Question

ApyhoLoaist Conductea) A study o stability (X) and The employees b pe,formaKV) and B stit e The Fite) Response fiaction An employee's emaioha Maxium ik estmas of B D) what is兀estiwatid pelability That emplyees will perform2 Bo and d) use仄likelihaJ Rat tutto determine onterroni obtain a 90% Ghfidene interval fo-

Answer 1

a) A binary logit model is denoted as:

$log\left ( \frac{p}{1-p} \right )=\beta_0+\beta_1X$

where p = P(Y=1) = P(employee will perform) and $\beta_0 \& \beta_1$ are the parameters.

From the tabulated results, it can be clearly seen that the Maximum Likelihood Estimates (MLEs) of the parameters $\beta_0 \& \beta_1$ are -10.3089 & 0.0189 respectively. Hence, the fitted response function can be written as:

$log\left ( \frac{p}{1-p} \right )=-10.3089+0.0189X$

b) For X = 550, the probability that the employee will perform is obtained using the formula:

$p=\frac{e^{\beta_0+\beta_1X}}{1+e^{\beta_0+\beta_1X}}=0.5215$

c) The joint 95% CI for $\beta_0 \& \beta_1$ using Bonferroni method can be written as:

$\hat\beta_0\pm B*S.E.(\hat\beta_0) \hspace{.1cm} and \hspace{.1cm} \hat\beta_1\pm B*S.E.(\hat\beta_1)$ respectively,

where $B=z_{(1-\alpha/2g)}=z_{(1-\alpha/4)}$ , g being the number of predictions.

d) For this question, $H_0:eta_0=0$ i.e. the intercept can be dropped.

To test this hypothesis using the Likelihood Ratio (LR) test, it is required to compute log L for the reduced model with 'covariates only', which is not given in the tabulated results.

However, the p-value for testing the significance of the above hypothesis using Wald's test is given, which is 0.0185 (refer to table 'analysis of MLEs'). Since this value is 0.05 (level of significance), Ho may be rejected, i.e. the intercept cannot be dropped.

e) The employee's job performance is denoted by Y, which takes value 1 with probability p and 0 with probability (1-p) such that,

The 90% confidence interval for Y is thus given by .

For X=550, p = 0.5215. Putting these values in the formula, the 90% CI is obtained as (-0.3002,1.342).