Two-tailed z test for normal population mean I am told that the mean of a sample follows a $N(\mu,\sigma^2/n)$ distribution if the sample is drawn from a $N(\mu,\sigma^2)$ distribution.
A random sample of 9 observations has a mean of 53, the variance of the normal population they come from is 16. Test the hypothesis that the mean is 50 using a two tailed test and a significance level of 5%
I know how to work out the answer, $z=(\bar X -\mu_0)/(\sqrt{\sigma^2/n}),$ then check the critical region.
My question is in the worked solution I was given I am told that this is a two tailed test using
$\bar X\sim N(50,\sqrt{16/9}).$ Why is $\sqrt{16/9}$ used rather than the variance?
So, in one tiny space I am told in notes somewhere that this is the standard error, I haven't been taught what it means, and I certainly don't know why it used rather than the variance.

For your question, it is clearly a two tailed test because
null hypothesis Ho: u=50
alternate hypothesis H1: u is not equal to 50 that is, it can either be greater than 50 or less than 50. So there are two possibilities for the test, that is why it is a two-tailed test.
You can try to think of an answer from the bell shaped graph of the normal distribution and can also check wikipedia for more clarification.
