Deriving a Confidence Interval for the Ratio of Two Variances

I derive the appropriate formula for a confidence interval for the ratio of two population variances (when we are sampling from normally distributed populations). I do not do any calculations or look at any examples in this video, I simply derive the appropriate confidence interval formula.

[@sGh0lybear]

I feel like the intervalls are wrong mate,
P(F(alpha/2) < U (statistic) < F(1-alpha/2)) seems more appropriate.

[@jbstatistics]

In the notation I use, F_{alpha/2} represents the value from the F distribution that has an area *to the right* of alpha/2. Many sources use this notation. Others define F_{alpha/2} to be the value with an area to the *left* of alpha/2. There are merits and disadvantages to both, and there isn't a universal notation. What I state in the video is correct, given my choice of notation.

[@Hauskkia]

yeah I would definitely get rid of those fractions that are in his conclusion. Since F_(alpha/2) = 1/[F_(1-alpha/2) you can swap their places and make a much (IMO) cleaner expression as a result

[@samarthumrao1877]

well explained video. 😃👍👍

[@whetstoneguy6717]

Professor --in calculating the p value of the F statistic, please amplify on why you take the smaller of the areas and multiply by two (2). I have an innate understanding but I would like to hear what your rationale is. Thank you.

[@vether7219]

You have just saved my life, thank you!

[@yurasereda7310]

1-alpha/2 and alpha/2 should be switched.

[@jbstatistics]

No, that's just different notation than what I use. Here, F_alpha/2 is the F value with an area of alpha/2 to the right. Some define it that way, some the other way.

[@andersonashok1]

Tq to the core... The best class I have ever seen....

[@varshahushare6981]

Good explanation

[@mchan1021]

is there a way to do this if the two population variances aren't equal?

[@jbstatistics]

We don't assume anything about the values of the population variances when constructing a confidence interval for their ratio.

[@mchan1021]

@@jbstatistics sorry, I meant to ask if this is possible when the two samples don't come from normal distributions?

[@mchan1021]

the context of my question is that I am trying to run a test for whether one group's variance is greater than another's (2 sample test) I was trying to use Levene's test, but that looks like a test that simply concludes whether variances of the two groups are equal. I saw that the F-test has a one tail version, but that requires normality....so I'm stuck. Or better yet, can I run Levene's test and conclude that if the variances aren't equal, then the group with the larger variance has statistically greater variance than the other group? Thank you so much!

[@abhishekdas2640]

Why the confidence Interval is finite?
Please comment

[@caamelot8802]

it is?