Amodel to explain the standarded outcase on a final exam (stndfril) in terms of percentage of classes attended (antrte), prior college grade point average (priGPA), and university entrance exam (ACT) score is stndfnl = Bo + B₁antrte+ B₂priGPA + B3priGPA² + P4ACT +BACT² + BopriGPA antrt Using the 680 observations for students in a course, the estimated equations are Model 1 Model 2 Model 3 0.2729 (0.6239) intercept 2.0502 (1.3603) 0.0296 (0.0379) *
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3- Amodel to explain the standarded outc are on a final exam (strdfril) in terms of percentage of classes
attended (antrte), prior college grade point average (priGPA), and university entrance exam (ACT) score is
stndfnl = Bo + B₁antrte+ P₂priGPA + PapriGPA²+ B4ACT +BACT² + BopriGPA * antrte
Using the 680 observations for students in a course, the estimated equations are
Model 1 Model 2
Model 3
2.0502
0.2729
(1.3603) (0.6239)
-0.0067
(0.0102)
--1.6285
intercept
C-
antrte
priGPA
priGPA^2
ACT
ACT^2
priGPA*antrte
-1.3436
(0.4810) (0.4650)
0.2959
0.3519
(0.1010)
(0.0885)
-0.1280)
(0.0984)
0.0045
0.0014
(0.0021) (0.0002)
0.0055
0.0296
(0.0379)
(0.0043)
R^2
0.2286
0.2170
0.0000
SSR
512.7624 520.4797 664.7634
Is the whole model 1 significant at a 5% significance level? Write the necessary hypotheses.
a-
b- How can you conclude, excluding or keeping the variables, antrte, ACT, and the interaction term
priGPA* antrte is good or not good for model 1 at a 5% significance level?
Show that the partial effect of attendance rate on final scores also relies on priGPA.
Id- Calculate the partial effect of attendance rate on final scores when the avarage priGPA is 2.59.
Interprete the result you have calculated.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf9316f6-0126-470d-bf73-3eb1f8220c4c%2F6afa4964-13e7-40cc-b0ca-2104b8cea2e2%2F89ovau_processed.jpeg&w=3840&q=75)
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Hello! As you have posted more than 3 sub parts, we are answering the first 3 sub-parts. In case you require the unanswered parts also, kindly re-post that parts separately.
a.
Null Hypothesis:
H0: The model 1 is not significant.
Alternative Hypothesis"
H1: The model 1 is significant.
Test Statistic:
From the given information,
Therefore,
P value:
df=(k, n-k-1)=(3, 680-3-1)=(3, 676).
P=0.0000, obtained from the excel function, =F.DIST.RT(69.2727,3,676).
Decision Rule:
If p-value ≤ α, then reject the null hypothesis.
Conclusion:
Let the level of significance is α=0.05
Here, the p-value is less than the level of significance.
From the decision rule, reject the null hypothesis.
It can be concluded that “The model 1 is significant”.
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