Chapter 8 - Multiple and Logistic Regression Graded: 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.

            Estimate        Std. Error     t value       Pr(>|t|)

(Intercept) 120.07 0.60 199.94 0.0000 parity -1.93 1.19 -1.62 0.1052

  1. Write the equation of the regression line.

\[\hat { baby.weight } =\quad 120.07\quad -\quad 1.93\quad \times \quad parity:not.first.born\]

  1. Interpret the slope in this context, and calculate the predicted birth weight of first borns and others.

All else being equal, children that are not first-born, tend to be 1.93 ounces lighter on average, than their corresponding first-born counterparts.

#infile<-getURL("https://raw.githubusercontent.com/jbryer/DATA606Spring2019/master/data/os3_data/Ch%208%20Exercise%20Data/babies.csv")
#babies<-read.csv(text=infile,header=TRUE)
#lm(babies$weight~babies$parity, data=babies)

(first.born.wt<-120.07)
## [1] 120.07
(not.first.born.wt<-120.07-1.93)
## [1] 118.14
  1. Is there a statistically significant relationship between the average birth weight and parity?

Given the high p-value associated with the slope of the parity variable, we fail to reject the null hypothesis which states that there is no significant relationship between first born status and estimated birth weight.

8.4 Absenteeism, Part I. 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)

        Estimate Std. Error   t value   Pr(>|t|)

(Intercept) 18.93 2.57 7.37 0.0000 eth -9.11 2.60 -3.51 0.0000 sex 3.10 2.64 1.18 0.2411 lrn 2.15 2.65 0.81 0.4177

  1. Write the equation of the regression line.
infile<-getURL("https://raw.githubusercontent.com/jbryer/DATA606Spring2019/master/data/os3_data/Ch%208%20Exercise%20Data/absenteeism.csv")
absenteeism<-read.csv(text=infile,header=TRUE)

(absent<-lm(absenteeism$Days~absenteeism$Eth+absenteeism$Sex+absenteeism$Lrn, data=absenteeism))
## 
## Call:
## lm(formula = absenteeism$Days ~ absenteeism$Eth + absenteeism$Sex + 
##     absenteeism$Lrn, data = absenteeism)
## 
## Coefficients:
##       (Intercept)   absenteeism$EthN   absenteeism$SexM  
##            18.932             -9.112              3.104  
## absenteeism$LrnSL  
##             2.154

\[\hat { days.absent } =\quad 18.93\quad -\quad 9.11\quad \times \quad ethnicity:not.aboriginal\quad +\quad 3.10\quad \times \quad sex:male\quad +\quad 2.15\quad \times \quad learner.status:slow\]

  1. Interpret each one of the slopes in this context.

All else being equal, students that are not aboriginal tend to be present for an additional 9.11 days above average, than their aboriginal counterparts.

All else being equal, male students tend to be absent for an additional 3.10 days above average, than their female counterparts.

All else being equal, slow learners tend to be absent for an additional 2.15 days above average, than the average learners.

  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.
absenteeism[1,]
##   Eth Sex Age Lrn Days
## 1   A   M  F0  SL    2
absent$residuals[1]
##         1 
## -22.19026
(residual.days.absent=(18.932+9.112*0+3.104*1+2.154*1)-2)
## [1] 22.19
  1. The variance of the residuals is 240.57, and the variance of the number of absent days for all students in the data set is 264.17. Calculate the R2 and the adjusted R2. Note that there are 146 observations in the data set.
(r.squared<-((264.17-240.57)/264.17))
## [1] 0.08933641
(adj.r.squared<-(1-((240.57/264.17)*((146-1)/(146-3-1)))))
## [1] 0.07009704
absent.summ<-summary(absent)
absent.summ$r.squared
## [1] 0.08933149
absent.summ$adj.r.squared
## [1] 0.07009202

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.

    Model              Adjusted R2

1 Fullmodel 0.0701 2 Noethnicity -0.0033 3 Nosex 0.0676 4 No learner status 0.0723

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

The learner status variable should be removed from the model first, since the model without this variable has a higher adjusted R-squared than the full model.

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.

Shuttle Mission 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Temperature 53 57 58 63 66 67 67 67 68 69 70 70 70 70 72 73 75 75 76 76 78 79 81 Damaged 5 1 1 1 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 Undamaged 1 5 5 5 6 6 6 6 6 6 5 6 5 6 6 6 6 5 6 6 6 6 6

  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.

Lower ambient temperatures are associated with more damaged O-rings. Ambient temperatures below 66 degrees were associated with at least 1 O-ring damaged, while ambient temperatures above 75 degrees were associated with no O-rings damaged. So there appears to be a significant relationship between these 2 variables that needs to be analyzed further.

  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.

        Estimate    Std. Error  z value   Pr(>|z|)

    (Intercept) 11.6630 3.2963 3.54 0.0004 Temperature -0.2162 0.0532 -4.07 0.0000

Based on the negative sign of the slope of the temperture variable, it can be stated that there is an inverse relationship between temperature and probability of O-ring damage. The log odds of O-ring damage decrease by -0.21 for each additional degree of temperature in fahrenheits.

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

\[\hat { log\quad (p/(1-p)) } =\quad 11.663\quad -\quad 0.2162\quad \times \quad temperature\]

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

Based on the p-value, the null hypothesis that the slope of the temperature variable is 0, can be rejected. This shows that there is a statistically significant relationship between temperature and O-ring damage and therefore the concerns are 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. 50 55 60 65 70 75 80 0.0 0.2 0.4 0.6 0.8 1.0 Probability of damage Temperature (Fahrenheit)

  1. The data provided in the previous exercise are shown in the plot. The logistic model fit to these data may be written as. where ˆp 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.
temperature<-c(51,53,55)

(odds.ratio=exp(11.663-0.2162*temperature))
## [1] 1.8904218 1.2267888 0.7961243
(p.damage=(odds.ratio/(1+odds.ratio)))
## [1] 0.6540297 0.5509228 0.4432456

The model-estimated probabilities for several additional ambient temperatures are provided below, where subscripts indicate the temperature: ˆp57 = 0.341 ˆp59 = 0.251 ˆp61 = 0.179 ˆp63 = 0.124 ˆp65 = 0.084 ˆp67 = 0.056 ˆp69 = 0.037 ˆp71 = 0.024

prob.damage<-function(temp) {
  odds.ratio=exp(11.663-0.2162*temp)
  prob.damage=(odds.ratio/(1+odds.ratio))
  
}
temperature<-c(51,53,55,57,59,61,63,65,67,69,71)
p.damage<-prob.damage(temperature)
p.damage
##  [1] 0.65402974 0.55092283 0.44324565 0.34064976 0.25109139 0.17869707
##  [7] 0.12372702 0.08393843 0.05612566 0.03715479 0.02443024
  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.
o.rings<-data.frame(temperature, p.damage)
ggplot(o.rings, aes(x=temperature,y=p.damage)) + geom_line() + geom_point()

  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.

The model assumes that o-ring damage is a binary event i.e. damaged or undamaged, and that once damaged, it is leading to shuttle failure. Extent of damage is not taken into consideration i.e. whether there is a possibility where a damaged o-ring can still function. Also the assumption is that observations are independent which may not be the case - for example, an entire batch of o-rings used for a specific launch may have issues. Lastly, there could be other variables that are collinear to temperature that might be more directly related to o-rings, which are not being considered here.

The assumptions required to accept the model’s validity are: - linear relationship between the independent variable i.e. temperature and the logit function that relates the probability of damage of o-ring to it. - near normal residuals i.e. independent outcomes. - constant variability of residuals.