Confidence interval and moment

Question

Suppose a random sample of size $n = 9$ is taken, where $X$ is normally distributed with unknown mean $\mu$ and unknown variance $\sigma^2$. Consider the following cases

(a) Before the sample is taken, it is noted that the endpoints of a two-sided $90$% confidence interval are random variables derived from the sample mean $\bar{X}$ and sample standard deviation $S$. If $L$ is the length of the interval, find the second moment of $L$ in terms of the variance $\sigma^2$.

(b) The sample is taken and we find that $\bar{x} = 1.56$ and $s^2 = 0.68$. Determine the approximate one-sided confidence interval for $\mu$.

My attempt:

(a) The population variance is unknown. So we find the t value. Using the tables or an online tool with $\alpha = 0.1 \implies \frac{\alpha}{2} = 0.05$ and $8$ degrees of freedom gives $t_{0.05} = 1.8595480$.

(If I am not wrong then) Length = $2 \cdot t_{0.05} \cdot \frac{s}{\sqrt{n}} = 2 \cdot 1.8595480 \cdot \frac{s}{\sqrt{9}} \approx 1.2397s$.

The second moment is given by $E(L^2) = E((1.2397s)^2) = 1.5369E(s^2)$

From this point I am not sure how to write this in terms of $\sigma^2$. I read of a similar problem that used unbiased estimators but I cannot see what conditions $E(ks^2)$ (where $k$ is a constant) needs to be considered an unbiased estimator of $\sigma^2$.

(b) We have an unknown mean and variance so we use $t$ rather than $z$. Note that $\alpha = 0.1$ and there are $8$ degrees of freedom.

The lower bound, $m$, is given by $m = \bar{x} - t_{0.1} \cdot \frac{s}{\sqrt{n}}$.

With the tables or the use of an online tool, we get that $t_{0.1} \approx 1.3968153$. Hence $m = 1.56 - 1.3968153 \cdot \sqrt{\frac{0.68}{9}} = 1.176052$.

Is this correct thus far? Any assistance you could provide with (a) is appreciated.

Answer 1

You have

• $X_1,\ldots,X_9\sim\operatorname N(\mu,\sigma^2),$
• $\overline X = (X_1+\cdots+X_9)/9,$
• $S^2 = 1/(9-1) \big( (X_1-\overline X)^2+\cdots +(X_9-\overline X)^2\big).$

A standard result says $\operatorname E(S^2)$ remains equal to $\sigma^2$ if $\sigma^2$ changes, and $S^2$ can be computed based on $X_1,\ldots,X_9$ without knowing $\mu$ or $\sigma^2,$ i.e. $S^2$ is an unbiased estimator of $\sigma^2.$

Since $S^2$ is proportional to $L^2,$ and you have identified the constant of proportionality, what you have done solves that problem.