Rpub Link: http://rpubs.com/ssufian/553164

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.

ans:

avg_birth_wt = -8.94*smoke + 123.05

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

ans:

slope = -8.94 => if mother smokes (independent variable) = 1, then birth weight of baby decreased by

8.94 units

# for smoker -> s = 1

s <- 1

m <- -8.94

intercept <- 123.05

birth_wt <- m*s + intercept

print(paste0("predicted birth wt of smokers: ", birth_wt))
## [1] "predicted birth wt of smokers: 114.11"
ns <-0

ns_birth_wt <- m*ns + intercept

print(paste0("predicted birth wt of non-smokers: ", ns_birth_wt))
## [1] "predicted birth wt of non-smokers: 123.05"
  1. Is there a statistically significant relationship between the average birth weight and smoking?

ans:

Based on this linear regression model, yes because the slope of -8.94 has a t-stat value > 0.05 or

p-value that is much much less than 0.05. This means that the slope is statistically significant

on a 95% confidence level

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.

ans:

abst = 2.15lrn + 3.1sex -9.11*eth + 18.93

intercept1 <- 18.93

eth_slope <- -9.11

sex_slope <- 3.1

lrn_slope <- 2.15
  1. Interpret each one of the slopes in this context.

ans:

eth_slope => aboriginal, yes or no

sex => female or male, yes or no

lrn => avg learner or slow learner, yes or no

NOTE: All the predictor variables are categorical (like dummy varaibles)

  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.
intercept1 <- 18.93

eth_slope <- -9.11

sex_slope <- 3.1

lrn_slope <- 2.15

eth1 <- 0

sex_1 <- 1

lrn_1 <- 1

abs_pred <- lrn_slope*lrn_1+lrn_slope*sex_1+eth_slope*eth1+intercept1 

print(paste0("Predicted days of absteeisim: ", abs_pred))
## [1] "Predicted days of absteeisim: 23.23"
residual <- abs_pred- 2

print(paste0("Residual: ", residual ))
## [1] "Residual: 21.23"
  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.
SSE <- 240.57

SST <- 264.17

n <- 146

Rsquare <- 1- SSE/SST

print(paste0("R-square: ", Rsquare  ))
## [1] "R-square: 0.0893364121588371"
Rsquare_adj <- 1-(1-Rsquare)*((n-1)/(n-2-1))

print(paste0("Adj R-square: ", Rsquare_adj))
## [1] "Adj R-square: 0.076599858482737"

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:

No Learner status should be removed becaue Adj R square results in lower R square and in this case,

it improves it, which is contracdictory

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.

ans:

Low temperature seems to cause O-ring damages.

  1. 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 probability of damaged o-rings decreases by \(\epsilon^{0.2162}\)

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

ans:

\(\ln(\frac{p}{1-p})\) = 11.633 - 0.2163*Temp

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

ans:

also indicates that lower temperatures result in greater probability of o-ring damage.

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:

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

ans:

#at temp = 51
P_hat51 = exp(11.663-0.2162*51)/(1+exp(11.663-0.2162*51))
P_hat51
## [1] 0.6540297
#at temp = 53
P_hat53 = exp(11.663-0.2162*53)/(1+exp(11.663-0.2162*53))
P_hat53
## [1] 0.5509228
#at temp = 55
P_hat55 = exp(11.663-0.2162*55)/(1+exp(11.663-0.2162*55))
P_hat55
## [1] 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.
T_F1 <- c(51,53,55,57,59,61,63,65,67,69,71)
P_model <- c(0.654,0.550,0.443,0.341,0.251,0.179,0.124,0.084,0.056,0.037,0.024)
P_meas <- c((5/6), (1/6), (1/6), (1/6), (0/6),(0/6),(0/6),(0/6),(0/6),(0/6),(1/6),(0/6),(1/6),(0/6),(0/6),(0/6),(0/6),(1/6),(0/6),(0/6),(0/6),(0/6),(0/6))
length(P_meas)
## [1] 23
T_F2 <- c(53,57,58,63,66,67,67,67,68,69,70,70,70,70,72,73,75,75,76,76,78,79,81)
length(T_F2)
## [1] 23
logistic_df <- data.frame(Temp = T_F1, P= P_model)
head(logistic_df)
meas_df <- data.frame(Temp = T_F2, P=P_meas)
head(meas_df)
suppressMessages(suppressWarnings(library(ggplot2)))
ggplot(NULL, aes(x=Temp,y=P)) + geom_line(data = logistic_df, colour = 'red')+geom_point(data=meas_df, colour='darkblue')

  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.

ans: