There's one thing I don't quite understand about part iii of this question. Parts i and ii are just an ANOVA test, and then part iii asks us to calculate a confidence interval for the common standard deviation among the 5 students assuming that the null hypothesis holds true. I would have thought that, assuming that the null hypothesis is true, we would then just treat the scores of the 5 students as one single sample as if they had come from one student. Therefore we would estimate our common variance by: SS_T/(n-1) rather than SS_R/(n-k) However the solutions use SS_R/(n-k) which gives a much smaller confidence interval than my approach. Can anyone see the flaw in my logic? Because right now I'm sure that if I got a question like this in an exam I would go for SS_T/(n-1) just because it makes intuitive sense to me to do it that way. The way I see it is that if we accept the null hypothesis then we are no longer treating the 5 samples separately anymore.
In null hypothesis, we state that there is no difference between sample means..which means that all sample means are equal and came from sam population. But that doesn't mean that all 5 samples are 1 n the same thing. As u know, infinite samples can be drawn from large population. And SSr/(n-k) is estimate of common variance (which is an assumption for ANOVA). So thats why we hav used SSr and not SSt
