Question

White's test for Ho: homoskedasticity
against Ha: unrestricted heteroskedasticity

chi2(45) = 68.26
Prob > chi2 = 0.0142

Cameron & Trivedi's decomposition of IM-test

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Source | chi2 df p
---------------------+-----------------------------
Heteroskedasticity | 68.26 45 0.0142
Skewness | 21.51 9 0.0106
Kurtosis | 2.56 1 0.1095
---------------------+-----------------------------
Total | 92.34 55 0.0012

is heteroscedasticity present this data? why?
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Answer 1

The actual p-value of chi-square = 0.0142

At 5% level of significance, the critical p-value is 0.05

Since, the actual p-value is less than critical p-value of 0.05, null-hypothesis is rejected and alternate hypothesis is not rejected.

Thus, at 5% level of significance, one would conclude that the available data indicate the presence of unrestricted heteroscedasticity.