Data 606 Chapter 8 Homework 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 (Intercept) 120.07 parity -1.93 Std. Error 0.60 1.19 t value 199.94 -1.62 Pr(>|t|) 0.0000 0.1052

1. Write the equation of the regression line.

Answer: \(Weight_{i} = 120.07 - 1.93*parity_{i}\)

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

Answer: The slope -1.93 means the the weight of non first born babies will be on average 1.93 less than the first born baby.

3. Is there a statistically significant relationship between the average birth weight and parity?

Answer: \(Null Hypothesis: \beta_\text{parity} = 0\) \(Alternative Hypothesis: \beta_\text{parity} \ne 0\)

Since the p value of the parity is 0.1052 > 0.05 at 5% level of significance we reject the null hypothesis and conclude that the relationship between the average birth weight and parity is not statistically significant.

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 sam- pled students in rural New South Wales, Australia, in a particular school year. Below are three observations from this data set.

eth sex lrn days 10112 2 0 1 1 11 . . . . . ….. 146 1 0 0 37 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).18 Estimate (Intercept) 18.93 eth -9.11 sex 3.10 lrn 2.15 Std. Error 2.57 2.60 2.64 2.65 t value 7.37 -3.51 1.18 0.81 Pr(>|t|) 0.0000 0.0000 0.2411 0.4177

1. Write the equation of the regression line.

Answer: \(\widehat{\text{days_absent}} = 18.93 - 9.11 * eth_{i} + 3.10 * sex_{i} + 2.15*lrn_{i}\) (b) Interpret each one of the slopes in this context.

Answer:Slope of eth -9.11 means that not aboritinials are on average 9.11 day less absent.

Slope of sex 3.10 means that males are on average 3.10 day more absent than females.

lrn slope 2.15 means slow learners are on average 2.15 day more absent than average learners.

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.

Answer:

predicted <- 18.93 + 3.10 + 2.15
2 - predicted

## [1] -22.18

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 R2 and the adjusted R2. Note that there are 146 observations in the data set.

Answer:

N <- 146 
k <- 3   
variance <- 240.57 
variance_all <- 264.17

R2 <- 1 - (variance/variance_all)  
R2

## [1] 0.08933641

R2_adjusted <- 1 - (variance/variance_all) * ((N-1) / (N-k-1)) 
R2_adjusted

## [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.

Model 1 Full model 2 No ethnicity 3 No sex 4 No learner status Adjusted R2 0.0701 -0.0033 0.0676 0.0723 Which, if any, variable should be removed from the model first?

Answer: No learner status should be removed from the 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.

ShuttleMission 1 2 3 4 5 6 7 8 9 10 11 12 Temperature 53 57 58 63 66 67 67 67 68 69 70 70 Damaged 511100000010 Undamaged 155566666656 ShuttleMission 13 14 15 16 17 18 19 20 21 22 23 Temperature 70 70 72 73 75 75 76 76 78 79 81 Damaged 10000100000 Undamaged 56666566666

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.

Answer:

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 <- 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)
par(mfrow = c(1,2))
plot(temp,damaged,main = "Temperatuer vs damaged")
plot(temp,undamaged,main = "Temperatuer vs undamaged")

There is a negative relationship between temperature and damaged o ring.

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. Estimate (Intercept) 11.6630 Temperature -0.2162 Std. Error 3.2963 0.0532 z value 3.54 -4.07 Pr(>|z|) 0.0004 0.0000

Answer:

When temperature is 0 on average there are 11.6630 damaged rings. For one unit increase in temperature o ring damage decreases by 0.2162 uints.

The Tempeartuer slope and intercept coefficiednts of this model is significant at 5% level of significance since p values of both are less than 0.05.

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

Answer: \(log(p/(1-p)) = 11.663 - 0.2162 * Temperature_{i}\)

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

Answer: Concerns regaridng O-rings are justified. At lower tempeartuer the rings are more damaged.

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.

1.0 0.8 0.6 0.4 0.2 0.0 (a) The data provided in the previous exercise are shown in the plot. The logistic model fit to these data may be written as 􏰄 pˆ 􏰅 log 1 − pˆ = 11.6630 − 0.2162 × T emperature where pˆ 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: pˆ57 = 0.341 pˆ59 = 0.251 pˆ61 = 0.179 pˆ63 = 0.124 pˆ65 = 0.084 pˆ67 = 0.056 pˆ69 = 0.037 pˆ71 = 0.024

Answer: The probabilities are

lapply(c(51,53,55),function(x){ exp(11.6630 - 0.2162 * x) / (1 + exp(11.6630 - 0.2162 * x))})

## [[1]]
## [1] 0.6540297
## 
## [[2]]
## [1] 0.5509228
## 
## [[3]]
## [1] 0.4432456

2. 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.

Answer:

library(ggplot2)

## Warning: package 'ggplot2' was built under R version 3.5.3

data <- data.frame(shuttle=seq(1:23),
                 temp,
                 damaged,
                 undamaged)
data$rating <- data$damaged / (data$damaged + data$undamaged)

ggplot(data,aes(x=temp,y=damaged)) + geom_point() +  stat_smooth(method = 'glm', family ='binomial')+theme_classic()

## Warning: Ignoring unknown parameters: family

3. 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.

Answer: There are outlier in the data. Damaged and un damaged has multpile categoris for this reason logistic regression will not correct in this case.