The Null Hypothesis - H0: There is no heteroscedasticity i.e. β_{1} = 0
The Alternative Hypothesis - H1: There is heteroscedasticity i.e. β_{1} 0
Reject H0 if |t | > t = 1.96155
MTB > Let c29=abs(c11)
MTB > let c30 = sqrt(c7)
MTB > let c31 = 1/(c7)
Regression Analysis: absolutness versus sqrttotexp, 1/totexp, sqtotexp
The regression equation is
absolutness = - 0.131 + 0.0153 sqrttotexp + 5.38 1/totexp - 0.000001 sqtotexp
Predictor Coef SE Coef T P VIF
Constant -0.13074 0.06910 -1.89 0.059
sqrttotexp 0.015309 0.005949 2.57 0.010 69.802
1/totexp 5.376 1.502 3.58 0.000 23.180
sqtotexp -0.00000105 0.00000055 -1.90 0.057 21.096
S = 0.0535194 R-Sq = 1.5% R-Sq(adj) = 1.3%
Analysis of Variance
Source DF SS MS F P
Regression 3 0.065252 0.021751 7.59 0.000
Residual Error 1498 4.290756 0.002864
Lack of Fit 26 0.062433 0.002401 0.84 0.702
Pure Error 1472 4.228323 0.002873
Total 1501 4.356008
2 rows with no replicates
Source DF Seq SS
sqrttotexp 1 0.016223
1/totexp 1 0.038664
sqtotexp 1 0.010364
Inverse Cumulative Distribution Function
Student's t distribution with 1498 DF
P( X <= x ) x
0.975 1.96155
Since the ltoexp 3.58 > 1.96155 (CV), there is evidence to suggest that H0 would be rejected indicating that there is heteroscedasticity.