"Suppose that $X_1,...,X_n$ is a sample on $N(\mu, \sigma)$ where $\sigma$ is known. You want to test the null hypothesis $H_0: \mu = 2$ against $H_1 : \mu < 2$ by using a suitable confidence interval. Which of the following confidence intervals should be used?
$A: I_{\mu} = (-\infty, \bar{x}+\frac{\sigma}{\sqrt{n}}\cdot \lambda_{\alpha})$
$B: I_{\mu} = (-\infty, \bar{x}+\frac{\sigma}{\sqrt{n}}\cdot \lambda_{\frac{\alpha}{2}})$
$C: I_{\mu} = (\bar{x}+\frac{\sigma}{\sqrt{n}}\cdot \lambda_{\alpha}, \infty)$
$D: I_{\mu} = (\bar{x}+\frac{\sigma}{\sqrt{n}}\cdot \lambda_{\frac{\alpha}{2}}, \infty)$ "
I don't really know where to start here, haven't been able to find any good resource on how to choose interval for hypothesis testing. Some hints on how to tackle this, and similar, problems would be greatly appreciated.
Hints:
You are considering whether an observed sample mean might be judged an extreme observation if the null hypothesis is true; if it is too extreme, you might reject the null hypothesis.
The alternative hypothesis gives an indication of which direction(s) count(s) as extreme in the sense of pointing away from the null hypothesis and towards the alternative hypothesis
The reverse approach of using confidence intervals produces similar results, in the sense that if the hypothesised population mean is in the confidence interval, this might be seen as a reason for not rejecting the null hypothesis
Given the alternative hypothesis, which of the confidence intervals would lead to the rejection of the null hypothesis when the sample mean is extreme in the same direction(s) as before
If the null hypothesis is true, which of the confidence intervals lead to the rejection of the null hypothesis with probability $\alpha$?