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.)

9.1(a)

Write the equation of the regression line.

# birth_weight = 123.05 - 8.94*smoke

9.1(b)

Interpret the slope in this context, and calculate the predicted birth weight of babies born to smoker and non-smoker mothers.

# The weight of a baby is reduced by 8.94 ounces if a mother is a smoker.

smoker_mom_bw <- 123.05 - 8.94*1

non_smoker_mom_bw <- 123.05 - 8.94*0

smoker_mom_bw

## [1] 114.11

non_smoker_mom_bw

## [1] 123.05

# Babies with smoker and non-smoker moms weigh 114.11 and 123.05 respectively.

9.1(c)

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

The p-value is 0 in both the cases which gives a strong reason to reject the null hypothesis as it is lower the significance level of 0.05. Therefore, there is a statistically significant relationship between the average birth weight and smoking.

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).

9.4(a)

Write the equation of the regression line.

# The equation of the regression line is as follows:
# days_absent = 18.93 - 9.11*eth + 3.10*sex + 2.15*lrn

9.4(b)

Interpret each one of the slopes in this context. The non-aboriginal female students with at least average learning skills tend to have the highest presence and aboriginal male students with below average learning skills tend to have the highest absence. For non-aboriginal males, the average days of absence increase by 3.10 days for average learners and an addition of another 2.15 days for slow learners.

9.4(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.

expected_days_absent <- 18.93 + 3.10 + 2.15

residual <- expected_days_absent - 2

residual

## [1] 22.18

# The residual is 22.18 days.

9.4(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 \(R^2\) and the adjusted \(R^2\). Note that there are 146 observations in the data set.

var_residuals <- 240.57

var_days <- 264.17

n_obs <- 146

k_val <- 3

sq_r <- 1 - (var_residuals/var_days)

adj_sq_r <- 1 - (var_residuals/(n_obs - k_val - 1)) / (var_days/(n_obs - 1))

sq_r

## [1] 0.08933641

adj_sq_r

## [1] 0.07009704

# The R-squared explains 8.933% of the variance.

# The adjusted R-squared explains 7.009% of the variance.

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.

9.8

Which, if any, variable should be removed from the model first?

Since the removal of learner status has actually improved the adjusted R-squared value, it should be removed from the model first. It shows that learner status is the least contributing factor to the 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.

9.16(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.

The o-rings are most likely to be damaged at lower temperatures in lower 50’s.

9.16(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.

The odds of an o-ring failing are decreased by 0.2162 by each degree increase in the temperature.

9.16(c)

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

Log(p_hat / (1 - p_hat)) = 11.6630 - 0.2162*Temperature

9.16(d)

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

The p-value is approaching 0 which is below the significance value of 0.05. Therefore, the null hypothesis is rejected and there is a high probability of link between lower temperature and failure of o-rings.

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.

9.18(a)

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:

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

# Probabilities of failure at 51, 53, and 55:

inter <- 11.663

slope <- 0.2162

prob_51 <- round((exp(inter - (slope * 51)) / (1 + exp(inter - (slope * 51)))),2)

prob_53 <- round((exp(inter - (slope * 53)) / (1 + exp(inter - (slope * 53)))),2)

prob_55 <- round((exp(inter - (slope * 55)) / (1 + exp(inter - (slope * 55)))),2)

prob_51

## [1] 0.65

prob_53

## [1] 0.55

prob_55

## [1] 0.44

# Probabilities of failure at 51, 53, and 55 are 0.65, 0.55, and 0.44 respectively.

9.18(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.

library(plotly)

## Loading required package: ggplot2

## 
## Attaching package: 'ggplot2'

## The following object is masked from 'package:openintro':
## 
##     diamonds

## 
## Attaching package: 'plotly'

## The following object is masked from 'package:ggplot2':
## 
##     last_plot

## The following object is masked from 'package:stats':
## 
##     filter

## The following object is masked from 'package:graphics':
## 
##     layout

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)
damaged_rings <- c(5,1,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,0)
shuttle_mission <- data.frame(temp, damaged_rings)
ggplot(shuttle_mission, aes(temp, damaged_rings)) + geom_point(colour = 'red') + stat_smooth(method = 'glm')

# The down trending line shows that the number of damaged o-rings falls with an increase in
# the temperature.

9.18(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 conditions for a logistic regression model are:

Logistic regression assumes linearity of independent variables and log odds, which requires a larger sample size. This sample is too small to test linearity 2. Independence of the occurences is the other condition which could be assumed in this case.

DATA606_HW9_RJM

RJM

2019-11-11