Chapter 8 - Multiple and Logistic Regression

Answer:

(a) Write the equation of the regression line.

\(\hat{y} = 120.07 - 1.93 \times parity\)

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

The slope indicates that for each one parsity increase, there will be a loss of 1.93 ounces in the baby’s weight.

Since the parity varies, below are the two functions for parity.

weight <- function(parity){
    yhat <- 120.07 - 1.93 * parity
    return(yhat)
}

• If baby is first born: parity = 0; the baby’s weight will be 120.07 ounces.

• If baby is NOT first born: parity = 1; the baby’s weight will be 118.14 ounces.

(c) Is there a statistically significant relationship between the average birth weight and parity?

Since the parity parameter’s \(p-value = 0.1052\); we can conclude that there is NOT a statistically significant relationship between average birth weight and parity.

Answer:

(a) Write the equation of the regression line.

\[\hat{y} = 18.93 - 9.11 \times eth + 3.10 \times sex + 2.15 \times lrn\]

(b) Interpret each one of the slopes in this context.

• The slope of eth indicates that, all else being equal, there is a \(9.11\) day reduction in the predicted absenteeism when the subject is NO aboriginal.

• The slope of sex indicates 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 indicates that, all else being equal, there is a \(2.15\) day increase in the predicted absenteeism when the subject is a slow learner.

(c) 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
missedActualDays <- 2

predictedDays <- 18.93 - 9.11 * eth + 3.1 * sex + 2.15 * lrn

residual <- missedActualDays - predictedDays

The residual for the above supplied information will be: -22.18.

(d) 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.

n <- 146 # Number os cases to fit the model
k <- 3   # number of predictor variables in the model
varResidual <- 240.57 # Variance of residual
varAllStudents <- 264.17      # Variance for all students

R2 <- 1 - (varResidual / varAllStudents)  # R2

adjustedR2 <- 1 - (varResidual / varAllStudents) * ( (n-1) / (n-k-1) ) # Adjusted R2

\[R^2 = 0.0893364\]

\[R^2_{Adjusted} = 0.070097\]

Answer:

Based on the Adjusted \(R^2 = 0.0723\) the lrn variable should be removed from the model first.

Answer:

(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)

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

plot(ShuttleMission)

By observing the above plot, we can find an interesting observation as follows:

• Higher number of damaged O-rings are observed when lower temperatures were recorded.

• Less number of damaged O-rings are observed when higher temperatures were recorded.

(b) 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.

This model is represented by two components: One, is the Intercept and the second one is the Temperature values. The Estimate identifies the parameter estimate for the model. The Z value and the P-value help with distinguishing important information from less significant parameters by telling us how good the variables predict this model.

(c) Write out the logistic model using the point estimates of the model parameters.

\[ log_e(\frac{p_i}{1 - p_i}) = 11.6630 - 0.2162 \times Temperature \]

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

Based on the collected data, we can deduct a high probability of damage to O-rings under \(50^o\). Also, since O-rings are “Critical” components, I do think concerns regarding the O-rings are justified.

Answer:

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

\[ log_e(\frac{\hat{p}}{1 - \hat{p}}) = 11.6630 - 0.2162 \times Temperature\]

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:

By solving the equation

\[ log_e(\frac{\hat{p}}{1 - \hat{p}}) = 11.6630 - 0.2162 \times Temperature\]

in terms of p, we obtain

\[\hat{p} = \frac{e^{11.6630 - 0.2162 \times Temperature}}{1 + e^{11.6630 - 0.2162 \times Temperature}} \]

p <- function(temp)
{
  damagedOring <- 11.6630 - 0.2162 * temp
  
  phat <- exp(damagedOring) / (1 + exp(damagedOring))
  
  return (round(phat*100,2))
}

# Testing final formula to provided results.
p57 <- p(57)
p59 <- p(59)
p61 <- p(61)

# Finding probabilities for 51, 53, and 55 temperatures.
p51 <- p(51)
p53 <- p(53)
p55 <- p(55)

• The probability that an O-ring will become damaged at 51 degrees Fahrenheit ambient temperatures is: 65.4%.

• The probability that an O-ring will become damaged at 53 degrees Fahrenheit ambient temperatures is: 55.09%.

• The probability that an O-ring will become damaged at 55 degrees Fahrenheit ambient temperatures is: 44.32%.

(b) 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.

ggplot(ShuttleMission,aes(x=temperature,y=damaged)) + geom_point() + 
  stat_smooth(method = 'glm', family = 'binomial')

(c) 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.)

Logistic regression conditions:

There are two key conditions for fitting a logistic regression model:

1. Each predictor \(x_i\) is linearly related to logit(\(p_i\) ) if all other predictors are held constant.

2. Each outcome \(Y_i\) is independent of the other outcomes.

Based on that definition, we have assumed that those conditions are met.