o. Consider the following information related to getting into graduate school. T
ID: 3229704 • Letter: O
Question
o. Consider the following information related to getting into graduate school. The dataset has a binary response (outcome, dependent) variable called admit. There are three predictor vari- ables: gre, gpa and rank. We will treat the variables gre and gpa as continuous. The variable rank takes on the values 1 through 4. Institutions with a rank of 1 have the highest prestige, while those with a rank of 4 have the lowest. mydata$rank -factor (mydata$rank) m -glm admit gre gpa rank, data mydata, family "binomial") summary (m E stimate Std. Error z value Pr( Iz (Intercept) 3.9900 1.1400 3.50 0.0005 gre 0.0023 0.0011 2.07 0.0385 gpa 0.8040 0.3318 2.42 Wo 0.0154 rank 2 0.6754 0.3165 2.13 0.0328 rank3 1.3402 0,3453 3,88 0.000 rank 4 1.5515 0.4178 3.71 0.000 Null deviance: 499.98 Residual deviance: 458.52 (a) (10 pts) Carry out likelihood-ratio test to check the model validity (clearly state your hypothesis, test statistic and conclusion)?Explanation / Answer
(a) Null Hypothesis H0: Reduced model (with only intercept) is true
Alternative Hypothesis HA: Current model (with all 5 parameters) is true
Chi-sq Test statistic = Null Deviance - Residual Deviance = 499.98 - 458.52 = 41.46
Degree of freedom = 5 (As 5 parameters are removed for the reduced model)
p-value if chi-sq stat = 41.46 with df=5 is < 0.001
So, we reject the null hypothesis and conclude that the current model is valid.
(b) With increase of 1 point in gpa, there is increase of 0.804 in the value of log(p/1-p) where p is the probability of getting into the school.
Increase in log(p/1-p) implies the increase in the probability of getting into the graduate school.
(c) For gre, estimate = 0.0023, standard error = 0.0011
For 95% confience interval, z = 1.96
Margin of error = 1.96 * 0.0011 = 0.002156
Confidence interval is (0.0023 - 0.002156, 0.0023 + 0.002156)
= (0.000144, 0.004456)
For intercept, estimate = -3.99, standard error = 1.14
For 95% confience interval, z = 1.96
Margin of error = 1.96 * 1.14 = 2.2344
Confidence interval is (-3.99 - 2.2344, -3.99 + 2.2344)
= (-6.2244, -1.7556)
(d) rank = 3, gpa = 3.5, gre = 600
The logistic regression equation is log(p/1-p) = -3.99 + 0.0023gre + 0.8040gpa -0.6754rank2 -1.3402rank3 - 1.5515rank4
log(p/1-p) = -3.99 + 0.0023*600 + 0.8040*3.5 -0.6754*0 -1.3402*1 - 1.5515*0
log(p/1-p) = -1.1362
=> p/1-p = 0.321
odds(success) = p/(1-p) = 0.321
odds(failure) = 1-p/p = 1/0.321 = 3.115
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.