Baby weights, Part I. (9.1, p. 350) The Child Health and Development Studies investigate a range of topics. One study considered all pregnancies between 1960 and 1967 among women in the Kaiser Foundation Health Plan in the San Francisco East Bay area. Here, we study the relationship between smoking and weight of the baby. The variable smoke is coded 1 if the mother is a smoker, and 0 if not. The summary table below shows the results of a linear regression model for predicting the average birth weight of babies, measured in ounces, based on the smoking status of the mother.

The variability within the smokers and non-smokers are about equal and the distributions are symmetric. With these conditions satisfied, it is reasonable to apply the model. (Note that we don’t need to check linearity since the predictor has only two levels.)

  1. Write the equation of the regression line.

Y = mX + b Y = baby_weight = 123.05 - 8.94*smoke

  1. Interpret the slope in this context, and calculate the predicted birth weight of babies born to smoker and non-smoker mothers.

the slope of -8.94 indicates that babies who were born to mothers that smoked were born 8.94oz less than babies born to non-smoking mothers.

The predicted births for mothers who smoke, and for those that don’t, is 114.11oz and 123.05oz respectively.

123.05 - 8.94 * 1
## [1] 114.11
123.05 - 8.94 * 0
## [1] 123.05
  1. Is there a statistically significant relationship between the average birth weight and smoking?

Because the slope = 0 , or the p-value, which is less than 0.05, the alpha level, we can confidently say there is a relationship between the average birth rate and smoking.

Absenteeism, Part I. (9.4, p. 352) 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.

absent_days = 18.93 - 9.11 * eth + 3.10 * sex + 2.15 * lrn.

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

It’s expected that by… i) changing aboriginal to not-aboriginal, we would decrease the number of days be 9.11 ii) changing male to female increases the number of days by 3.1 iii) changing average > slow learner increases the number of days by 2.1

  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.
eth=0
sex=lrn=1
days=2

y=18.93-9.11*eth+3.1*sex+2.15*lrn

residual = days - y
residual
## [1] -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_squared <- round(1 - (240.57 / 264.17), 3)
r_squared
## [1] 0.089
adj_r_squared <- round(1 - (240.57 / 264.17) * (146 - 1) / (146 - 3 - 1), 3)
adj_r_squared
## [1] 0.07

Absenteeism, Part II. (9.8, p. 357) Exercise above 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.

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

The variable “no learner status” should be removed from the model first because it is the highest \(R^2\) value.

Challenger disaster, Part I. (9.16, p. 380) 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.

Between temperatues and damaged O-rings, we can see that there is a positive correlation

  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.

The key components of this summary table are that for every increase of 1 degree, the chances of a damaged O-ring decreases by 21.62%. Because the p-value is close to 0, there is a statistically significant relationship between temperature and damaged O-rings.

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

Y = 11.6630 - 0.2162 * temperature

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

According to this model, concerns regarding O-rings are justified because temperature is statistically significant, and that lower temperatures have a higher chance of damaging O-rings.

Challenger disaster, Part II. (9.18, p. 381) Exercise above 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.

\begin{center} \end{center}

  1. The data provided in the previous exercise are shown in the plot. The logistic model fit to these data may be written as \[\begin{align*} \log\left( \frac{\hat{p}}{1 - \hat{p}} \right) = 11.6630 - 0.2162\times Temperature \end{align*}\]

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:

\[\begin{align*} &\hat{p}_{57} = 0.341 && \hat{p}_{59} = 0.251 && \hat{p}_{61} = 0.179 && \hat{p}_{63} = 0.124 \\ &\hat{p}_{65} = 0.084 && \hat{p}_{67} = 0.056 && \hat{p}_{69} = 0.037 && \hat{p}_{71} = 0.024 \end{align*}\]

  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.

We can say that the probability an O-ring will endure damage is 65.4%, 55.1%, 44.3%, for temperatures at 51f, 53f, and 55f respectively

deg_51 <- round(exp(11.6630 - 0.2162 * 51) / (1 + exp(11.6630 - 0.2162 * 51)), 3)

deg_53 <- round(exp(11.6630 - 0.2162 * 53) / (1 + exp(11.6630 - 0.2162 * 53)), 3)

deg_55 <- round(exp(11.6630 - 0.2162 * 55) / (1 + exp(11.6630 - 0.2162 * 55)), 3)

deg_51
## [1] 0.654
deg_53
## [1] 0.551
deg_55
## [1] 0.443
temp_2 <- c(seq(51, 71, 2))
probability <- exp(11.6630 - 0.2162 * temp_2) / (1 + exp(11.6630 - 0.2162 * temp_2))
plot(data.frame(temp_2, probability), type = "b", pch = 19)

  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 assume that the variables are independent, and normally distritued. A conern is that the sample size is small, however understandably so because this is a rare event. Although with todays technology, this can be simulated n-number of times, therefore obtaininig a more accurate model.