The Null Hypothesis - H0: There is no heteroscedasticity i.e. β1 = 0
The Alternative Hypothesis - H1: There is heteroscedasticity i.e. β1 0
Reject H0 if |t | > t = 1.96153
MTB > let c44 = abs (c11)
MTB > let c45 = sqrt (c7)
MTB > let c46 = 1/c7
C44 = absolutness
C45 = sqrttoexp
C46 = 1/totexp
C11 = RESI1
C7 = totexp
Regression Analysis: absolutness versus sqrttotexp, 1/totexp, sqtotexp
The regression equation is
absolutness = - 0.0432 + 0.00697 sqrttotexp + 3.96 ltotexp - 0.000000 sqtotexp
Predictor Coef SE Coef T P
Constant -0.04319 0.06231 -0.69 0.488
sqrttotexp 0.006965 0.005250 1.33 0.185
1/totexp 3.962 1.403 2.82 0.005
sqtotexp -0.00000003 0.00000045 -0.07 0.945
S = 0.0553218 R-Sq = 1.8% R-Sq(adj) = 1.6%
Analysis of Variance
Source DF SS MS F P
Regression 3 0.083167 0.027722 9.06 0.000
Residual Error 1515 4.636654 0.003060
Total 1518 4.719821
Source DF Seq SS
sqrttotexp 1 0.011248
l/totexp 1 0.071905
sqtotexp 1 0.000015
Since the ltoexp 2.82 > 1.96153 (CV), there is evidence to suggest that H0 would be rejected indicating that there is heteroscedasticity.