Assignment 8
Nnaemezue Obi-Eyisi
November 25, 2017
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
- 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 means that if the child is not first born then there is a 1.93 units decrease in baby weight.
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
- Is there a statistically significant relationship between the average birth weight and parity?
when P-value is less than 0.05 we conclude that the relationship is statistically significant. but in this case p???value=0.1052; we can conclude that there is NOT a statistically significant relationship between average birth weight and 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.
- Write the equation of the regression line. \[\hat{y} = 18.93 - 9.11 \times eth + 3.10 \times sex + 2.15 \times lrn\]
- 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 NOT 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.
- 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
y <- 2
Y_hat <- 18.93 - 9.11 * eth + 3.1 * sex + 2.15 * lrn
residual <- y - Y_hat
residual## [1] -22.18The residual for the above supplied information will be: -22.18.
- 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. /
n <- 146 # Number os cases to fit the model
p <- 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-p-1) ) # Adjusted R2
R2## [1] 0.08933641adjustedR2## [1] 0.07009704\[R^2 = 0.0893364\] \[R^2_{Adjusted} = 0.070097\]
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.
ans Based on the Adjusted R2=0.0723 the lrn variable should be removed from the model first for No learner status. We chose that because that is what gave the highest increase in R2 adjusted, since a higher R2 is a better model
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.
- Each column of the table above represents a di???erent shuttle mission. Examine these data and describe what you observe with respect to the relationship between temperatures and damaged O-rings.
I observe that the higher the temperature the less the damages
- 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 slope means that for a per unit increase in Temperature, there will be a 0.2162 decrease in number of damaged o-rings. The pvalue of the temperature which is less than 0.05 shows a strong relationship and statistical significance to the damaged o-rings
- Write out the logistic model using the point estimates of the model parameters?
\[log_e(\frac{p_i}{1 - p_i}) = 11.6630 - 0.2162 \times Temperature\]
- Based on the model, do you think concerns regarding O-rings are justified? Explain Yes, they are because we can see that at low temperature 54 a lot of rings get damaged. SInce this is a critical component we should be attentive to the ambient temperatures when launching space shuttle
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.
a)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:
\[\hat{p} = \frac{e^{11.6630 - 0.2162 \times Temperature}}{1 + e^{11.6630 - 0.2162 \times Temperature}} \]
p <- function(temp)
{
damagedOring <- 11.6630 - 0.2162 * temp
phat <- exp(damagedOring) / (1 + exp(damagedOring))
return (round(phat*100,2))
}
# 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%.
c)Add the model-estimated probabilities from part (a) on the plot, then connect these dots using a smooth curve to represent the model-estimated probabilitie
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)
require(ggplot2)## Loading required package: ggplot2ShuttleMission <- data.frame(temperature, damaged, undamaged)
ggplot(ShuttleMission,aes(x=temperature,y=damaged)) + geom_point() +
stat_smooth(method = 'glm', family = 'binomial')## Warning: Ignoring unknown parameters: family
Logistic regression conditions:
There are two key conditions for fitting a logistic regression model:
Each predictor xi is linearly related to logit(pi) if all other predictors are held constant.
Each outcome Yi is independent of the other outcomes.
Based on that definition, we have assumed that those conditions are met.