Homework Chapter 8

8.2 Baby weights, Part II.

Baby weights, Part II. Exercise 8.1 introduces a data set on birth weight of babies. Another variable we consider is parity, which is 0 if the child is the first born, and 1 otherwise. The summary table below shows the results of a linear regression model for predicting the average birth weight of babies, measured in ounces, from parity.

Part (a) Write the equation of the regression line.

\(\widehat{baby\ weight} = 120.07 - 1.93 \times parity\)

Part (b) Interpret the slope in this context, and calculate the predicted birth weight of first borns and others.

The estimated weight of babies that are not first born is 1.93 ounces lower than the estimated weight of first born babies.

Not first born: 120.07 - 1.93 * 1 = 118.14

First born: 120.07 - 1.93 * 0 = 120.07

Part (c) Is there a statistically significant relationship between the average birth weight and parity?

Accoding to the summary table, the p???value is about 0.1 and greater than 0.05. We fail to reject the null hypothesis that the slope parameter is different than 0. There is no significant relationship between the average birth weight and parity.

8.4 Absenteeism, Part I.

Absenteeism, Part I. 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

Part (a) Write the equation of the regression line.

\(\widehat{absenteeism} = 18.93 - 9.11 \times eth + 3.10 \times sex + 2.15 \times lrn\)

Part (b) Interpret each one of the slopes in this context.

eth: The model predicts that the average number of days absent by non-aboriginal students is 9.11 days lower than by aboriginal students.

sex: The model predicts that the average number of days absent by male students is 3.1 days higher than by female students.

lrn: The model predicts that the average number of days absent by slow learners is 2.15 days higher than by average learners.

Part (c) 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.

The residual for the first observation is -22.18.

\(\hat{y}_1 = 18.93 - 9.11 * 0 + 3.10 * 1 + 2.15 * 1 = 24.18\)

\(e_1 = y_1 - \hat{y}_1 = 2 - 24.18 = -22.18\)

Part (d) 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 R2 and the adjusted R2. Note that there are 146 observations in the data set.

var_residual  = 240.57
var_birth_Weights = 264.17
n = 146
k = 3
R2 = 1 - (var_residual/var_birth_Weights)
Adi_R2 = 1 - ((var_residual * (n-1)) / (var_birth_Weights * (n-k-1)))
              
R2   
## [1] 0.08933641
Adi_R2
## [1] 0.07009704

\(Var(y_i) = 264.17\)

\(R^2 = 1 - \frac{Var(e_i)}{Var(y_i)} = 1 - \frac{240.57}{264.17} \approx 0.0893\)

\(R^2_{adj} = 1 - \frac{Var(e_i)}{Var(y_i)} \times \frac{n-1}{n-k-1} = 1 - \frac{240.57}{264.17} * \frac{146 - 1}{146-3-1} \approx 0.0701\)

\(R^2_{adj} = 1 - \frac{Var(e_i)}{Var(y_i)} \times \frac{n-1}{n-k-1} = 1 - \frac{240.57}{264.17} * \frac{146 - 1}{146-3-1} \approx 0.0701\)

8.8 Absenteeism, Part II.

Absenteeism, Part II. Exercise 8.4 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. Model Adjusted R2 1 Fullmodel 0.0701 2 No ethnicity -0.0033 3 No sex 0.0676 4 No learner status 0.0723 Which, if any, variable should be removed from the model first?

The fourth model (no learner status) has the highest adjusted $R^2_{adj} = 0.0723$, and it is higher than the adjusted $R^2_{adj} = 0.0701$ of the full model. Variable learner status should be eliminated from the model first.

8.16 Challenger disaster, Part I.

Challenger disaster, Part I. 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.

Part (a) 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.

Out of 23 shuttle missions, all but 1 had either no damaged O-rings or just 1 damaged O-ring. The mission with the most damaged O-rings (5) is also the mission that launched on the coldest day (53 F). All missions launched in temperature below 65 F had at least one damaged O-ring. Out of 13 missions launched in temperature 70 F or higher, only 3 show damaged O-rings. There may be some significant relationship between two variables and further analysis is necessary.

Part (b) 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.

Based on the summary table, there is a negative relationship between temperature and O-ring failures. Increase in temperature decreases the probability of an O-ring failure. Additionally, the p???value is close to 0, so the relationship between temperature and O-ring failure has significance.

Part (c) Write out the logistic model using the point estimates of the model parameters.

\(log(\frac{\hat{p}}{1 - \hat{p}}) = 11.6630 - 0.2162 \times Temperature\)

Part (d) Based on the model, do you think concerns regarding O-rings are justified? Explain.

As mentioned in part b, the p???value shows that the model parameter is not due to chance and has significance. It is justified to be concerned.

8.18 Challenger disaster, Part II.

Exercise 8.16 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.

Part (a) 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.

temp <- c(53, 57, 58, 63, 66, 67, 67, 67, 68, 69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 81,
          53, 57, 58, 63, 66, 67, 67, 67, 68, 69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 81,
          53, 57, 58, 63, 66, 67, 67, 67, 68, 69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 81,
          53, 57, 58, 63, 66, 67, 67, 67, 68, 69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 81,
          53, 57, 58, 63, 66, 67, 67, 67, 68, 69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 81,
          53, 57, 58, 63, 66, 67, 67, 67, 68, 69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 81)
failure <- c(1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
             1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
             1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
             1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
             1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
             0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

glm.missions <- glm(failure ~ temp, family = binomial)

summary(glm.missions)
## 
## Call:
## glm(formula = failure ~ temp, family = binomial)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.2646  -0.3395  -0.2472  -0.1299   3.0216  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept) 11.66299    3.29616   3.538 0.000403 ***
## temp        -0.21623    0.05318  -4.066 4.77e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 76.745  on 137  degrees of freedom
## Residual deviance: 54.759  on 136  degrees of freedom
## AIC: 58.759
## 
## Number of Fisher Scoring iterations: 6
prob <- data.frame(temp = c(50:85), prob = rep(0, length(c(50:85))))
prob[, 2] <- predict(glm.missions, newdata = data.frame(temp = prob[, 1]), type = "response")

prob[prob$temp == 51, 2]
## [1] 0.6536388
prob[prob$temp == 53, 2]
## [1] 0.5504788
prob[prob$temp == 55, 2]
## [1] 0.4427862

The probability that O-ring will become damaged at 51 F is 65.36%. At 53 F it is 55.05%; and, at 55 F it is 44.28%. Alternatively, I could have plugged the temperatures into the model equation and solved for \(\hat{p}\).

Part (b) 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.

plot(prob$temp, prob$prob, xlab = "Temperature", ylab = "Probability of Damage")
curve(predict(glm.missions, data.frame(temp=x), type="response"), add=TRUE)

Part (c) 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 model assumes that an O-ring damage is a binary event - it is either damaged (1) or not (0). However, it may be the case that degree of damage carries significant information. It also assumes that observations are independent. However, they may not be independent due to the manufacturing process of O-rings and other factors or if the O-rings are reused for different missions. There may be some connection between failures on the same mission. We have 138 observations with 11 failures, but they come from only 23 missions. The sample size may not be large enough. Finally, there may be other significant factors that contribute to O-ring damage, so the model could be improved by adding other significant variables.