42. Recall that in exercise 39, the admissions officer for Clearwater College de
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Question
42. Recall that in exercise 39, the admissions officer for Clearwater College developed the following estimated regression equation relating final college GPA to the student’s SAT mathematics score and high-school GPA.
yˆ = -1.41 + .0235x1 + .00486x2
where
x1 = high-school GPA
x2 = SAT mathematics score
y = final college GPA
A portion of the Minitab computer output follows.
a. Complete the missing entries in this output.( please make the numbers clear for me to plug in)
b. Use the F test and a .05 level of significance to see whether a significant relationship is present.
c. Use the t test and = .05 to test H0: f31 = 0 and H0: f32 = 0.
d. Did the estimated regression equation provide a good fit to the data? Explain.
The regression equati on is 1.41 SE Coef Predict01 Coef Constant 1.4053 R-sq(adi) R sq Analysis of Variance SOURCE Regression 1.76203 Residual EMT or TotalExplanation / Answer
a) From the given information n-1=9
n=10
So MSR=0.8810
MSE =0.0168
Hence F= MSR/MSE
=0.8810/0.0168
=52.44
It is clear that F~F=(2,7)
From the F distribution P value=0.000061
Thus the ANOVA table as follows
soure of
variation
degrees
freedom
sum
of squires
mea sum
of squires
significance
F
For the value of intercept t=-1.4053/0.4848
=-2.8987
The value of P using t (-2.8987) =0.017624
For the value of X1
t=0.0235/0.0087
=2.7011
The value of P using t( 2.7011) =0.02435
For the value of X2
t=0.0049/0.0011
=4.4545
The value of P using t(4.4545) =0.00159
b)State the hypothesis
From the above test statistic value is 52.44 and corresponding P value is =0.000061
Use the level of significance =0.05
0.000061<( =0.05)
Here P value is less than the significance value , reject the null hypothesis,
i.e. there is a significant difference i the given model
c) State the Hypothesis
Null hypothesis
Ho:1=0
. There is no significant difference in the given model
Alternative Hypothesis
HA:10 .
there is a significant difference in the given model
Rejection Rule
P-value < (=0.05), the reject is null hypothesis
From the given output the test statistic is 2.7011and the corresponding p-value is 0.02435
Here the P-value (0.02435) <( =0.05)
Then reject the null hypothesis
i.e. there is a significant difference i the given model
d) For the above information we can clearly say that the estimated regression equation provides a good fit for the data
soure of
variation
degrees
freedom
sum
of squires
mea sum
of squires
Fsignificance
F
Regression 2 1.7621 0.8810 52.44 .0.000061 Residuel 7 0.1179 0.0168 Total 9 1.88Related Questions
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