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(- e s://owl.uwo.ca/portal/site/afa98a5e-b4d0-4684-9286-51192800680/tool/c2ba5653-c550-4980-b6b4-e2b80e2ba53 Part 12 of 20-Q12 Question 12 of 20 1.0 Points 1 . reg EARNINGS EXP Source I Model 1575.59016157.59016 Total I Number of obs 540 F( 1, 538)= * Prob > F R-squared =0.0088 Adj R-squared e.0069 Root MSE Residual 5 : 0.0296 333.485069 18 . 198 EARNINGS! Coef. Std. Err, t p>It] [ 95% Conf. Interval] EXP .3807076 cons I 2.18 .0300378405 7235747 3.066503464 0.0008.202738 20.25031 The estimated variance of the regression model is: Mark for Review Part 13 of 20-Q13 Question 13 of 20 1.0 Points

Explanation / Answer

Yes, so whatever is given in the table, will be enough to take out the remaining figures:

The F(1,538) is actually has df1 and df2 provided.
So, dfresidual = 538
Rsquare = SSmodel/SStotal = 1575.59016/(1575.59016+SSmodel) = .0088
SSmodel = 177468.7

So, SStotal = SSmodel+SSresidual = 179044.3
dftotal = dfmodel+dfresidual = 1+538 = 539

_cons' coeffieicnt = ?
We know that : Coeff/SE = t
Coeff/3.066503 = 4.64
Coefficient of _cons =3.066503*4.64 = 14.2286

Similarly, for EXP,
Coefficient of _EXP / SE for EXP = 2.18
.3807076/SE for EXP= 2.18
SE for EXP = .3807076/2.18 = .1746

F(1,538) = 3.8588 at alpha = .05 for df1 = 1, df2 = 538

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