Chapter 9 - Multiple and Logistic Regression
Sin Ying Wong
11/23/2019
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.)
Write the equation of the regression line.
Ans:
y = -8.94*x + 123.05, while y stands for
weightand x stands forsmokeInterpret the slope in this context, and calculate the predicted birth weight of babies born to smoker and non-smoker mothers.
Ans:
The negative slope is to separate the birth weight into two levels by the smoke status. For mothers being smokers, their babies would be 8.94oz lighter than non-smokers’ babies.
smoke = 123.05 - 8.94*1 = 114.11 (oz)
Non-smoke = 123.05 - 8.94*0 = 123.05 (oz)
Is there a statistically significant relationship between the average birth weight and smoking?
Ans:
As the p-valus is too small (being approximately 0), we would reject null hypothesis (no difference). Therefore, there is a statistically significant relationship between the average birth weight and smoking.
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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).
Write the equation of the regression line.
Ans:
Absent days = 18.93 -9.11*eth + 3.10*sex + 2.15*lrn
Interpret each one of the slopes in this context.
Ans:
slope of eth: The average number of absent days by “not aboriginal” students is 9.11 lower than “aboriginal” students.
slope of sex: The average number of absent days by “male” students is 3.10 higher than “female” students.
slope of lrn: The average number of absent days by “slow learner” students is 2.15 higher than “average learner” students.
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.
Ans:
predicted number of absent days = 18.93 -9.11*0 + 3.10*1 + 2.15*1 = 24.18
residual = observed - expected = 2 - 24.18 = -22.18 (days)
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.
Ans:
variance of the residuals = SS_error = 240.57 -> unexplained variability
variance of the number of absent days = SS_total = 264.17 -> total variability
n = 146, p = 3
R^2 = explained variability in y / total variability in y
= SS_model / SS_total
= (264.17 - 240.57) / 264.17
= 0.089336
R^2_adj = 1 - (SS_error / SS_total)*((n-1)/(n-p-1))
= 1 - (240.57/264.17)*(145/142)
= 0.070097
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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?
Ans:
Our adjusted R^2 of full model is 0.0701.
The only model in the table that has a higher adjusted R^2 is the model without learner status (0.0723).
Therefore, learner status should be removed to improve the model.
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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.
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.
Ans:
The lower the temperature, the higher the rate of having damaged O-rings.
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.
Ans:
The coefficient of Temperature is negative, which means the higher the temperature, the lower the chance of failure. The p-value = 0 also proves there is relationship between temperature and failure.
Write out the logistic model using the point estimates of the model parameters.
Ans:
failure = logit(p) = log(p/(1-p)) = 11.630 - 0.2162*temperature
Based on the model, do you think concerns regarding O-rings are justified? Explain.
Ans:
Yes, the model shows that the lower the temperature, the higher the value of failure we will obtain, which means O-rings are more likely to fail at low temperature.
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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}
- 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:
\[\begin{align*} &\hat{p}_{57} = 0.341 && \hat{p}_{59} = 0.251 && \hat{p}_{61} = 0.179 && \hat{p}_{63} = 0.124 \\ &\hat{p}_{65} = 0.084 && \hat{p}_{67} = 0.056 && \hat{p}_{69} = 0.037 && \hat{p}_{71} = 0.024 \end{align*}\]
Ans:
The failure probabilities at temperature (51,53,55) are (65.40%, 55.09%, 44.32%).
p <- function(temp)
{
p_hat <- 11.6630 - 0.2162 * temp
return (exp(p_hat) / (1+exp(p_hat)))
}
p(c(51,53,55))## [1] 0.6540297 0.5509228 0.4432456# List of failure rate from 51 to 81 degrees Fahrenheit.
df <- data.frame(seq(51,81,1), p(seq(51,81,1)))
names(df) <- c('Temp', 'Failure Rate')
dfAdd the model-estimated probabilities from part~(a) on the plot, then connect these dots using a smooth curve to represent the model-estimated probabilities.
Ans:
plot((orings$damage/6)~jitter(orings$temp), xlab = "Temperature in degrees Fahrenheit", xlim = c(50, 82), ylab = "Probability of damage", ylim = c(0,1), main = "Failure Rate of O-Rings by Temperature", col = "royalblue1", pch = 21) points(df$Temp, df$`Failure Rate`, col="red", pch=2) lines(df$Temp, df$`Failure Rate`, col="red", lty=1) legend(x=75, y=1, legend=c("Actual", "Prediction"), cex = 0.8, pch = c(21,2), lty = c(NA, 1), col = c("royalblue1", "red"), text.col=c("royalblue1", "red"))
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.
Ans:
One major condition for logistic regression is that each x value is linearly related to logit(p) when other predictors are held constant. The sample size here is too small with only one predictor (temperature) in the model. There may be other factors other than temperature also affecting the o-rings.
If we are keeping only one x value (temperature), I would suggest look at each of the 6 o-rings separately with assumption that they are all independent of each other, which we may see which o-rings have high correlation with temperature instead of looking at their average failure rate.