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 = 1.96155

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.