:

temp <- seq(from = 51, to = 71, by = 2)
damage <- c(round(p(51)), round(p(53)), round(p(55)), 0.341, 0.251, 0.179, 0.124, 0.084, 0.056, 0.037, 0.024)
plot(temp, damage, 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.

Answer: This dataset of 23 data points is too small to verify whether: predictor temperature is linearly related to logit p and each outcome damaged o ring is independent of other outcomes.