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.