bweight = 120.07 - 1.93 * parity

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.

The t-value and P value is -.162 and 0.1052 respectively. Given that the P value is greater than 0.05, we state that there is NO statistically significant relationship between the average birth weight and parity.

days_absent = 18.93 - 9.11 * eth + 3.10 * sex + 2.15 * lrn

The ethnic background slope predict that a non-aboriginal student would have a higher average number of days absent.

The sex slope predict that a male student would have a higher average number of days absent.

The learner status slope predict that a slow learner student would have a higher average number of days absent.

Absent <- 18.93 - 9.11 * (0) + 3.10 * (1) + 2.15 * (1)
Residual <- 2 - Absent
paste("Residual for this student: ", Residual)
## [1] "Residual for this student:  -22.18"
var_res <- 240.57
var_num <- 264.17
n <- 146
k <- 3

R2 <- 1 - (var_res / var_num); R2
## [1] 0.08933641

Having the highest adjusted R2 of 0.0723, learner statu (lrn) should be removed from the model first.

From visual inspection, it appears that colder temperatures had more damaged O-rings than at average or even higher temperatures. This effect starts to become noticable around the 50 degree range, particularly at 53 degrees temperature. Therefore, my preliminary conclusion is that the colder the ambient temperature, the more likely the O-rings were going to be damaged.

An intercept of 11.6630 suggests the probability of damaged O-rings at 0 temperature. The negative temperature coefficient suggests that an increase of 1 degree in temperature reduces the number of damaged O-rings by 0.2162

The logistic regression formula here is: logit(pi) = log((pi)/(1-pi)) = 11.6630 - 0.2162 * (temperature), where if the final result nears 1, this would indicate the more likelhiood of a damaged O-ring.

0Yes, ambient temperature contributes to the number of damaged O-rings and critical to the success/failure of shutlle mission launch

c_temp <- c(51,53,55)
damaged_prob <- exp(11.6630-(0.2162*c_temp))/(1+exp(11.6630-(0.2162*c_temp)))
damaged_prob
## [1] 0.6540297 0.5509228 0.4432456
temps <- seq(from = 51, to = 81)
predicted_prob <- exp(11.6630-(0.2162*temps))/(1+exp(11.6630-(0.2162*temps)))
dtemp <- as.data.frame(cbind(temps, predicted_prob))
plot(dtemp$temps, dtemp$predicted_prob,col = "red")