Chapter 9 - Multiple and Logistic Regression

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.)

(a)

Write the equation of the regression line.

Answer

Since this is linier model we can write the equation as .

Answer

babyweight=123.05− 8.94*smoke

(b)

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

Answer

Since this is a categorical variable, theirs means slope means the weight of babies whose mother are smokers are expected to be 8.94 ounces less than those whose mother are non-smokers.

#Predicted weight of babies born to smoker: 
#babyweight = 123.05 − 8.94 ∗ smoke
babyweightS=123.05-8.94*1
babyweightS

## [1] 114.11

#Predicted weight of babies born to non - smoker: 
#babyweight = 123.05 − 8.94 ∗ smoke
babyweightNS =123.05-8.94*0
babyweightNS

## [1] 123.05

(c)

Is there a statistically significant relationship between the average birth weight and smoking?

Answer

The p-value of both the slope and intercept is 0, which means both two coefficients are statistically significant, and therefore there is statistically significant relationship between the average birth weight of babies with smoking/non smoking moms.

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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).

        & Estimate  & Std. Error  & t value   & Pr

(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 \end{tabular} \end{center}

(a)

Write the equation of the regression line.

Answer

Absenteeism = 18.93 − 9.11∗eth + 3.10∗sex + 2.15∗lrn

(b)

Interpret each one of the slopes in this context.

Answer

Eth slope (−9.11): Assuming that other demographic characteristics are held constant, The average number days of absenteeism will decrease by 9.11 if the ethnic background of the student is not aboriginal.

Sex slope(3.10): Assuming that other demographic characteristics are held constant.The average number of days of absenteeism will increase by 3.10 if the student is male compared to when the student is female.

lrn slope(2.15): Assuming that other demographic characteristics are held constant, The average number of days of absenteeism will increase by 2.15 if the student is a slow learner compared to an average learner.

(c)

Answer

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 = 1
lrn = 1
Absenteeism= 18.93 - (9.11*eth) +(3.10*sex) + (2.15*lrn)
Absenteeism

## [1] 24.18

first_observation= 2
residual=first_observation-Absenteeism
residual

## [1] -22.18

Answer

The residual for a student who is aboriginal, male, a slow learner, and missed 2 days of school is: 24.18.

(d)

Answer

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 = 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.

SSE <- 240.57
SST <- 264.17
SSR <- SST - SSE
n = 146
p = 3

Rsquare <- SSR / SST
round(Rsquare,4)

## [1] 0.0893

Rsquare_adj <- 1- SSE / SST * (n-1) / (n-p-1)
round(Rsquare_adj,4)

## [1] 0.0701

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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.

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

Answer

The learner status should be removed from the model first, since we get the best adjusted R2.

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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.

(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.

As temperature become lower, occurrence of damaged oring increase.

Answer

(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.

Answer

Based on the table their is a negative relationship between Temperature and O ring failure.The table also shows low p value which means there is significant relationship between the two variables

(c)

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

y = log((p)/(1-p)) = 11.6630 - 0.2162 * (temperature)

Answer

(d)

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

Answer

Yes, since P value os <.05. There is statistical significane between the two variables.

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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:

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

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.

temp0 = 0
y <- 11.6630 - 0.2162*temp0
oring_prob <- exp(y)/(1 + exp(y))

# Temperature 51 degrees Fahrenheit
degrees51 <- 51
y51 <- 11.6630 - 0.2162*degrees51
oring_fail51 <- exp(y51)/(1 + exp(y51))
round(oring_fail51,4)

## [1] 0.654

degrees53 <- 53
y53 <- 11.6630 - 0.2162*degrees53
oring_fail53 <- exp(y53)/(1 + exp(y53))
round(oring_fail53,4)

## [1] 0.5509

degrees55 <- 55
y55 <- 11.6630 - 0.2162*degrees55
oring_fail55 <- exp(y55)/(1 + exp(y55))
round(oring_fail55,4)

## [1] 0.4432

temp.x = seq(from = 51, to = 71, by = 2)
y = c(oring_fail51, oring_fail53, oring_fail55, 0.341, 0.251, 0.179, 0.124, 0.084, 0.056, 0.037, 0.024)
plot(temp.x, y, type = "o")

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.

The condition of Logistic regression are hard to check due to Presence of outliers and the size of the dataset.