Please help me derive the formula for upper bound for one sided confidence interval $\bar{x} + z_{\alpha}(\frac{\sigma}{\sqrt{n}})$?

Question

I want to derive for myself the known formula for the upper bound for one sided confidence interval $\bar{x} + z_{\alpha}(\frac{\sigma}{\sqrt{n}})$ for mean $\mu$ for a sample of size $n$ from a normal distribution $N(\mu, \sigma^{2})$ where $\sigma$ is known, at $1-\alpha$ confidence level. Using standard terminology, I start with

$$P\left(\frac{\sqrt{n}(\bar{X} - \mu)}{\sigma} \geq z_{\alpha}\right) = 1 - \alpha$$

and rearranging this gives

$$P(\mu \leq \bar{X} - z_{\alpha}\left(\frac{\sigma}{\sqrt{n}}\right)) = 1 - \alpha$$

So according to me the upper bound is $\bar{X} - z_{\alpha}\left(\frac{\sigma}{\sqrt{n}}\right)$, i.e. has a -ve sign instead of +ve sign between the two terms. Where have I made a mistake?

Answer 1

What you have written down does not work because by definition of $z_{\alpha}$: $P(X>z_{\alpha}) = \alpha$ not $1-\alpha$.

What can possibly, and in fact will work, is that since $\sqrt{n}(\bar{X} - \mu)/\sigma$ is $N(0,1)$ and in particular it is symmetric around zero (real proof of symmetry here) we can conclude that (from the definition of $z_{\alpha}$)

$$P\left(\frac{\sqrt{n}(\bar{X}-\mu)}{\sigma}<-z_{\alpha}\right) = P\left(\frac{\sqrt{n}(\bar{X}-\mu)}{\sigma}>z_{\alpha}\right) = \alpha$$ and hence also

$$P\left(\frac{\sqrt{n}(\bar{X}-\mu)}{\sigma} \geq -z_{\alpha}\right) = 1-\alpha$$

Solving for $\mu$ gives the desired result.

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My comments above about the two sided results is just another way of reaching the last equality above. If you know that $X$ is in particular, $N(0,1)$, and so symmetric about zero then $$ (1 - 2\alpha) + \alpha = P\left(-z_{\alpha} \leq X\leq z_{\alpha}\right) + P(z_{a}<X) = P(-z_{\alpha} \leq X)$$ and you reach the same conclusion as above.