library(ggplot2)

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.

  1. Write the equation of the regression line.

\(y=120-1.93x\)

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

The predicte birth weight of first borns is 120 ounces. Each successive child is estimated to be 1.93 ounces less than the previous child born to the same parents.

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

No, the p-value is \(.1052 > .05\), which is not a statistically significant p-value.

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 South Wales, 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

  1. Write the equation of the regression line.

\(y=18.93-9.11eth +3.10sex + 2.15lrn\)

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

Ethinicity has a statistically significant p-value, and the slope indicates that, on average, aboriginal students miss 9 more days of school than non-aboriginal students. Sex and learning ability do not have statistically significant p-values, but they imply that male students miss 3 more days on average and slow learning students miss 2 more days on average.

  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.

\(y=18.93-9.11(0) +3.10(1) + 2.15(1)=24.18\) is the estimated days missed. So the residual would be \(2-24.18=-22.18\).

  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 \(R^2\) and the adjusted \(R^2\). Note that there are 146 observations in the data set.

\(R^2=1-\frac{240.57}{264.17}=.08933641\)

adjusted: \(R^2_{adjusted}=1-\frac{240.57}{264.17}*\frac{(n-1)}{(n-k-1)}=1-\frac{240.57}{264.17}*\frac{145}{142}= 0.070097\)

where n=number of observations and k=number of variables.

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 \(R^2\) 1 Full model 0.0701 2 No ethnicity -0.0033 3 No sex 0.0676 4 No learner status 0.0723

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

The model would improve if the learner status variable was removed, because the adjusted \(R^2\) value when this variable is removed is higher than the original model’s \(R^2\) value.

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 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.

At a glance, it appears the damaged o-rings occur at lower temperatures. This could be the contrast between the shuttle and the air temperature, among many other situations.

  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

When the ambient temperature is 0 degrees Farenheit, there will be an estimated 11 damaged O-rings. As temperature increases by 1 degree Farenheit, the number of damaged O-rings will decrease by .2162. These are both statistically significant variables.

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

\(y=11.6630-.2162x\)

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

Yes, since the P-value is <.01, there is evidence that there is a relationship between the damaged O-rings and the change in temperature. This is something that should be fixed, since the temperature drops the further you get from Earth.

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. The data provided in the previous exercise are shown in the plot. The logistic model fit to these data may be written as \(log(\frac{p}{1-p}) = 11.6630-0.2162*Temperature\)

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: \(p_{57}=0.341\) \(p_{59}=0.251\) \(p_{61} = 0.179\) \(p_{63} = 0.124\) \(p_{65} = 0.084\) \(p_{67} = 0.056\) \(p_{69} = 0.037\) \(p_{71} = 0.024\)

Temperature <- c(51, 53, 55)
p <- exp(11.6630-0.2162*Temperature)/(exp(11.6630-0.2162*Temperature)+1)
p
## [1] 0.6540297 0.5509228 0.4432456

So we can see that \(p_{51}=0.6540297\) \(p_{53}=0.5509228\) \(p_{55}=0.4432456\)

  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.
temp <- c(51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71)
prob <- c(0.6540297, 0.5509228, 0.4432456, 0.341, 0.251, 0.179, 0.124, 0.084, 0.056, 0.037, 0.024)
Oringdata <-as.data.frame(cbind(temp, prob))
Oringdata
##    temp      prob
## 1    51 0.6540297
## 2    53 0.5509228
## 3    55 0.4432456
## 4    57 0.3410000
## 5    59 0.2510000
## 6    61 0.1790000
## 7    63 0.1240000
## 8    65 0.0840000
## 9    67 0.0560000
## 10   69 0.0370000
## 11   71 0.0240000
ggplot(Oringdata, aes(x=temp, y=prob)) + geom_point() + stat_smooth(method = 'glm')

  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.

We are assuming the observations are independent, and we don’t have enough observations to be confident that this model is accurate.