T
Tree
Member
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.
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.