SUMMARY OUTPUT |
||||||
Regression Statistics |
||||||
Multiple R |
0.9011 |
|||||
R Square |
0.8119 |
|||||
Adjusted R Square |
0.7884 |
|||||
Standard Error |
0.1786 |
|||||
Observations |
10 |
|||||
ANOVA |
||||||
df |
SS |
MS |
F |
Significance F |
||
Regression |
1 |
1.10 |
1.10 |
34.53 |
0.00 |
|
Residual |
8 |
0.26 |
0.03 |
|||
Total |
9 |
1.36 |
||||
Coefficients |
Standard Error |
t Stat |
P-value |
Lower 95% |
Upper 95% |
|
Intercept |
0.4676 |
0.4718 |
0.9912 |
0.3506 |
-0.6203 |
1.5555 |
ACT |
0.1079 |
0.0184 |
5.8765 |
0.0004 |
0.0656 |
0.1503 |
a) GPA = 0.4676+0.1079*ACT
b) As the value of the intercept is low and the p-value is less than 0.05, it is not significant and a chance factor
c) GPA predicted increases by 0.5395=(0.1079*5)
d) Yes, the residuals sum to zero
Student |
GPA |
ACT |
Predicted |
Residuals |
1 |
2.8 |
21 |
2.7 |
-0.1 |
2 |
3.4 |
24 |
3.1 |
-0.3 |
3 |
3 |
26 |
3.3 |
0.3 |
4 |
3.5 |
27 |
3.4 |
-0.1 |
5 |
3.6 |
29 |
3.6 |
0.0 |
6 |
3 |
25 |
3.2 |
0.2 |
7 |
3.2 |
25 |
3.2 |
0.0 |
8 |
3.5 |
28 |
3.5 |
0.0 |
9 |
2.5 |
20 |
2.6 |
0.1 |
10 |
3.7 |
30 |
3.7 |
0.0 |
Sum |
0.1 |
e) GPA =0.4676+0.1079*20 = 2.63. yes as predicted value is close to actual value
f) We need to use R square value to understand the variation in GPA. As the value is 0.8119, 81.19 percent of variation in GPA is explained by ACT.
4. [35 points] The following table contains the ACT scores and the GPA for ten college...
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