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.
 A: 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.
A: If you're testing $H_0: \mu = 50$ against $H_a: \mu \ne 50,$ using a z test, then your test statistic involves
$\bar X = \frac{1}{n}\sum_{1=1}^n X_i = \frac{1}{9}\sum_{i=1}^9 X_i = 53.$  The z statistic is $Z = \frac{53-\mu_0}{\sigma/\sqrt{n}} 
\sim \mathsf{Norm}(0,1).$
That is, according to $H_0,$ we need to subtract the mean of $\bar X$ for the numerator of $Z$ and divide by the standard deviation
of $\bar X$ in the denominator. If $\bar X$ is the mean of a random
sample from population $\mathsf{Norm}(\mu_0, \sigma)\equiv
\mathsf{Norm}(50, 4)$ has $\bar X \sim\mathsf{Norm}(50, 4/3)$
because $E(\bar X) = 50,$ $Var(\bar X) = \sigma^2/n,$ and
$SD(\bar X) = \sqrt{\sigma^2/n} = \sqrt{16/9} = 4/3.$
Putting it all together,
$$Z = \frac{\bar X-\mu_0}{\sigma/\sqrt{n}} = \frac{53 - 50}{4/3}= 2.25.$$
Then we reject $H_0: \mu-50$ in favor of $H_a: \mu \ne 50,$
at the 5% level of significance if $|Z| \ge 1.96.$
Here is output from Minitab software, showing result of this
z two-sided z test. We do reject $H_0$ because $|Z| = |2.25| = 2.25 > 1.96.$
One-Sample Z 

Test of μ = 50 vs ≠ 50
The assumed standard deviation = 4

N   Mean  SE Mean      95% CI         Z      P
9  53.00     1.33  (50.39, 55.61)  2.25  0.024

Instead of the critical values $\pm 1.96,$ many statistical
software programs give a P-value. In the figure below,
the critical values are indicated by vertical dotted
lines. The area under the standard normal curve outside
these lines is $0.05 = 5\%.$
The observed value of the test statistic $Z = 2.25$ is
indicated by a heavy vertical black line. The value $-2.25$
is equally 'extreme' (thin black line), that is 'equally far from $0.$'. For a two-sided test, the P-value $(0.024)$ is the area
under the under the standard normal curve outside the two
black lines. Using the P-value, we reject $H_0$ at the 5% level of significance because the P-value is 0.024 < 0.05 = 5%.$
2 * pnorm(-2.25)
[1] 0.02444895   # P-value


R code for figure above:
hdr="Density of Standard Normal Dist'n"
curve(dnorm(x), -4, 4, lwd=2, col="blue", ylab="PDF", xlab="z", main=hdr)
 abline(h = 0, col="green2")
 abline(v = 0, col="green2")
 abline(v = c(1.96, -1.96), col="maroon", lty="dashed")
 abline(v = 2.25, lwd = 3)
 abline(v = -2.25)

