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From Hogg & Tanis, 8th ed., p. 291:

Let $X_1, X_1, \dots X_n$ by a random sample of size $n$ from the normal distribution $N(\mu, \sigma^2)$. Calculate the expected length of a 95% confidence interval for $\mu$, assuming that $n = 5$ and variance is (a) known, (b) unknown.

For (a), I let $L = 2z_{\alpha / 2}(\sigma / \sqrt{n})$. Since everything is constant, $E[L] = L$; just plug in all the numbers and out comes an expression in terms of $\sigma$.

For (b), though, I'm not as sure. I let $L = 2t_{\alpha / 2}(n - 1)\cdot(S / \sqrt{n}) = 2t_{0.025}(4)\cdot(S / \sqrt{5})$. Then $E[L] = \frac{5.552}{\sqrt{5}}E[S]$. By an earlier result (to which a hint for this question refers),

$$E[S] = \frac{\sigma}{\sqrt{n-1}}\cdot\frac{\sqrt{2}\Gamma(n/2)}{\Gamma((n-1)/2)}$$

Plugging in all the relevant values, I get

$$E[L] = \frac{4.164\sqrt{\pi}}{\sqrt{10}}\sigma$$

which looks a bit like the previous result. The result for (a) makes sense to me because $\sigma$ is known; plug in $\sigma$ and out comes the length of the interval. For (b), though, $\sigma$ is unknown. So — if this is correct — how would that form of $E[L]$ be at all informative? Can anyone offer any intuition on what's going on here?

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up vote 1 down vote accepted

Stack Exchange awarded me with a Tumbleweed badge for this one a day or two ago; I'd forgotten that it was still out there. I figure if this is good enough, maybe I can accept my own answer and fluff up my rating.

Anyway, for (a), $L= 2 z_{\alpha / 2} (\sigma / \sqrt{n}) = 2(1.96)(\sigma / \sqrt{5}) = 3.92 \sigma / \sqrt{5} \approx 1.753 \sigma$. As I wrote above, everything is constant, and so this is the expected value of $L$ as well.

For (b), we find that $E[L] = \frac{5.552}{\sqrt{5}}E[S]= \frac{5.552}{\sqrt{5}}\cdot \frac{\sigma}{\sqrt{5-1}}\cdot \frac{\sqrt{2}\Gamma(5 / 2)}{\Gamma((5 - 1)/2)} = \frac{4.164 \sqrt{\pi}}{\sqrt{10}}\sigma \approx 2.334 \sigma$. So the above was correct after all.

As for my real question — what this second part tells us — I asked my professor: The expression for $E[L]$ when $\sigma$ is unknown doesn't give us, like, a formula for calculating the length of the interval; after all, as soon as we have a value of $\sigma$ to plug in, it becomes the wrong formula. Rather, we look at the ratio of the two numbers, $\left(\frac{3.92}{\sqrt{5}} \sigma \right) / \left( \frac{4.164\sqrt{\pi}}{\sqrt{10}}\sigma \right) \approx 1.331$. This tells us that, whatever $\sigma$ is, if we don't know it, our confidence interval will be about $1.331$ times wider than it would be if we did know it — at least for $n = 5$.

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