Problem 8.2 - Baby weights, Part II

baby weight = 120.07 - 1.93(parity)

  1. Slope is the weight estimate of first born babies

Born first = 120.07 - 1.93(0) => 120.07 otherwise = 120.07 - 1.93(1) => 118.14

  1. p-value is more than 0.05 and hence the relationship is insignificant between weight and parity.

Problem 8.4

  1. absenteeism = 18.93 - 9.11 (eth) + 3.10(sex) + 2.15(lrn)

eth: It says that avg number of absent days by non-aboriginal students is 9.11 days lower as compared with aboriginal students. sex: As per the number, average number of absent days by mail students is 3.10 days higher than female students. lrn: Average numer of absent days by slow learner students is 2.15 days higher than the average learners.

  1. Y1 = 18.93 - 9.11(0) + 3.10(1) + 2.15(1) y1= 24.18

e= y1 - y1(cap) = 2 - 24.18 e = -22.18

var_resids <- 240.57
var_absents <- 264.17
n <- 146
k <- 3

r2 <- 1 - (var_resids/var_absents)
r2
## [1] 0.08933641
adj.r2 <- 1 - (var_resids/var_absents)*((n-1)/(n-k-1))
adj.r2
## [1] 0.07009704

Problem 8.8

No learner status should be removed from the model first as it has highest value.

Problem 8.16

  1. Visually, it shows that as the temperature goes up, there are less chances of damages and vice versa. First mission had most of the damages i.e. 5 at 53 degrees.
  2. With the 1 degree Fahrenheit in temperature, chances of being damages will go down by 0.2162.
  3. logit(pi) = log(pi)/(1-pi) = 11.6630 - 0.2162(Temperature)
  4. Yes, p-value is less than 0.05, hence it has significant impact on the damages.

Problem 8.18

temp51 <- 51
log51 <- 11.6630 - 0.2162*temp51
failprob_51 <- exp(log51)/(1+exp(log51))
failprob_51
## [1] 0.6540297
temp53 <- 53
log53 <- 11.6630 - 0.2162*temp53
failprob_53 <- exp(log53)/(1+exp(log53))
failprob_53
## [1] 0.5509228
temp55 <- 55
log55 <- 11.6630 - 0.2162*temp55
failprob_55 <- exp(log55)/(1+exp(log55))
failprob_55
## [1] 0.4432456
temp51_81 <- seq(from =51, to = 81)
prediction <- exp(11.630-(0.2162*temp51_81))/(1+exp(11.630-(0.2162*temp51_81)))
dftemp51_81 <- as.data.frame(cbind(temp51_81, prediction))
plot(dftemp51_81$temp51_81, dftemp51_81$prediction, col="blue")

  1. The assumptions are not confidently said to be met. Although all the observations seem to be independent but since the manufacturers were using previous data for remanufacturing, it could actually affect their next processes. Although the predictor seem to have significant relationship with outcome.