Problem 1

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. birth_weight = 123.05 + (-8.94 * smoker)

  2. Interpret the slope in this context, and calculate the predicted birth weight of babies born to smoker and non-smoker mothers. For a Mom that smokes the babies predicted weight is almost 9 oz less than the weight of a non-smoking Moms’ babies weight. The predicted weights are:
    non-smoking Mom 123 oz (7.7 lbs)
    smoking Moms babies weight: About 9 oz less at 114 oz (7.13 lbs)

  3. Is there a statistically significant relationship between the average birth weight and smoking? Given the high t-value of -8.65 and the associated p-value of 0, which is less than 0.05. Based on this there is a statistically significant relationship; baby weights of Mom’s who smoke is significantly less than weights of Moms that don’t smoke.

Problem 2

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. absent = 18.93 + (-9.11 * eth) + (3.10 * sex) + (2.15 * lrn)

  2. Interpret each one of the slopes in this context.
    eth: an aboriginal student has 9 more absentees than a non-aboriginal student

sex: a male student has 3 more absentees than a female student

lrn: a slow learner student has 2 more absentees than an average learner

  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.
pred_absent = 18.93 + (-9.11 * 0) + (3.10 * 1) + (2.15 * 1)
cat('Predicted absences:',pred_absent,'\n')
## Predicted absences: 24.18
actual_absent = 2

cat('Residual is:', pred_absent - actual_absent)
## Residual is: 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.
res_var <- 240.57
act_var <- 264.17
n <- 146
k <- 3

rsquared <- 1 - (res_var/act_var)
cat('R_squared is:', rsquared, '\n')
## R_squared is: 0.08933641
#adj_rsquare = 1 - ((res_var/act_var) * ((n-1)/(n-k-1)))

adj_rsquared <- rsquared * ((n-1)/(n-k-1))
cat('Adjusted R_squared is:', adj_rsquared)
## Adjusted R_squared is: 0.0912238

Problem 3

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? Using backward elimination the “New Learner Status” is removed first because doing so results in the highest R2 value.

Problem 4

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. More bad-orings seem to occur at temperatures lower than 66 and it appears the damaged number of rings jumps up a lot once the temperature drops below 57 degrees.

  2. 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. Temperature is the predictor for the response variable of o-rings damaged. The z-value and corresponding p-value indicate how well the predictor predicts. In this case the p-value is < 0.05 so temperature is a significant predictor. The temperature’s coefficient is -0.21 meaning that for each degree drop in temperature the likelihood of a damaged o-ring moves closer to 0 (bad o-ring). The high intercept of 11.66 implies that there is a lot of variance in the data points, so it would not be practical to use this model to predict individual o-rings going bad.

  1. Write out the logistic model using the point estimates of the model parameters. log e(pi/1-pi) = 11.6630 - (0.2162 * temperature)

  2. Based on the model, do you think concerns regarding O-rings are justified? Yes. Explain. A visual review of the data and the negative coefficient for temperature supports that as the temperature drops the probability of an o-ring being bad increases. Fortunately Florida doesn’t tend to get extremely low temperatures, but since temperatures in the 50’s are not unusual, particularly during the winter, concerns about the o-rings is justified.

Problem 5

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:

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

#at 51
temp = 51
damaged = 11.6630 - (0.2162 * temp)

prob = exp(damaged)/(1 + (exp(damaged)))
cat('Probability at', temp, 'degrees of being damaged is', prob, '\n')
## Probability at 51 degrees of being damaged is 0.6540297
#at 53
temp = 53
damaged = 11.6630 - (0.2162 * temp)

prob = exp(damaged)/(1 + (exp(damaged)))
cat('Probability at', temp, 'degrees of being damaged is', prob, '\n')
## Probability at 53 degrees of being damaged is 0.5509228
#at 55
temp = 55
damaged = 11.6630 - (0.2162 * temp)

prob = exp(damaged)/(1 + (exp(damaged)))
cat('Probability at', temp, 'degrees of being damaged is', prob, '\n')
## Probability at 55 degrees of being damaged is 0.4432456
  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.
library(ggplot2)
## 
## Attaching package: 'ggplot2'
## The following object is masked from 'package:openintro':
## 
##     diamonds
temperature <- c(53, 57, 58, 63, 66, 67, 67, 67, 68, 69, 70, 70)
damaged <- c(5, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0)
undamaged <- c(1, 5, 5, 5, 6, 6, 6, 6, 6, 6, 5, 6)

orings = data.frame(temperature, damaged, undamaged)

ggplot(orings,aes(x=temperature,y=damaged)) + geom_point() + 
  stat_smooth(method = 'glm')

  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. I have no concerns: Values obtained must be independent - temperature can be assumed independent from o-rings Predictor/explanatory variable must be linearly related to the Response feature - per the plot above there appears to be a linear relationship, so no concerns with this critera.