Chapter8: 8.2, 8.4, 8.8, 8.16, 8.18

8.2 (a)Write the equation :

 #weight = 120.07 - 1.93*parity
  1. Interpret: the model estimates one unit second born baby weight to parity is 1.93 ounces lower than first born baby weight.
#First baby:  120.07 - 1.93*0=120.07
#others baby: 120.07 - 1.93*1=118.14
  1. p-value is greater than 5% (t = -1.62) which locates in 95% CI, so it’s failed to reject that there is no different between the average birth weight and parity.

8.4 (a) Write the equation : avgDayAbsent = 18.93 ???9.11???eth +3.10???sex +2.15???lrn

(b)Interpret: The model estimates that one unit not-aboriginal lower 9.11 in the average daily absent, one unit male increases 3.10 in the average daily absent, one unit slow learner increases 2.15 average daily absent.

  1. Residual of the first observation: 2=18.93 ???9.11???0 +3.10???1 +2.15???1+e -> e=-22.18

  2. x be the variance of the residuals and X is the variance of the number of absent days for all students

#R-Sqare: 1 - var(x)/var(X) = 1 - 240.57/264.17 = 8.9336%
#adjusted R-Sqare: 1 - var(x)/var(X) * (146 -1)/(146 -3 -1) = 20.9252%

8.8 Answer: remove “No learner status” first since it has the largest p-value.

8.16 (a) Examine these data damaged O-rings: There are 23 shuttle missions and one of each present number of degree for the temperature, number of damaged and undamaged O-rings. Total number of O-rings for each shuttle missions is 6. The data shows lower temperature (< 63 degree) causes more damaged O-rings.

  1. Describe the key components of the summary table The model estimates that one unit temperature decreases 0.2163 unit of damaged O-ring. This model has a standar deviation 0.0532 and z value at -4.07, that means temperature is a significan statistic factor since p-value approximates to 0.

  2. Write out the logistic model:

#log(p/(1???p)) = 11.663 ???0.2162???temperature
  1. Yes, p-value approximates to 0, we can reject that there is no differencee in temperature and number of damaged O-ring.

8.18 (a)

#log(p/(1???p)) = 11.663 ???0.2162???temperature where temperatures are 51, 53, 55
#p51=0.65405   p53=0.550923   p55=0.443246
prob<-c(0.65405,0.550923,0.443246,0.341,0.251,0.179,0.124,0.084,0.056,0.037,0.024)
temp<-c(51,53,55,57,59,61,63,65,67,69,71)
scatter.smooth(temp, prob, span = 2/3, degree = 1, evaluation = 50, col = 'red')

  1. Describe any concerns and note any assumptions to accep model’s validity Anser: The scatter plot shows non colinear reltionship between temperatures and probabilities.