MSDS Spring 2018

DATA 606 Statistics and Probability for Data Analytics

Jiadi Li

Chapter 8: Multiple and Logistic Regression

HW 8: 8.2, 8.4, 8.8, 8.16, 8.18

8.2 Baby weights, Part II.
Exercise 8.1 introduces a data set on birth weight of babies. Another variable we consider is parity, which is 0 if the child is the first born, and 1 otherwise. The summary table below shows the results of a linear regression model for predicting the average birth weight of babies, measured in ounces, from parity.

(a) Write the equation of the regression line.
weight = 120.07 -1.93\(\times\)parity

  1. Interpret the slope in this context, and calculate the predicted birth weight of first borns and others.
    If the child is the first born, parity will be 0, the baby’s weight will be 120.07 ounces; otherwise, partify will be 1, the baby’s weight will be 118.14 ounces.

  2. Is there a statistically significant relationship between the average birth weight and parity?
    No, the p-value is 0.1052 for parity which is larger than 0.05 and therefore no statistical significance.


8.4 Absenteeism.
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).


(a) Write the equation of the regression line.
absent_days = 18.93 - 9.11 \(\times\) eth + 3.10 \(\times\) sex + 2.15 \(\times\) lrn

  1. Interpret each one of the slopes in this context.
    If the student is aboriginal, eth will be 0, the absent days will not be impacted; otherwise, eth will be 1, there will be 9.11 less days.
    If the student is female, sex will be 0, the absent days will not be impacted; otherwise, sex will be 1, there will be 3.10 additional days.
    If the student is average learner, lrn will be 0, the absent days will not be impacted; otherwise, lrn will be 1, there will be 2.15 additional days.

  2. 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.

eth <- 0 #aboriginal
sex <- 1 #male
lrn <- 1 #slow learner

predicting <- 18.93 - 9.11 * eth + 3.10 * sex + 2.15 * lrn

residual <- 2 - predicting

residual
## [1] -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.
var_residual <- 240.57
var_allstudents <- 264.17
1 - (var_residual / var_allstudents) #R^2
## [1] 0.08933641
n <- 146 # number of cases
k <- 3   # number of predictors
1 - (var_residual / var_allstudents) * ( (n-1) / (n-k-1) ) #adjusted R^2
## [1] 0.07009704



8.8 Absenteeism, Part II.
Exercise 8.4 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?
R-squared evaluates the scatter of the data points around the fitted regression line. It is also called the coefficient of determination, or the coefficient of multiple determination for multiple regression. For the same data set, higher R-squared values represent smaller differences between the observed data and the fitted values.
Based on the adjusted \(R^2\) table, No ethnicity variable, with only -0.0033, should be removed from the model first during backwards elimination process.


8.16 Challenger disaster, Part I.
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.
(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.

Temperature <- 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 <- c(5,1,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,0)
Undamaged <- c(1,5,5,5,6,6,6,6,6,6,5,6,5,6,6,6,6,5,6,6,6,6,6)

Shuttle_Mission <- data.frame(Temperature,Damaged,Undamaged)

hist(Shuttle_Mission$Temperature,main = 'Histogram: Temperature')

plot(Shuttle_Mission$Temperature,Shuttle_Mission$Damaged,main = 'Temperature vs. Damaged',xlab = 'Temperature',ylab = 'Damaged')

plot(Shuttle_Mission$Temperature,Shuttle_Mission$Undamaged,main = 'Temperature vs. Undamaged',xlab = 'Temperature',ylab = 'Undamaged')

cor(Shuttle_Mission$Temperature,Shuttle_Mission$Damaged)
## [1] -0.6384574
cor(Shuttle_Mission$Temperature,Shuttle_Mission$Undamaged)
## [1] 0.6384574

Temerature is correlated with both Damaged and Undamaged with correlation coefficients of -63.85% and 63.85%.
Damaged and Undamaged contain the exactly same information.
Higher temperature is associated with less damaged O-rings.

  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.

    There are two estimates which are similar to a linear model. When there is an additional unit of temperature, the \(log\) value will be decreased by 0.2162.

  2. Write out the logistic model using the point estimates of the model parameters.
    \(log_e(\frac{\hat{p}}{1-\hat{p}})\) = 11.6630 - 0.2162 \(\times\) Temperature

  3. Based on the model, do you think concerns regarding O-rings are justified? Explain.
    With the level of significance and the association between Temerature and Damaged, the model should be justified.


8.18 Challenger disaster, Part II.
Exercise 8.16 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.

(a) The data provided in the previous exercise are shown in the plot. The logistic model fit to these data may be written as

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:

get_y <- function(temperature){
  logit <- 11.6630 - 0.2162 * temperature
  p <- exp(logit)/(1+exp(logit))
  return(p*100)
}


p.51 <- get_y(51)
p.53 <- get_y(53)
p.55 <- get_y(55)
p.57 <- get_y(57)
p.59 <- get_y(59)
p.61 <- get_y(61)

c(p.51,p.53,p.55,p.57,p.59,p.61)
## [1] 65.40297 55.09228 44.32456 34.06498 25.10914 17.86971
  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.
x <- seq(51,61,2)
y <- c(p.51,p.53,p.55,p.57,p.59,p.61)
scatter.smooth(x,y)

  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.
    For an logistic regression, two conditions need to be met which are: linearity (related to logit) and independence (of the outcomes). Both are met.