Solved – How to derive a confidence interval from an F distribution

confidence intervaldistributionsself-study

So, this is the question I'm working on:

Suppose we observe a random sample of five measurements: 10, 13, 15, 15, 17, from a normal distribution with unknown mean $µ_1$ and unknown variance $σ_1^2$. A second random sample from another normal population with unknown mean $µ_2$ and unknown variance $σ_2^2$ yields the measurements: 13, 7, 9, 15, 11.

b. Use the pivotal method (and a pivotal statistic with F distribution) to derive a 95% confidence interval for $σ_2/σ_1$. Work it out for these data. And test the null hypothesis that $σ_2 = σ_1$ at the 5% level of significance. [6] (recall that $F_{\nu_1,\nu_2,a} = 1/F_{\nu_2,\nu_1,1-a}$).

So, I'm completely at a loss as to how I can use the pivotal method on the F distribution. Please help me.

Best Answer

You're working the question from the wrong end.

Forget the F; it doesn't come into it yet. First find a nice pivotal quantity, then worry about its distribution.

Check the actual definition of a pivotal quantity here (read as far as the '[1]').

(What important piece of information did you leave out of the definition?)

Now, just think about trying to estimate $\sigma_1/\sigma_2$. How might you estimate it?

Would the distribution of that estimate depend on $\sigma_1/\sigma_2$? Does the estimator contain $\sigma_1$ or $\sigma_2$?

How might you fix both issues at the same time to turn it into a pivot? That is, what do you do to the estimator in order to make it also a function of $\sigma_1$ and $\sigma_2$, while its distribution no longer depends on them?

If you get that far, you'll get a pivot, but it won't necessarily have an F-distribution. [If that happens, the only thing left then is to understand the relationship between what you have and something that does have an F distribution.]

Related Question