Homework Answers
a) The correct degrees of freedom of the Wald statistic is one.
b) Now 1% quantile(upper) of a chi square random variable with DF unity is 6.634897. Thus Wald statistic is less than the 1% upper quantile of a chi square random variable with DF unity.
Actually, Wald test rejects the null if Wald statistic exceeds 1% quantile(upper) of a chi square random variable with DF unity.
Here the Wald statistic does not exceed the 1% upper quantile of a chi square random variable with DF unity and hence, we fail to reject the null hypothesis.
For query in above, comment.
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The following ANOVA model is for a multiple regression model
with two independent variables:
Degrees
of
Sum
of
Mean
Source
Freedom
Squares
Squares
F
Regression
2
60
Error
18
120
Total
20
180
Determine the Regression Mean Square (MSR):
Determine the Mean Square Error (MSE):
Compute the overall Fstat test statistic.
Is the Fstat significant at the 0.05 level?
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2
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Error
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Total
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