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: The equation of the regression line, Predicted Average Birth 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.

Answer: For each unit increase in smoke, the average birth weight is expected to decrease by 8.94, all else held constant. (1) The predicted birth weight of babies born to smoker mother = 123.05 - 8.94(1) = 114.11 ounces. (2) The predicted birth weight of babies born to non-smoker mother = 123.05 - 8.94(0) = 123.05 ounces.

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

Answer: The is a statistically significant relationship between the average birth weight and smoking because the p-value of test statistic is less than 0.05.

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 the regression line, Predicted Average Number of Days Absent = 18.93 - 9.11 * (ethnic background) + 3.10 * (sex) + 2.15 * (learner status)

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

Answer: (1) For the ethnic background slope, for each unit increase in ethnic background, the average number of days absent is expected to decrease by 9.11, all else held constant. (2) For the sex slope, for each unit increase in sex, the average number of days absent is expected to increase by 3.10, all else held constant. (3) For the learner status slope, for each unit increase in learner status, the average number of days absent is expected to increase by 2.15, all else held constant.

  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: The residual for the first observation in the data set is 2 - [18.93 - 9.110 + 3.101 + 2.15*1] = 2 - 24.18 = -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: (1) R-squared = 1 - (variance of residuals)/(variance in outcome) = 1 - 240.57/264.17 = 1 - 0.9107 = 0.0893. (2) R-squared Adjusted = 1 - (variance of residuals)/(variance in outcome)(n-1)/(n-k-1) = 1 - (240.57/264.17)((146-1)/(146-3-1)) = 1 - 0.9299 = 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: The variable “No Learner Status” should be removed from the model first of the backwards elimination process because it will lead to improve the adjusted R-Square adjusted to 0.0723 as compared to 0.0701 from the full 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.

Answer: There are higher chance of o-ring damaged at lower temperatures compared to higher temperatures. We can see that there is higher frequency of damaged O-rings occured below 66 Fahrenheit and only one O-rings damaged above 66 Fahrenheit. So, we can say that O-rings will be stable at temperature above 66 Fahrenheit.

  1. Failures have been coded as 1 for a damaged O-ring and 0 for an undamaged O-rings, 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 key components of this summary table are the estimated coeffficient for the intercept, coefficient for temperature and the p-value. The estimated coefficient for the intercept is the log odds of an O-Rings with a temperature of zero being in a shuttle launch. The coefficient for temperature is the difference in the log odds. In other words, for a one unit increase in temperature, the expected change in log odds is -0.2162. There is a strong association between temperature and O-rings damage because the p-value for temperature is significantly less than 0.05.

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

Answer: The logistic model using the point estimates of the model parameters is \(\log _{ e } (\frac { { P }_{ i } }{ 1-{ P }_{ i } } ) = 11.6630 - 0.2162*Temperature.\)

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

Answer: I do think concerns regarding O-rings are justified because the p-value for the temperature is significantly less than 0.05, thus making this very statistically significant.

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: Rearrange the logistic model to get P = exp(11.6630-0.2162Temperature)/(1+exp(11.6630-0.2162Temperature)). SO, for temperature at 51, P = exp(11.6630-0.216251)/(1+exp(11.6630-0.216251)) = 0.6540. For temperature at 53, P =exp(11.6630-0.216253)/(1+exp(11.6630-0.216253)) = 0.5509. For temperature at 55, P =exp(11.6630-0.216255)/(1+exp(11.6630-0.216255)) = 0.4433.

  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: The model-estimated probabilities plot as below.

library(ggplot2)
## 
## Attaching package: 'ggplot2'
## The following object is masked from 'package:openintro':
## 
##     diamonds
probability <- c(0.6540, 0.5509, 0.4433, 0.341, 0.251, 0.179, 0.124, 0.084, 0.056, 0.037, 0.024)

temperature <- c(51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71)

model_est_prob <- data.frame(temperature, probability)

ggplot(model_est_prob, aes(temperature, probability)) + geom_point() + geom_smooth() + ylab("Estimated Probabilities")
## `geom_smooth()` using method = 'loess' and formula 'y ~ x'

  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.