$1-\alpha$ confidence interval

Question

Let $Y$ be a r.v following $N(\mu_0,1)$ for some quantile $q\in(0,1)$ and let $z_q$ denote the $q$-quantile of $N(0,1)$ normal distribution.

Question?!

For a $\mu_0$ would the 0.95-confidenceinteval be $$[Y-z_{(0.05)},Y-z_{(0,95)}]$$

Thanks.

Answer 1

Let $U$ have standard normal distribution.

If I am well informed (I am no statistician, so please check me on it) then $P(U\leq z_q)=q$ for $q\in(0,1)$, so that: $$P(z_{0.05}<U<z_{0.95})=0.9$$

Applying that on $U=Y-\mu_0$ we find:$$P(z_{0.05}<Y-\mu_0<z_{0.95})=0.9$$or equivalently:$$P(\mu_0\in [Y-z_{0.95},Y-z_{0.05}])=0.9$$

Answer 2

To understand the problem, a drawing is a self evident explanation...

Of course this picture assumes you are deriving a symmetrical confidence interval, that is the optimal interval (say, the minimum size 95% confidence interval). Of course you can build many 95% confidence intervals, discarding a different error% in the two tails.