Z confidence interval

Question

$W_1, \ldots, W_5$ be iid $Normal(3.1, 5.2^2)$. What is the probability that $\left(\frac{\bar W - 3.1}{\hat \sigma / \sqrt{5}} \geq 2.1\right)$?

I am not sure how to start. Is this equivalent to the probability that $Z \leq 2.1$?

Answer 1

You're close, but it's actually $1-P[t \le 2.1]$.

The distribution of $\bar{W} = \frac{1}{5}\sum_i^5 {W_i}$ is

$$ \bar{W} \sim \mathscr{N} \left( 3.1, \frac{5.2^2}{5} \right) . $$

Normally, $ (\bar{W} - 3.1)/\left( \frac{5.2^2}{\sqrt{5}}\right) $ would be the standardization of $\bar{W}$ into a standard normal $Z \sim \mathscr{N}(0,1)$. Because of the use of $\hat{\sigma}$ in the question, this is presumably coming from a sample of $5$ observations of $W$, and we are using a $t$-distribution with $4$ degrees of freedom, not a standard normal.

So, we can rewrite the probability we want in terms of $t$:

$$ P \left[ \frac{\bar{W} - 3.1}{\frac{\sigma}{\sqrt{5}}} \ge 2.1 \right] $$ $$ = 1 - P \left[ \frac{\bar{W} - 3.1}{\frac{\sigma}{\sqrt{5}}} \le 2.1 \right] $$ $$ = 1 - P \left[ t \le 2.1 \right] . $$ $$ \approx 0.052 .$$