Graded: 8.2, 8.4, 8.8, 8.16, 8.18

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.

Estimate Std. Error t value Pr(>|t|) (Intercept) 120.07 0.60 199.94 0.0000 parity -1.93 1.19 -1.62 0.1052

  1. Write the equation of the regression line.

\(Avg\_weight = 120.07 - 1.93 * parity\)

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

The estimated average second born baby weight to parity is 1.93 unit lower than first born baby.

120.07 - 1.93 * 0 = 120.07 for first born 120.07 - 1.93 * 1 = 118.14 for first born

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

Since the P-value is greater than 0.05, we rejected there is no different between first born and second born on average baby weight to parity.

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 SouthWales, 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

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(>|t|)

(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

  1. Write the equation of the regression line.

\(Avg\_Dayabsent = 18.93 - 9.11 * eth + 3.10 * sex + 2.15 * lrn\)

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

The model predicts 9.11 days decrease in average day absent for one not aboriginal increase, 3.1 days increase in average day absent for one male increase, 2.15 days increase in average day absent for one slow learner increase.

  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_i-Avg_Dayabsent = 2-(18.93-9.110 + 3.101 +2.15*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 R2 and the adjusted R2. Note that there are 146 observations in the data set.

\(R^{2}\) = \(1- 204.57/264.17 =8.9336\%\) \(R_{adj}^{2}\) = \(1- 204.57/264.17* (146-1)/(146-3-1) = 20.9252\%\)

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 R2

1 |Fullmodel |0.0701

2 |Noethnicity | -0.0033

3 |Nosex |0.0676

4 |No learner status | 0.0723

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

Removed from the highest p-value frist, therefore, ‘No learner status’ will be removed first.

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.

Shuttle Mission 1 2 3 4 5 6 7 8 9 10 11 12 Temperature 53 57 58 63 66 67 67 67 68 69 70 70 Damaged 5 1 1 1 0 0 0 0 0 0 1 0 Undamaged 1 5 5 5 6 6 6 6 6 6 5 6 Shuttle Mission 13 14 15 16 17 18 19 20 21 22 23 Temperature 70 70 72 73 75 75 76 76 78 79 81 Damaged 1 0 0 0 0 1 0 0 0 0 0 Undamaged 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 damaged O-rings.

In the data set, there are a set of outcomes for every 6 damaged or undamaged O-rings testing in constant increasing temperatures. The result shows either 6 undamaged O-rings, or 1 damaged and 5 undamaged O-rings. It could be consider as a categorial data.

  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

The statistic smmary result predicted 0.2162 number of damaged O-ring decrease for every 1 F temperature increase. The model assumed a constant 11.663 number of damaged o-ring for missing temperature.

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

\(log(p/(1 - p)) = 11.663 - 0.2162 * temperature\)

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

Yes.P-value is less than 0.05, which means we rejected there is no different in temperature to the number of damaged O-ring.

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 takeo??? 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

\(log(^p/(1 - ^p)) = 11.6630 - 0.2162 * Temperature\)

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:

^p57 = 0.341 | ^p59 = 0.251 | ^p61 = 0.179 | ^p63 = 0.124

^p65 = 0.084 | ^p67 = 0.056 | ^p69 = 0.037 | ^p71 = 0.024

\(log(^p/(1 - ^p)) = 11.6630 - 0.2162 * Temperature\) => \(p_{51} = 0.65405\) \(p_{53} = 0.550923\) \(p_{55} = 0.443246\)

  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.
probabilities<-c(0.65405,0.550923,0.443246,0.341,0.251,0.179,0.124,0.084,0.056,0.037,0.024)
temperatures<-c(51,53,55,57,59,61,63,65,67,69,71)
scatter.smooth(temperatures, probabilities, span = 2/3, degree = 1, evaluation = 50, col = 'blue')

  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.

The plot shows temperature doesn’t have colinear relation to the model-estimated probability of O-ring will become damaged.