Question

NEED TO DO IN PROGRAM R

Wage EDUC EXPER AGE Male 40 39 38 53 59 36 45 37 37 43 32 40 49 43 31 45 31 37.85 21.72 14.34 21.26 24.65 71 25.65 815.45 9 20.39 10 29.13 11 27.33 12 18.02 1320.39 15 12 1 0 12 14 18 1424.18 1517.29 16 15.61 1 10 17 35.07 18 1920.39 20 21 40.33 14 16.61 16.33 30 28 Ch17 009 Data File ype here to search 1 0 1010100101010000110 0288072 703131755760 534335345536 5 3 934605F 644 566 4669574644659 1952 6340 6453991 .9 .3 .4 4838 113 988 7 696 211 8 2 2111 2 0123 3,333 0-78 10010110100 0548 6_3-322444456 32762 05041027251 44956794148 91 56754 60 6834 736 2 3634 2122 2 /9012 345678 123 4-4-4 567 4-4.4 4:55-5-5 64 Click Q☆ Read the Excel file into R > mydata fitlogit summary(frogeの which of the two variables is significant (useful n te model) at a 10% significance level? Cick to select) O Type here to search の >fitlogit predict glm(fitiogit, type="response.. data frame(X14, X2 X3e# X4")) c. What is the p-valkue of the predictor variable that is most useful in the logistic regression mode? (Round your answer to 4 decimal places) d. Run another logistic regression model, this time using only Wage and EDUC to predict the probability that an employee is male Х1 + X2 data:mydata, family-binomíal(link#7ogir) summarythtogicz) which of the two variables is significant (useful in the model) at a 10% significance | (Cick to select) e. Use the logistic regresion model from part (d) to predicdt the probability that an employee s maleifthe employee eams $30 per hour and has 8 years of tigher education > predict glmift ogt2 type*Tesponse. dato frame(X1zB X2a#1) alllch:4.3 pata,F Ch17.009 pata,F O Type here to search

Answer 1

a) Run the following code:

fitlogit <- glm(Male~Wage+EDUC+EXPER+AGE, data=mydata, family=binomial(link="logit"))
summary(fitlogit)

Results are:

Coefficients:
Estimate Std. Error z value
(Intercept) -5.18674 2.09966 -2.470
Wage 0.11319 0.07883 1.436
EDUC 0.32020 0.17132 1.869
EXPER 0.02064 0.06772 0.305
AGE 0.01471 0.03478 0.423
Pr(>|z|)
(Intercept) 0.0135 *
Wage 0.1510
EDUC 0.0616 .
EXPER 0.7605
AGE 0.6724

Hence, Logit(Male) = -5.187 + 0.113(Wage) + 0.320(EDUC) + 0.020(EXPER) + 0.015(AGE)

b) Run the following code:

predict.glm(fitlogit,type="response",data.frame(Wage=30, EDUC = 8, EXPER = 20, AGE = 50))

Result: 0.872

c) Most important predictor from the above table (which has the lowest p-value) = 0.0616

d) Run the following code:

fitlogit2 <- glm(Male~Wage+EDUC, data=mydata, family=binomial(link="logit"))
summary(fitlogit2)

Results:

Coefficients:
Estimate Std. Error
(Intercept) -4.57524 1.70025
Wage 0.13429 0.07161
EDUC 0.28704 0.15712
z value Pr(>|z|)   
(Intercept) -2.691 0.00713 **
Wage 1.875 0.06073 .
EDUC 1.827 0.06772 .

Both the variables are significant at 10% significance level as p-value of both < 0.1

e) Run the following code:

predict.glm(fitlogit2,type="response",data.frame(Wage=30,EDUC=8))

Answer: 0.852