Question

The director of a training program for a large insurance company has the business objective of
determining which training method is best for training underwriters. The three methods to be
evaluated are classroom, online, and courseware app. The 30 trainees are divided into three randomly
assigned groups of 10. Before the start of the training, each trainee is given a proficiency exam that
measures mathematics and computer skills. At the end of the training, all students take the same endof-
training exam. The results are organized and stored in Underwriting.
Develop a multiple regression model to predict the score on the end-of-training exam, based on the
score on the proficiency exam and the method of training used. For (a) through (j), do not include an
interaction term.
a. State the multiple regression equation.
b. Interpret the regression coefficients in (a).
c. Predict the mean end-of-training exam score for a student with a proficiency exam score of
100 who had courseware app-based training.

d. Is there a significant relationship between the end-of-training exam score and the
independent variables (proficiency score and training method) at the 0.05 level of
significance?
e. At the 0.05 level of significance, determine whether each independent variable makes a
contribution to the regression model. Indicate the most appropriate regression model for this
set of data.
f. Construct and interpret a 95% confidence interval estimate of the population slope for the
relationship between the end-of-training exam score and the proficiency exam score.
g. Construct and interpret 98% confidence interval estimates of the population slope for the
relationship between the end-of-training exam score and the online method.
h. Compute and interpret the adjusted r2.
i. Compute the coefficients of partial determination and interpret their meaning.
j. What assumption about the slope of proficiency score with end-of-training exam score do
you need to make in this problem?
k. Add interaction terms to the model and, at the 0.05 level of significance, determine whether
any interaction terms make a significant contribution to the model.
l. On the basis of the results of (e) and (k), which model is most appropriate? Explain.

End-of-Training	Proficiency	Method
14	94	classroom
19	96	classroom
17	98	classroom
38	100	classroom
40	102	classroom
26	105	classroom
41	109	classroom
28	110	classroom
36	111	classroom
66	130	classroom
38	80	online
34	84	online
43	90	online
43	97	online
61	97	online
63	112	online
93	115	online
74	118	online
76	120	online
79	120	online
55	92	courseware app
53	96	courseware app
55	99	courseware app
52	101	courseware app
35	102	courseware app
46	104	courseware app
57	107	courseware app
55	110	courseware app
42	111	courseware app
81	118	courseware app

Answer 1

Using R programming:

mydata = read.csv ("F://pdata.csv") head(mydata) #to see the first 6th rows ## End_of_Training Proficiency Method ## 1 classroom ## 2 classroom ## 3 classroom ## 4 100 classroom ## 5 102 classroom ## 6 classroom 94 96 98 105

model1 = Im(End_of_Training Proficiency+Method, data = mydata) summary(modeli) 7.138Ma ## ## Call: ## Im(formula = End_of_Training Proficiency + Method, data = mydata) ## ## Residuals: ## Min 10 Median 30 Max ## -18.980 -5.841 -2.559 7.126 19.428 ## ## Coefficients: ## Estimate Std. Errort value Pr>t) ## (Intercept) -86.2700 17.0341 -5.065 2.83e-05 ## Proficiency 1.1258 0.1589 7.087 1.59-07 *** ## Methodcourseware_app 22.2887 4.3154 5.165 2.18e-05 * ## Methodonline 30.3767 4.3230 7.027 1.84e-07 * ## ## Signif. codes: ****' 0.001 '**' 0.01 '*' 0.05 '.' 0.1'' 1 ## ## Residual standard error: 9.635 on 26 degrees of freedom ## Multiple R-squared: 0.7857, Adjusted R-squared: 0.761 ## F-statistic: 31.77 on 3 and 26 DF, p-value: 7.529e-09 * * * * * *

(a):

From above output: Our Multiple Regression equation is:

End of training = -86.2700 + 1.1258 * Proficiency + 22.2887 * MethodCourseware_app + 30.3767 * MethodOnline

(b): Interpretation:

When Proficiency increase by one unit then End of training will increase by 1.1258.

(c):

Given, Proficiency = 100

MethodCourseware_app = 1, MethodOnline = 0

End of training = -86.2700 + 1.1258 * 100+ 22.2887 * 1

End of training = 48.60

(d): All varibable are signficant because they are less than 0.05, you can see p value(Pr>|t|) of both variables in above output.

(e): Yes, Indicate the most appropriate regression model for this set of data because all variable are highly significant

(f):

confint(modeli, level = 0.95) ## ## (Intercept) ## Proficiency ## Methodcourseware app ## Methodonline 2.5 % 97.5 % -121.2839511 -51.255955 0.7992475 1.452316 13.4181779 31.159167 21.4906726 39.262766

(g):

confint(modeli, level = 0.98) ## ## (Intercept) ## Proficiency ## Methodcourseware_app ## Methodonline 1 % 99 % -128.4910656 -44.048841 0.7320353 1.519528 11.5923176 32.985027 19.6616111 41.091828

(i):

First make a model 2 withou method variable, then apply partial R square, to check which model is good.

model2 = Im(End_of_Training Proficiency, data = mydata) summary(model2) ## ## Call: ## Im(formula = End_of_Training Proficiency, data = my data) ## ## Residuals: ## Min 1Q Median 3Q Max ## -26.575 -12.241 3.669 11.616 33.273 ## Coefficients: ## Estimate std. Error t value Pr( t) ## (Intercept) -58.7754 27.8764 -2.108 0.044070 * ## Proficiency 1.0305 0.2659 3.876 0.000585 *** ## --- ## Signif. codes: '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1''1 ## ## Residual standard error: 16.18 on 28 degrees of freedom ## Multiple R-squared: 0.3492, Adjusted R-squared: 0.326 ## F-statistic: 15.02 on 1 and 28 DF, p-value: 0.0005853

library(asbio) partial. R2(model2, modell) ## [1] 0.6707124

Here we can partial R square 0.67 that less than from model1 R-sqaure 0.78, model 1 is better than model 2.

(k): model 3 with interaction term.

model3 = Im(End_of_Training-Proficiency+Proficiency: Method, data = mydata) summary(model) ## ## Call: ## Im(formula = End_of_Training ## data = mydata) Proficiency + Proficiency:Method, ## ## Residuals: ## Min 10 Median ## -18.774 -6.725 -1.749 30 8.186 Max 18.484 ## * * * * * * * * ## Coefficients: ## Estimate Std. Error t value Pr(t) ## (Intercept) -64.92633 16.97740 -3.824 0.000738 ## Proficiency 0.16219 5.713 5.18e-06 ## Proficiency: Methodcourseware app 0.20590 0.04193 4.911 4.25e-05 *** ## Proficiency: Methodonline 0.28600 0.04184 6.836 2.96e-07 *** ## --- ## Signif. codes: "***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1''1 ## ## Residual standard error: 9.83 on 26 degrees of freedom ## Multiple R-squared: 0.7769, Adjusted R-squared: 0.7512 ## F-statistic: 30.19 on 3 and 26 DF, p-value: 1.26e-08

All interaction term are significant at 0.05.

(l):  On the basis of the results of (e) and (k), model1 is most appropriate because model1(without interaction term) R square is greater than model3(with interaction term).