What is the pivotal quantity/ formula to calculate the confidence interval of the Common standard deviation of samples from two different populations
Refer tables page number 22 & 23 If you meant for pooled standard deviation, then use t-distribution Otherwise , chi-square distribution for two identical population.
you just modify the standard variance formula: \(\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}\) to: \(\frac{(n_1 + n_2 -2) S_P^2}{\sigma^2}\sim\chi^2_{n_1 + n_2 -2}\)
I don't think you'll be asked to do this. Confidence intervals (CI) for the population variance can be asked in these ways: (1) given a single sample, use sample variance to find CI for population variance (use chi-square) (2) given two samples, use ratio of sample variances to find CI for the ratio of population variances (use F) (3) in a regression question, use estimate of error variance (SS(RES)/n-2) to find a CI for the error variance (use chi-quare with n-2 degrees of freedom) (4) in an ANOVA question, use estimate of error variance (SS(RES)/n-k) to find a CI for the error variance (use chi-square with n-k degrees of freedom).
Thanks, Please provide the C.I. formula. I have tried...but unable to modify. Such questions are very rarely asked!
Ok, Thanks, but that is for Single sample. The question asks to find C.I for common S.D. of two different independent samples.
The limits for a 95% will be: [(n1+n2-2)Sp^2]/chi-square (0.025, n1+n2-2) and [(n1+n2-2)Sp^2]/chi-square (0.975, n1+n2-2). Compared with the single sample, you just change the n-1 to n1+n2-2 and the single sample variance to the pooled sample variance.
