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.

  1. Write the equation of the regression line.

y^=120.07−1.93×parity

  1. Interpret the slope in this context, and calculate the predicted birth weight of babies born to smoker and non-smoker mothers.
120.07 - 1.93
## [1] 118.14
  1. Is there a statistically significant relationship between the average birth weight and smoking? Since the parity parameter’s p−value=0.1052; we can conclude that there is not a statistically significant relationship between average birth weight and parity.

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.

  1. Write the equation of the regression line.

y^=18.93−9.11×eth+3.10×sex+2.15×lrn

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

When student is not aboriginal, average absent days is 9.11 less than 18.93 (statistically significant).

When student is male, average absent days is 3.10 more than 18.93.

When student is a slow learner, average absent days is 2.15 more than 18.93.

  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

predicted <- 18.93 - eth*9.11 + 3.10*sex + 2.15*lrn

2 - predicted
## [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.
var_residual <- 240.57
var_outcome <- 264.17
n <- 146 
k <- 3

R2 <- 1 - (var_residual/var_outcome)
R2_adjusted <- 1 - (var_residual/var_outcome) * ((n-1)/(n-k-1))
R2
## [1] 0.08933641
R2_adjusted
## [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?

Based on the Adjusted R2=0.0723 the 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.
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)

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

plot(ShuttleMission)

According to the plot we see:

-Higher number of damaged O-rings are observed when lower temperatures were recorded.

-Less number of damaged O-rings are observed when higher temperatures were recorded.

  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.

This model is represented by two components: One, is the Intercept and the second one is the Temperature values. The Estimate identifies the parameter estimate for the model. The Z value and the P-value help with distinguishing important information from less significant parameters by telling us how good the variables predict this model.

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

log(pi1−pi)=11.6630−0.2162×Temperature

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

According to the collected data, we can deduct a high probability of damage to O-rings under 50o. Also, since O-rings are “Critical” components, I do think concerns regarding the O-rings are justified.

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
logModel = function(temperature){
  oRing  = 11.6630 - 0.2162 * temperature
  answer = 100*(exp(oRing) / (1+exp(oRing)))
  return(answer)
}

temps = c(57,65,59,67,61,69,63,71,51,53,55)
probabilites = sapply(temps,logModel)
probabilites
##  [1] 34.064976  8.393843 25.109139  5.612566 17.869707  3.715479 12.372702
##  [8]  2.443024 65.402974 55.092283 44.324565
  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
plot(y=probabilites, x=temps)
curve(logModel(x), from=40,to=80,add=TRUE,xlab='o-ring damage',ylab='temperature')

  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.

There are two key conditions for fitting a logistic regression model:

-Each predictor xi is linearly related to logit(pi ) if all other predictors are held constant. -Each outcome Yi is independent of the other outcomes. Based on that definition, we have assumed that those conditions are met.