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I am facing the following study:

In the 1980's, Tennessee conducted an experiment in which kindergarten students were randomly assigned to regular and small classes, and given standardized tests at the end of the year. (Regular classes contained approximately 24 students and small classes contained approximately 15 students.) Let SC denote a binary variable equal to 1 if the student is assigned to a small class and equal to 0 otherwise. The standard deviation of test scores in the sample is 75. A regression of TS (testscore) on SC yields TS =918 +13.9*CS (5)

R2 = 0.01 ; SER = 74.6

Now I am asking myself: Am I able to conclude that the error term is homo- or heteroscedastic? I would guess that at smaller test levels, they are smaller. And hence not homoscedastic.

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  • $\begingroup$ Without testing? How about looking at the plot of the model (residuals)? $\endgroup$ – user2974951 Dec 17 '18 at 9:11
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With only two groups, it would be more natural to just do a hypothesis test (but linear regression gives the same results, only presented differently). But; just looking at the numbers, 75 and 74.6, there cannot be any reason from the data alone to mistrust equal variance.

But think at the situation. Before random assignment, all the kindergartners were supposedly equal (in a statistical sense exchangeable.) The only difference is that some random process choose different groups for them, and so they were given different treatments (different class sizes). So, why should the variances differ? It must, if they differ, be because the treatment not only changed the mean, it also changed the variance. But then a change of variance in itself would be a proof of treatment effect (but maybe not the wanted effect).

So for a null hypothesis test, assuming equal variances should be OK. After all, the basis for a null hypothesis test is the null distribution. If the variances could differ under the alternative should be irrelevant (this argument is based on randomization to groups actually being performed. For an observational study, such a conclusion do not follow).

But, if you want a confidence interval, then no longer only the null hypothesis distribution is relevant, so the logic differ ...

As for your data, you should start with visualization, maybe parallel boxplots could be useful.

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