8.2, 8.4, 8.8, 8.16, 8.18

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.

Answer : Equation is y = b0 + b1x Weight = 120.07 - 1.93 * Parity

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

Answer Slope of the regression line in this context says that, for every child who is not a first born will predicted to be have 1.93 less than birth weight of the first child. So the firstborn weighs 120.07 ounces and every other child afterwords would be 120 - 1.93 * 1 = 118.07

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

Answer
The ‘p value’ of the parity is 0.1052 which is greater than 0.05. So there is no statistically significant relationship between avarage birth weight and parity.

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 SouthWales, Australia, in a particular school year. Below are three observations from this data set.

_ eth sex lrn days
1 0 1 1 2
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).

_ 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.

Answer : Equation is y = b0 + b1x eth + b2 * sex + b3 * lrn = 18.93 + (- 9.11 * (ethnic background) ) + 3.10 * (sex) + 2.15 * (learner status)

  1. Interpret each one of the slopes in this context. Answer if we keep other constant and check : Ethnic background: The model predicts a 9.11 absent days decrease in non-aboriginal children.

Sex: The model predicts a 3.10 absent days increase in males over females.

Learner status: The model predicts a 2.15 absent days increase in slow learners over average learners, Slow learners miss 2.15 more days

  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.

ANswer

Absent <- 18.93 - 9.11 * (0) + 3.10 * (1) + 2.15 * (1)
Residual <- 2 - Absent
paste("Residual for this student: ", Residual)
## [1] "Residual for this student:  -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 R2 and the adjusted R2. Note that there are 146 observations in the data set.
# R-squared = 1 - (variance of residuals)/(variance in outcome)
# R-squared adjusted. 1 - (variance of residuals)/(variance in outcome)*(n-1)/(n-k-1), where k is predictor in variables in model.

R2.Ab <- 1 - (240.57)/(264.17)
R2.Ab.adj <- 1 - (240.57/264.17)*((146-1)/(146-3-1))
paste("R-squared: ", round(R2.Ab,4)) ;paste("R-squared adjusted: ", round(R2.Ab.adj,4))
## [1] "R-squared:  0.0893"
## [1] "R-squared adjusted:  0.0701"

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?

Answer Since adjusted R2 improves when learner status is removed, learner status should be removed first.

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 Temperature 53 57 58 63 66 67 67 67 68 69 70 70 Damaged 5 1 1 1 0 0 0 0 0 0 1 0 Undamaged 1 5 5 5 6 6 6 6 6 6 5 6

Shuttle Mission 13 14 15 16 17 18 19 20 21 22 23 Temperature 70 70 72 73 75 75 76 76 78 79 81 Damaged 1 0 0 0 0 1 0 0 0 0 0 Undamaged 5 6 6 6 6 5 6 6 6 6 6

  1. Each column of the table above represents a di???erent shuttle mission. Examine these data and describe what you observe with respect to the relationship between temperatures and damaged O-rings.

Answer There are total 11 damaged o-ring, 8 damaged o-rings at temperature ???63oF. There are 3 damaged o-rings above that temperature. It does seem that low temperatures contribute to o-ring damage.

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

data <- data.frame(temperature = temperature, damaged = damaged, undamaged = undamaged)

ggplot(data,aes(x=temperature,y=damaged)) + geom_point()

  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

Answer Given that this is a logistic regression, it behaves fairly similarly to a multiple regression. Therefore, if the numerical variable temperature increases, in this case 1 degree, then the likelihood of being damaged decreases by .2162. The final result will need to be transformed to make sense of the information.

The slope mean that for every 10F above zero, the probability of damaged o-rings decreases by 0.2162 in the exponential term.

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

Answer ln(phat/(1-phat)) = 11.6630 - 0.2162*T

Where p^ is the probability of damaged o-rings and T is temperature (F).

  1. Based on the model, do you think concerns regarding O-rings are justified? Explain. Answer Yes, the result is statistically significant beyond the 95% confidence level at p=0.0000. The slope also indicates that lower temperatures result in greater probability of o-ring damage. ***

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 takeo??? 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

log(phat/(1 ??? phat)) = 11.6630 ??? 0.2162 * Temperature

where ^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: ^p57 = 0.341 ^p59 = 0.251 ^p61 = 0.179 ^p63 = 0.124 ^p65 = 0.084 ^p67 = 0.056 ^p69 = 0.037 ^p71 = 0.024

P_hat51 = exp(11.663-0.2162*51)/(1+exp(11.663-0.2162*51))
P_hat51
## [1] 0.6540297
P_hat53 = exp(11.663-0.2162*53)/(1+exp(11.663-0.2162*53))
P_hat53
## [1] 0.5509228
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.

Answer

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)
##   Temp     P
## 1   51 0.654
## 2   53 0.550
## 3   55 0.443
## 4   57 0.341
## 5   59 0.251
## 6   61 0.179
meas_df <- data.frame(Temp = T_F2, P=P_meas)
head(meas_df)
##   Temp         P
## 1   53 0.8333333
## 2   57 0.1666667
## 3   58 0.1666667
## 4   63 0.1666667
## 5   66 0.0000000
## 6   67 0.0000000
suppressMessages(suppressWarnings(library(ggplot2)))
ggplot(NULL, aes(x=Temp,y=P)) + geom_line(data = logistic_df, colour = 'red')+geom_point(data=meas_df, colour='lightblue')

  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

Answer Conditon must meet for Logistic Regression: a. Each Predictor Xi must be linearly related to logit(pi), if all other predictor held constant b. each out come of Yi is indipendent .

The sample size 23 is fairly small to varify the 1st condition. Regarding the second condition, AS the shuttlemission is very complicated, the reason of failure could be because of some other cause. More investigation is required. Its assumed that the events are independent.