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I have a part of exercise solution, could someone explain me why here chi-square was used while mean and variance are unknown? I thought it can be used with known mean and unknown variance. Is it possible to use it?

CaptainBlack said: That solution shown is unrelated to the Exercise (or at most only obliquely). You should be looking at the ratio: $$r=\frac{\frac{S_1^2}{n_1 \sigma_1^2}}{\frac{S_2^2}{n_2 \sigma_2^2}}$$ which assuming I have done this right has an $F$ distribution. So compute the required confidence interval for $r$ and then convert that to one for the ratio of variances. You would then be using the results that the ratio of two $\chi^2$ random variables each divided by their number of degrees of freedom has an $F$ distribution. Click to expand... Thank you!
Leona said: I have a part of exercise solution, could someone explain me why here chi-square was used while mean and variance are unknown? I thought it can be used with known mean and unknown variance. Is it possible to use it? View attachment 44683 Click to expand... That solution shown is unrelated to the Exercise (or at most only obliquely). You should be looking at the ratio: $$r=\frac{\frac{S_1^2}{n_1 \sigma_1^2}}{\frac{S_2^2}{n_2 \sigma_2^2}}$$ which assuming I have done this right has an $F$ distribution. So compute the required confidence interval for $r$ and then convert that to one for the ratio of variances. You would then be using the results that the ratio of two $\chi^2$ random variables each divided by their number of degrees of freedom has an $F$ distribution.