Title: CUNY SPS MDS DATA606_HW9”

Author: Charles Ugiagbe

Date: 11/6/2021

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.
  2. Interpret the slope in this context, and calculate the predicted birth weight of babies born to smoker and non-smoker mothers.
  3. Is there a statistically significant relationship between the average birth weight and smoking?

Solution 1:

1a) The equation of the regression line is given by:

babyweight = 123.05 − 8.94 ∗ smoke

1b)

Because smoke is a categorical variable with value 1 for smoker and 0 for non-smoker, the slope means the weight of babies whose mother are smokers are expected to be 8.94 ounces less than those whose mother are non-smokers.

  Predicted weight of babies born to smoker: 
  babyweight = 123.05 − 8.94 ∗ smoke
  babyweight = 123.05 − 8.94 ∗ 1 = 133.99

  Predicted weight of babies born to non - smoker: 
  babyweight = 123.05 − 8.94 ∗ smoke
  babyweight = 123.05 − 8.94 ∗ 0 = 123.05

1c)

The p-value of both the slope and intercept is 0, which means both two coefficients are statistically significant, and therefore justified a statistifically siginificant 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).

  1. Write the equation of the regression line.
  2. Interpret each one of the slopes in this context.
  3. 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.
  4. 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.

Solution 2:

2a)

The equation of the regression line is given by:
Absenteeism = 18.93 − 9.11∗eth + 3.10∗sex + 2.15∗lrn

2b)

First slope(−9.11): The average number days of absenteeism will decrease by 9.11 if the ethnic background of the student is not aboriginal assuming that other demographic characteristics are held constant.

Second slope(3.10): The average number of days of absenteeism will increase by 3.10 if the student is male compared to when the student is female assuming that other demographic characteristics are held constant.

Third slope(2.15): The average number of days of absenteeism will increase by 2.15 if the student is a slow learner compared to an average learner assuming that other demographic characteristics are held constant.

2c)

eth <- 0 
sex <- 1
lrn <- 1
Predicted_value <- 18.93 − 9.11∗eth + 3.10∗sex + 2.15∗lrn
Observed_value <- 2
residual <- Observed_value - Predicted_value
residual
## [1] -22.18

The residual for a student who is aboriginal, male, a slow learner, and missed 2 days of school is: -22.18

2d)

\(R^{2} = 1− \frac{Var(residuals)}{Var(outcome)}\)

\(R^{2}_{adj} = 1− \frac{Var(residuals)}{Var(outcome)}∗ \frac{n−1}{n−k-1}\)

SSE <- 240.57
SST <- 264.17
SSR <- SST - SSE
n = 146
p = 3

r_sq <- SSR / SST

r_sq_adj <- 1- SSE / SST * (n-1) / (n-p-1)

r_sq
## [1] 0.08933641
r_sq_adj
## [1] 0.07009704

The \(R^2\) is 0.08933641 and the adjusted \(R^2\) is 0.07009704.

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?

Solution 3:

We expected to increase the adjusted \(R^2\) through removing variables. Using the process of backward elimination, we should remove the variable with the lowest adjusted R2. Hence, remove the “No ethnicity” variable first since it has the lowest adjusted R2.

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.
  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.
  1. Write out the logistic model using the point estimates of the model parameters.
  2. Based on the model, do you think concerns regarding O-rings are justified? Explain.

Solution 4:

4a)

According to the given data, there is a negative relationship between temperature and damaged o-ring. The lower the temperature, the higher the chances that the O-rings got damage.

4b)

The summary includes 6 components:

Intercept: 11.6630 value of the intercept is the value of the damaged o-ring when the value of temperature is equal to zero(0)

The slope: which is the coefficients for the temperature, is -0.2162. It means that for 1 Fahrenheit degree increase / decrease on the temperature, the probability of damage on the O-rings is expected to be decrease / increase by 0.2162 unit, which justified my observation in part (a).

Standard error: This mean that the observed value falls 3.2963 from the Regression line for the intercept and 0.0532 for the slope(temperature)

z-value mean the value of deviation from the mean for the estimate are -4.07 for the temperature and 3.54 for the intercept.

4c)

The logit model using the point estimates of the model is given by:

      P(x) = 1 / (1+ exp(-(11.6630-0.2162 * x)))

4d)

\(H_{0}:β_{1} = 0\)

\(H_{0}:β_{1} ≠ 0\)

The p-value is approximately 0 and less than 0.05. which means the model is statistically significant. We therefore reject the null hypothesis. Hence, the concerns regarding O-rings are justified based on the given data.

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.

  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:

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

  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.
  2. 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.

Solution 5:

5a)

logi_model <-  function(x){
  return (1/(1+ exp(-(11.6630-0.2162 * x))))}

logi_model(51)
## [1] 0.6540297
logi_model(53)
## [1] 0.5509228
logi_model(55)
## [1] 0.4432456

P(51) = 0.654, P(53) = 0.551, P(55) = 0.443

5b)

library(tidyverse)
## -- Attaching packages --------------------------------------- tidyverse 1.3.1 --
## v ggplot2 3.3.5     v purrr   0.3.4
## v tibble  3.1.5     v dplyr   1.0.7
## v tidyr   1.1.4     v stringr 1.4.0
## v readr   2.0.2     v forcats 0.5.1
## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()
x<- seq(51,82,2)

data <- data.frame(x, logi_model(x)) %>%
  rename(temp = x, P = logi_model.x.)

plot(orings[,1] + 
       c(rep(0, 5), c(-0.1, 0, 0.1), 0, 0, -0.07, -0.07, 0.07, 0.07, 
         rep(0, 4), -0.07, 0.07, 0, 0, 0), 
     orings[,2]/6, 
     xlab = "", ylab = "Probability of damage", 
     xlim = c(50, 82), ylim = c(0,1), 
     col = COL[1,2], pch = 19)
mtext("Temperature (Fahrenheit)", 1, 2)
points(data$temp, data$P, col = 'deeppink', pch = 19)
lines(data$temp,data$P, col = 'deeppink', lty = 3)
legend(x = 75, 
       y = 1, 
       legend = c('Observation','Prediction'), 
       cex = 0.7, 
       pch = c(19,19), 
       lty = c(NA,3),
       col = c('lightblue3', 'deeppink'),
       text.col = c('lightblue4', 'deeppink'))

5c)

The major concern I will have in applying logistic regression in this application is the sample size.The model shows statistically significant, however the size of the data is too small to get an optimal model and the attributes are too few to justify the practical significance of the model.