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I have a linear model $y=ax+b$. I have fit the relations using two data sets. I have found that estimate of $a$ from the two data sets are similar, but that the uncertainty of $a$ is much larger in one of the data sets. I am using the $\chi^2$ test of variance to show this. The test statistic is $$T=(N-1)(s/\sigma_0)^2$$ where $N$ is the number of points, $s$ is the standard deviation of $a$ from one of the data sets, and $\sigma^2$ is the standard deviation of $a$ from the other data set. I calculate $T$ and compare it to the $\chi^2$ distribution and find a p-value of $1e\mathrm{-}5$.

However, I know that $y=ax+b$ is heteroskedastic. Does this impact the test?

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    $\begingroup$ Almost certainly. Are these datasets from the same units (eg, patients), or are they unmatched? The $\sigma_0$ in your formula is supposed to be an s-priori value, not one estimated w/ error. This isn't likely the best way to test this. Can you say more about your study, your data, & your goals? $\endgroup$
    – gung - Reinstate Monica
    Commented Jan 3, 2017 at 20:42
  • $\begingroup$ @gung I think the answer is that it tests for homoskedasticity rather than rely. To say it relies on it means that homoskedasticity is assumed. That woulld be nonsensical. $\endgroup$
    – Michael R. Chernick
    Commented Jan 3, 2017 at 21:44
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    $\begingroup$ @rhombidodecahedron, hmmm. This seems like a different situation that I thought you were describing. Can you say more about your situation, the data, the models, & your goals? What are the variables, eg? Are they correlated? Can you paste in the model output that shows what you are referring to? If you added another variable, the residuals shouldn't be able to shrink, but there could be some collinearity or some other issue. $\endgroup$
    – gung - Reinstate Monica
    Commented Jan 3, 2017 at 22:19
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    $\begingroup$ You appear to have misapplied that test. It compares a sample variance to a known variance. It does not compare two sample variances. $\endgroup$
    – whuber ♦
    Commented Jan 4, 2017 at 5:30
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    $\begingroup$ You need to explain what you mean by "homoscedasticity," because it could have several distinct meanings in this context. After all, the F test is testing for one kind of heteroscedasticity! $\endgroup$
    – whuber ♦
    Commented Jan 4, 2017 at 17:40

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