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Question:Suppose you fit the first-order model y=β0+β1x1+β2x2+β3x3+β4x4+β5x5+εto n=30 data points and obtain SSE = 0.33 and R2=0.92

(A) Do the values of SSE and R2suggest that the model provides a good fit to the data? Explain.

(B) Is the model of any use in predicting Y ? Test the null hypothesis H0:β1=β2=β3=β4=β5=0 against the alternative hypothesis {H}at least one of the parameters β1,β2,...,β5 is non zero.Useα=0.05 .

Short Answer

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(A) Low value of SSE and high value of R2 indicate that the model is a good fit to the data.

(B) At 95% confidence interval, it can be concluded that Hα:β1=β2=β3=β4=β5=0

Step by step solution

01

Step-by-Step Solution Step 1: Good fit to the data

A low sum of squares of error values like 0.33 indicate that the there is low variability in the set of observations. It means that the data points are close to the model fitted. Also, high value like 0.92 forR2indicates that around 92% of the variation in the data is explained by the model which is a good thing.

Hence, we can say that SSE = 0.33 andis a good measure of good fit to the data.

02

Overall significance of the model

H0:β1=β2=β3=β4=β5=0

Ha:At least one of the parameters β1,β2,...,β5is non zero

Here, F test statistic =SSEn-(k+1)=0.3330-6=0.01375

Value ofF0.05,24,24is 1.984

H0is rejected if F statistic > F0.05,24,24For α=0.05, since F <F0.05,24,24

Do not have sufficient evidence to reject H0 at 95% confidence interval.

Therefore,β1=β2=β3=β4=β5=0

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