Dr. Jayanthi Sankaran
Well-Known Member
Hi David,
I am having problems in computing the F statistic for the above:
2. Following is an estimated regression, with includes standard errors:
Price (caret) = 119.2 + 0.485 BDR + 23.4 Bath + 0.156 Hsize + 0.002 Lsize - 0.090Age - 48.8 Poor
(23.9) (2.61) (8.94) (0.011) (0.00048) (0.311) (10.5)
R square = 0.72, SER = 41.5
Note that the above regression has six regressors. The F-statistic for omitting BDR and Age from the regression is 3.31. The following four critical values are provided to you: critical F(2df, infinite @5%) = 3.00; F(2df, infinite @1%) = 4.61; F(6df, infinite@5%) = 2.01, F(6df,infinite@1%) = 2.64. Are the coefficients BDR and Age statistically different from zero, at respectively, the 5% and 1% level?
My answer is:
SER = SQRT((SSRunrestricted)/(n - k - 1))
SER^2 = SSRunrestricted/(n - k - 1)
41.5^2 * (213) =SSRunrestricted = 366,839.25
SER^2 = SSRrestricted/217
41.5^2 = SSRrestricted/217
SSRrestricted = 41.5^2 * 217 = 373,728.25
Using the forumula for the F statistic
F = (SSRrestricted - SSRunrestricted)/q
Your answer for the F statistic is 3.31. Can you please explain how you get this number?
However, I must add that my calculated figure of 4.0 leads to the same conclusion as yours:
Although the regression has six regressors, we are restricting only two such that the correct critical values are F(2df, infinite)
Thanks!
Jayanthi
I am having problems in computing the F statistic for the above:
2. Following is an estimated regression, with includes standard errors:
Price (caret) = 119.2 + 0.485 BDR + 23.4 Bath + 0.156 Hsize + 0.002 Lsize - 0.090Age - 48.8 Poor
(23.9) (2.61) (8.94) (0.011) (0.00048) (0.311) (10.5)
R square = 0.72, SER = 41.5
Note that the above regression has six regressors. The F-statistic for omitting BDR and Age from the regression is 3.31. The following four critical values are provided to you: critical F(2df, infinite @5%) = 3.00; F(2df, infinite @1%) = 4.61; F(6df, infinite@5%) = 2.01, F(6df,infinite@1%) = 2.64. Are the coefficients BDR and Age statistically different from zero, at respectively, the 5% and 1% level?
My answer is:
SER = SQRT((SSRunrestricted)/(n - k - 1))
SER^2 = SSRunrestricted/(n - k - 1)
41.5^2 * (213) =SSRunrestricted = 366,839.25
SER^2 = SSRrestricted/217
41.5^2 = SSRrestricted/217
SSRrestricted = 41.5^2 * 217 = 373,728.25
Using the forumula for the F statistic
F = (SSRrestricted - SSRunrestricted)/q
SSRunrestricted /n - kunrestricted - 1)
= (373,728.25 - 366,839.25)/1.0 = 4.00
= (373,728.25 - 366,839.25)/1.0 = 4.00
366,839.25/213
Your answer for the F statistic is 3.31. Can you please explain how you get this number?
However, I must add that my calculated figure of 4.0 leads to the same conclusion as yours:
Although the regression has six regressors, we are restricting only two such that the correct critical values are F(2df, infinite)
- As computed F-statistic of 4.00 is GREATER THAN critical F(2df, infinite)@5% = 3.00; we reject the null; that is they are significant with 95% confidence
- As computed F-statistic of 4.00 is LESS than F(2df, infinite)@1% = 4.61, we fail to reject null; ie they are not significant with 99% confidence.
Thanks!
Jayanthi
Last edited: