8.2 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.

  1. Write the equation of the regression line.
  2. Interpret the slope in this context, and calculate the predicted birth weight of first borns and others.
  3. Is there a statistically significant relationship between the average birth weight and parity?
babies <- read.csv('https://raw.githubusercontent.com/jbryer/DATA606Fall2018/master/data/os3_data/Ch%208%20Exercise%20Data/babies.csv')

#a ^y = 120.0684 - 1.9287 * parity
summary(lm(babies$bwt ~ babies$parity))
## 
## Call:
## lm(formula = babies$bwt ~ babies$parity)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -65.068 -11.068   0.396  10.932  57.860 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   120.0684     0.6005 199.942   <2e-16 ***
## babies$parity  -1.9287     1.1895  -1.621    0.105    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 18.22 on 1234 degrees of freedom
## Multiple R-squared:  0.002126,   Adjusted R-squared:  0.001317 
## F-statistic: 2.629 on 1 and 1234 DF,  p-value: 0.1052
#b Predicted weight of first born is 120.0684
# Predicted weight of second born and so forth is 118.1397
120.0684 - 1.9287 * 0
## [1] 120.0684
120.0684 - 1.9287 * 1
## [1] 118.1397
#c No there is not statistically significant relationship between the average weight and parity because p-value is greather than 0.05.

8.4 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.
  2. Interpret each one of the slopes in this context.
  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.
  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.
#a ^y = 18.93 ??? 9.11 * eth + 3.10 * sex + 2.15 * lrn

#b The slope of eth says that, all else being equal, there is a 9.11 day decrease in the predicted absenteeism when the subject is not aboriginal.
# The slope of sex says that, all else being equal, there is a 3.10 day increase in the predicted absenteeism when the subject is male.
# The slope of lrn says that, all else being equal, there is a 2.15 day increase in the predicted absenteeism when the subject is a slow learner.

#c The residuals is -22.18
eth <- 0 #aboriginal
sex <- 1 #male
lrn <- 1 #slow learner
missed_days <- 2
prediction <- 18.93 - 9.11 * eth + 3.1 * sex + 2.15 * lrn
missed_days - prediction #residuals
## [1] -22.18
#d R2: 0.08933641, Adjusted R2: 0.07009704
n <- 146 #number of observations
k <- 3   #number of explanatory variables
variance_residual <- 240.57 
varriance_all_students <- 264.17    
1 - (variance_residual / varriance_all_students)  # R2
## [1] 0.08933641
1 - (variance_residual / varriance_all_students) * ( (n-1) / (n-k-1) )
## [1] 0.07009704

8.8 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.

# No learner status variable should be removed from the model first because it has the highest adjusted R2.

8.16 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.
  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.
  3. Write out the logistic model using the point estimates of the model parameters.
  4. Based on the model, do you think concerns regarding O-rings are justified? Explain.
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)
orings <- data.frame(temperature, damaged, undamaged)

#a When the temperature is low, there is more damage to the o-rings
plot(orings$temperature, orings$damaged)

#b The summary of this model tells the number of damaged O-rings, with 11.66 as the y-intercept and number of O-rings damaged to decrease by -0.2162 as temperature increases. This is statistically significant because of z-value is -4.07 and the p-value is less than 0.05.

#c log(pi/1-pi) = y-intercept + b1(temperature) 
#  log(pi/1-pi) = 11.6630 - .2172 * temperature

#d Based on the model, I do think concerns regarding o-rings are justified because temperature drops as we go higher. Therfore, its essential to have o-rings that can withstand the temeperature drop.

8.18 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. The data provided in the previous exercise are shown in the plot. The logistic model fit to these data may be written as log(pi/1-pi) = 11.6630 - .2172 * 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:
  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.
  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.
temperature <- c(51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71)
p <- function(temperature){
  regression_line <- 11.6630 - 0.2162 * temperature
  p_hat <- exp(regression_line) / (1 + exp(regression_line))
  return (round(p_hat*100,2))
  }

#a Probablity of O-rings becoming damaged at each of the following temperature:
p(51)
## [1] 65.4
p(53)
## [1] 55.09
p(55)
## [1] 44.32
p(57)
## [1] 34.06
p(59)
## [1] 25.11
p(61)
## [1] 17.87
p(63)
## [1] 12.37
p(65)
## [1] 8.39
p(67)
## [1] 5.61
p(69)
## [1] 3.72
p(71)
## [1] 2.44
#b 
probability <- c(p(51), p(53), p(55), p(57), p(59), p(61), p(63), p(65), p(67), p(69), p(71)) 
model_plot <- data.frame(temperature, probability)

library(ggplot2)
ggplot(model_plot, aes(temperature, probability)) +
  geom_point() +
  geom_smooth()

#c There are two key conditions for fitting a logistic regression model: 
#  1. Each predictor xi is linearly related to logit(pi) if all other predictors are held constant.
#  2. Each outcome Yi is independent of the other outcomes.
# Condition 1 is hard to say because we do not have that many observations. But we can assume that the observations we have are independent.