URGENT PLEASE HELP!
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 end-of-training exam. 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. Complete parts a through f below.
THE BLANKS ARE JUST "INCREASE" OR "DECREASE"
(a)
y | X1 | X2 | X3 |
14 | 94 | 1 | 0 |
18 | 96 | 1 | 0 |
18 | 97 | 1 | 0 |
39 | 101 | 1 | 0 |
40 | 101 | 1 | 0 |
26 | 105 | 1 | 0 |
41 | 110 | 1 | 0 |
28 | 110 | 1 | 0 |
36 | 111 | 1 | 0 |
65 | 130 | 1 | 0 |
37 | 79 | 0 | 1 |
35 | 85 | 0 | 1 |
43 | 90 | 0 | 1 |
44 | 97 | 0 | 1 |
61 | 97 | 0 | 1 |
62 | 112 | 0 | 1 |
93 | 115 | 0 | 1 |
74 | 117 | 0 | 1 |
76 | 120 | 0 | 1 |
79 | 119 | 0 | 1 |
55 | 92 | 0 | 0 |
53 | 95 | 0 | 0 |
54 | 100 | 0 | 0 |
52 | 102 | 0 | 0 |
34 | 102 | 0 | 0 |
47 | 104 | 0 | 0 |
57 | 106 | 0 | 0 |
55 | 109 | 0 | 0 |
41 | 111 | 0 | 0 |
80 | 117 | 0 | 0 |
(b) using software the regression alalysis showed following information
Analysis of Variance | |||||
Source | DF | Sum of | Mean | F Value | Pr > F |
Squares | Square | ||||
Model | 3 | 8689.38517 | 2896.46 | 30.91 | <.0001 |
Error | 26 | 2435.98149 | 93.6916 | ||
Corrected Total | 29 | 11125 | |||
Root MSE | 9.67944 | R-Square | 0.781 | ||
Dependent Mean | 48.56667 | Adj R-Sq | 0.7558 | ||
Coeff Var | 19.93022 | ||||
Parameter Estimates | |||||
Variable | DF | Parameter | Standard | t Value | Pr > |t| |
Estimate | Error | ||||
Intercept | 1 | -63.47383 | 17.0023 | -3.73 | 0.0009 |
X1 | 1 | 1.12017 | 0.16112 | 6.95 | <.0001 |
X2 | 1 | -22.20429 | 4.33744 | -5.12 | <.0001 |
X3 | 1 | 8.38412 | 4.33025 | 1.94 | 0.0638 |
regression equation y=-63.47+1.12X1-22.2X2+8.38X3
(b and c) X1 has significant positive impact and X2 has significant negative impact, X3 has non-significant positive impact at 5% level of significance
(d) (1-alpha)*100% confidence interval for beta=beta±t(alpha/2,error df)*SE(beta)
95% confidence interval for X1=1.12±t(0.05/2, 26)*0.1611=1.12±2.06*0.1611=1.12±0.33=(0.79,1.55)
95% confidence interval for X2=-22.2±t(0.05/2, 26)*4.34=-22.2±2.06*4.34=
95% confidence interval for X3=8.38±t(0.05/2, 26)*4.33=1.12±2.06*4.33
(e) after adding the interaction term the following information is generated, and none of the interaction term is significant at alpha=0.05 so adding interaction term is not required
Analysis of Variance | |||||
Source | DF | Sum of | Mean | F Value | Pr > F |
Squares | Square | ||||
Model | 5 | 8847.69985 | 1769.54 | 18.65 | <.0001 |
Error | 24 | 2277.66682 | 94.90278 | ||
Corrected Total | 29 | 11125 | |||
Root MSE | 9.74181 | R-Square | 0.7953 | ||
Dependent Mean | 48.56667 | Adj R-Sq | 0.7526 | ||
Coeff Var | 20.05863 | ||||
Parameter Estimates | |||||
Variable | DF | Parameter | Standard | t Value | Pr > |t| |
Estimate | Error | ||||
Intercept | 1 | -9.32082 | 45.52686 | -0.2 | 0.8395 |
X1 | 1 | 0.59847 | 0.4376 | 1.37 | 0.1841 |
X2 | 1 | -88.94048 | 55.9612 | -1.59 | 0.1251 |
X3 | 1 | -52.5468 | 50.60616 | -1.04 | 0.3095 |
X12 | 1 | 0.64098 | 0.53458 | 1.2 | 0.2422 |
X13 | 1 | 0.58745 | 0.48635 | 1.21 | 0.2389 |
X12 is interaction of X1 and X2
X13 iis interaction of X1 and X2
( f) first model is appropriate i.e. without interaction model
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URGENT PLEASE HELP! 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 method...
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