SOLUTIONS:
A.
Given model:
log( p/(1-p) ) = -3.77714 + 0.14486*X , where p=P(Y=1) so for X=26.0 in model we have,
log( p/(1-p) ) = -3.77714 + (0.14486*26) or, log( p/(1-p) ) = -0.01078
or, p/(1-p) = exp(-0.01078) or, p/(1-p) = 1.010838
or, p = 1/(1+1.010838) = 0.4973051 So, P(Y=1) = 0.4973051 = 0.50(appr)
B.
As beta is the slope coeff = 0.14486, so when value of X increases by 1 unit , the value of odds ratio incrases by,
exp(0.14486) = 1.1558 = 1.16 (appr).
C.
Over the range of values the P(y=1) increases after taking X=18. Before that the probability that Y=1 was almost zero.
D.
As
log( p/(1-p) ) = -3.77714 + 0.14486*X
d( p/(1-p) ) * (1-p)/p = -3.77714 + 0.14486*X
or, d( p/(1-p) ) /dx = exp(-2.61826) * 0.14486 = 0.0099
E.
The LI effect is significant at 0.05 level as the p-value is 0.014. Moreover, the sum of squares explained by model is 34.372 higher than sum of squares due to residuals 26.073.
Pake R
###Question 1(a) Estimate the percentage of labeled cells at which the probability of remission is 0.50.
-as.numeric(coefficients(fit)[1]/coefficients(fit)[2])
## [1] 26.07384
###Question 1(b) Estimate the effect of a one-unit increase in LI on the odds of remission.
exp(coefficients(fit)[2])
## LI
## 1.155881
Question 1(c) How does the estimated probability of remission change from the lower to upperquartile values of labeling index, that is, fromLI= 13 toLI= 25?
quantile(LI)
## 0% 25% 50% 75% 100%
## 8 13 18 25 38
