I'm in a statistics class and am doing a problem for homework about confidence intervals. I don't really know what it's asking though or when I've even reached a valid solution.
The problem says:
Suppose you have a random sample of size $n = 100$. The random variables $X_1,...,X_{100}$ are continuous with the mean $E(X_i) = \mu$ and $Var(X_i)=\sigma^2$. You want to have an interval estimate for $\mu$. Assume the true value of $\sigma^2$ is given.
a. In this hypothetical situation (knowing $\sigma^2$), you can calculate the length of a $95\%$ confidence interval (CI) without knowing the observed sample mean $\bar{x}$. What is the length in terms of $\sigma$?
b. Suppose we want to have a new random sample of size m from the same population and construct a $99\%$ CI of the same length as a $95\%$ CI. What is the sample size $m$?
For part a, I'm going off the fact that a confidence interval of $95\%$ is $\bar{x} \pm 1.960\frac{\sigma}{\sqrt{100}}$ My issue is, what does this technically equal, so that I can solve for $\bar{x}$ and what does the "length" even mean in this case? Just the length of the interval?
The way I tried to solve it from here was (probably wrong):
$\bar{x} \pm 1.960\frac{\sigma}{\sqrt{100}} = 0.95$ and then somehow attempting to solve for $\sigma$? So how can I do this, or how do I approach this problem?
For part b I'm assuming I need the answer from part a.. but would I just end up doing something like:
$\bar{x} \pm 1.960\frac{\sigma}{\sqrt{100}} = \bar{x} \pm 2.576\frac{\sigma}{\sqrt{m}}$ and then somehow solve for m?
Well, you have largely solved it. The length of an interval $[a, b]$ is just $b-a$. So if the 95% CI for $\mu$ is $$ \left[ \bar x - 1.96 \frac{\sigma}{\sqrt{100}}, \bar x + 1.96 \frac{\sigma}{\sqrt{100}} \right] $$
the length is just $$ 2 \cdot 1.96 \frac{\sigma}{\sqrt{100}} $$ Now for b the length of the 99% CI is going to be $$ 2 \cdot 2.58 \frac{\sigma}{\sqrt{m}} $$
And as you said, what remains is to equate and solve for $m$