Question

Suppose the athletic director at a local university would like to develop a regression model to predict the point differential for games played by its men's baskotbal team. A point differential is the difference between the final points scored by two competing teams. A postive differential is a win for the local university, and a negative differential is a loss. For a random sample of games, as given in the accompanything table, the point differential was calculated for the university, along with the number of assists, rebounds, tumovers, and personal fouls. Complete parts a through d below EEl Click the icon to view the data table
Game   Point_Differential   Assists   Rebounds   Turnovers   Personal_Fouls
1   15   15   38   11   9
2   36   20   43   8   13
3   16   21   29   7   13
4   45   22   46   11   11
5   12   11   40   7   22
6   -10   10   31   13   26
7   11   19   45   11   7
8   12   16   32   16   14
9   3   16   27   18   15
10   19   9   34   17   17
11   40   16   41   9   17
12   44   12   29   9   22
13   16   17   36   13   20
14   12   21   21   9   22
15   11   14   34   19   23
16   14   20   36   18   12
17   8   12   42   14   17
18   -2   8   31   8   18
19   11   15   26   10   10
20   -15   10   48   11   26
21   -3   12   20   9   17
22   5   13   28   12   18
23   16   12   26   8   18
24   -1   16   27   12   19
25   -3   6   33   12   23
26   3   13   36   20   17
27   2   7   44   13   13
28   -5   14   30   12   15
29   -13   17   23   12   21
30   -21   18   19   9   18
31   -10   4   22   8   21
32   -1   6   21   9   22
33   -4   10   33   8   19
34   23   16   39   8   22
35   38   19   43   15   22
36   1   9   42   16   26
37   6   12   38   13   17
38   12   8   45   22   19
39   18   17   27   13   26
40   19   14   39   13   15
41   9   19   42   14   24
42   8   16   39   16   25
43   -10   12   24   9   19
44   38   19   44   10   19
45   34   15   39   12   17
46   2   13   31   7   17
47   27   19   38   14   19
48   2   17   24   12   20
49   8   10   34   21   33
50   5   16   36   11   27
51   26   14   49   12   20
52   10   10   34   20   20
53   18   20   30   11   23
54   10   15   43   13   16
55   -13   11   26   17   38
56   7   11   37   18   24
57   11   18   41   15   23
58   -9   9   26   9   30
59   -5   8   30   13   26
60   25   9   33   10   17
61   -18   15   36   16   25
62   4   14   28   12   22
63   -2   16   32   15   24
64   -4   13   27   14   17
65   3   9   26   16   19
66   -7   9   31   7   18

*ANSWERS ALREADY IN THE BOXES ARE INCORRECT*

Answer 1

The step-by-step instr uctions to obtain Multiple Regression utput using Minitab18 * Store the dependent variable (y), (A), (X,), (X3) and (X ) in the columns of the Minitab 18 worksheet *Choose Stat -Regression Regression -Fit Regression Model * Select the depen dent variable (V) in the Responses box and then select the vari ables (X), (X,).(X,) and (X,) in the continuous predictors box * Click ok. The Multiple Regression Analysis Output:

SUMMARY OUTPUT Rearession Statistics MultipleR R Square Adjusted R Square Standard Error Observations ANOVA 0.970472691 0.941817244 0.895271039 2.959139201 10 MS 177.179369 20.23403 0.002739171 8.75650481 Significance F Regression Residual Total 4 708.7174759 543.78252405 752.5 Standard Error 9 Coefficients P-value Lower 95% Uper 95% Lower 90.0% r90.0% -24.4770510412.732001631.922482557 0.11257 -57.20570316 8.251601072-50.13265021 1.178548126 0.9513285660.3190790512.981482372 0.030747 0.131109753 1.771547378 0.308368843 1.594288288 1.301241566 0.225104088 5.780621662 0.00218 0.722593087 1.879890045 0.84764594 1.754837192 0.39349973 0.513313817 0.629601-0.809534411 1.2135121080.5909321430.99490984 0.27149222.537937967 0.052024 -1.386923281 0.008862556-1.236100279-0.141960446 t Stat Intercept Assist_(X1) Rebounds_(X2) Turnovers_(X3) Personal Fouls (X4) 0.201988848 0.689030363

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ct The multiple linear regression equation is -24.47705104+0.951328566X, +1.301241566 X2 0.201988848X3 -0.689030363X Step1 The null and alternative hypotheses are as follows Ho: The overall model is not significant. versus H1: The overall model is significant Step The level of significance is c 0.10 Step The formula for calculating the F-test statistic is MSR MSE The values of MSR 177.1794, MSE 8.756505 in the equation (1), we get 177.1794 8.756505 20.23402863 F 20.2340 Step4 f thep-value <a-0.10. We reject the null hypothesis Ho Otherwise, we do not reject the null hypothesis Ho Step5 The critical values are: F 3.520196 Conclusion Since, the test stati sticis F 20.23403> F-3.520196 We reject the null hypothesis H Since, the P- value -0.002739171 <a 0.10. We reject the null hypothesis Ho There is sufficient evidence that the overall model is significant variable Test stati stic P-valueT-critácał values 2.981482 0.030747 5.780622 0.00218 0.513314 0.629601 -2.53794 0.052024 2.015048 2.015048 2.015048 2.015048

Independent variable.X, Since, the test stati sti c is t-2.9815>0.105 2.015048 We reject the null hypothesis H,o Since, the P- value - 0.030747 <a 0.10. We reject the null hypothesis Ho There is sufficient evidence that the overall model is significant. The independent vari able (X,) is a significant predi ctor of the dependent variable (Y) Independent variable,X2 Since, the test stati sti c is t- 5.780622> 1010,52.015048 We reject the null hypothesis Ho Since, the P- value -0.00218 <a 0.10. We reject the null hypothesis H,o There is sufficient evidence that the overall model is significant The independent variable (X2) is a significant predictor of the dependent variable (F Independent variable-X Since, the test stati stic ist0.513314 <to ns 2.015048 We do not reject the null hypothesis H0 Since, the P- value 0.629601> 0.10 We do not reject the null hypothesis H0 There is not sufficient evidence that the overall model is significant The independent variable (X,) is not a significant predictor of the dependent variable (F Independent variable-X, Since, the test stati stic is t 2.53794 > to 105 2.015048 We do not reject the null hypothesis H0 Since, the P- value 0.052024 <a0.10 We reject the null hypothesis H,o There is sufficient evidence that the overall model is significant. The independent variable (X4) is a significant predictor of the dependent variable (Y)

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The 90% confidence interval for the coefficient of personal fouls independent variable is to From the above multiple regression output, we note that b-0.689030363,S -0.2714922,40.0,5 2.015048 The lower 90% confidence interval is 4-t0.10, 15 -0.689030363-2.015048(0.2714922) -0.689030363-2.015048(0.2714922) -0.689030363- 0.547069916 1.236100279 -1.2361 The upper 90% confidence interval is +40.105 -0.689030363+2.015048(0.2714922 -0.689030363+2.015048(0.2714922) -0.689030363+0.547069916 -0.141960446 01419 Interpretatıon We can be 90% confident that the slope coefficient of the independent variable personal fouls is between -1.2361 and - 0.1419 all other variables are held constant, one can be 90% confident that each personal foul will change the point differential by an amount between the lower and upper confi dence interval