8.2 a, Predicted baby weight = 120.07 - 1.93 * parity

b, Parody 1 = 120.07 - 1.93 or 118.14 / Parody 0 = 120.07 / There is a 1.93 birthweight difference between first borns and latters.

c, Given the t value of -1.62, there is not a significant relationship to the typical .05 alpha level.

8.4 a, y???=18.93???9.11???eth+3.1???sex+2.15???lrn

b, All else equal, the child can be expected to be absent 9.11 less days when not aboriginal, 3.1 more days when male, and 2.15 more days when a slow learner.

c,

predicted <- 18.93 + 3.1 + 2.15
predicted

## [1] 24.18

residual <- predicted - 2
residual

## [1] 22.18

d, R2= 1 ??? 240.57/264.17= 0.0893364

R2(adj) = 1 ??? 240.57/264.17 x 145/(145???3) = 0.070097

8.8 Answer: The adjusted R squared method would have us remove learner status. Adjusted R squared increases without it. It would also be removed with the P value method.

8.16 a, t appears there is an inverse relationship between damaged rings and temperature. The four lowest temperatures occurr when the there is at least one broken ring.

b, The intercept serves mostly to center the line. The lowest recorded temperature is 53, so a temperature of 0 isn’t very reasonable. The log odds of a ring failure will decrease by .2162 for each additional degree in farenheight. This parameter is easily statiscially signficant.

c, Log[(p^/ (1???p^)] = 11.6630???0.2162×Temperature

d, The temperature parameter is easily staistically signficant. It is also practically significant, as the high point of a 95% confidence interval still have a parameter of about -.1. With the point estimates, low temperatures indicate high probability of failure, which is what is observed in the data.

8.18 Challenger disaster, Part II.

a,

logModel = function(temperature){
  oRing  = 11.6630 - 0.2162 * temperature
  answer = 100*(exp(oRing) / (1+exp(oRing)))
  return(answer)
}

temps = c(57,65,59,67,61,69,63,71,51,53,55)
probabilites = sapply(temps,logModel)
probabilites

##  [1] 34.064976  8.393843 25.109139  5.612566 17.869707  3.715479 12.372702
##  [8]  2.443024 65.402974 55.092283 44.324565

b,

plot(y=probabilites, x=temps)
curve(logModel(x), from=40,to=80,add=TRUE,xlab='o-ring damage',ylab='temperature')

c, As explained before, as temperature rises, the technology used on incremental missions increases, but we do not account for that. It may be that temperature has NO affect on o-rings; but we wouldn’t know with this little information.

Untitled

8.2

a, Predicted baby weight = 120.07 - 1.93 * parity

b, Parody 1 = 120.07 - 1.93 or 118.14 / Parody 0 = 120.07 / There is a 1.93 birthweight difference between first borns and latters.

c, Given the t value of -1.62, there is not a significant relationship to the typical .05 alpha level.

8.4

a, y???=18.93???9.11???eth+3.1???sex+2.15???lrn

b, All else equal, the child can be expected to be absent 9.11 less days when not aboriginal, 3.1 more days when male, and 2.15 more days when a slow learner.

c,

d, R2= 1 ??? 240.57/264.17= 0.0893364

R2(adj) = 1 ??? 240.57/264.17 x 145/(145???3) = 0.070097

8.8

Answer: The adjusted R squared method would have us remove learner status. Adjusted R squared increases without it. It would also be removed with the P value method.

8.16

a, t appears there is an inverse relationship between damaged rings and temperature. The four lowest temperatures occurr when the there is at least one broken ring.

b, The intercept serves mostly to center the line. The lowest recorded temperature is 53, so a temperature of 0 isn’t very reasonable. The log odds of a ring failure will decrease by .2162 for each additional degree in farenheight. This parameter is easily statiscially signficant.

c, Log[(p^/ (1???p^)] = 11.6630???0.2162×Temperature

8.18 Challenger disaster, Part II.

a,

b,

c, As explained before, as temperature rises, the technology used on incremental missions increases, but we do not account for that. It may be that temperature has NO affect on o-rings; but we wouldn’t know with this little information.