Showing that the given critical value (?) gives a test with significance level 0.05 So I have a maths problem that I'm struggling to understand... The original language isn't English, but I have done my best to translate the necessary background information into English.
Assume that we have observed the following values:
0.50, -0.15, 2.82, 0.68, 0.19, 1.23, -1.65, 2.38, -0.49, 1.59, 0.66, -0.12, 1.20, 1.08, 0.82
from a sample $X_1, X_2,...,X_{15}$ where $X\sim N(\mu,1)$ for $i=1,2,...,15$ and $\mu$ is an unknown parameter. We are interested in the following two hypotheses:
$H_0:\mu=0\\
H_1:\mu=0.5$
Let $W=\frac{1}{15}(X_1,X_2,...,X_{15})$ be the average of the sample, and for $c>0$ observe the test that accepts the null hypothesis if $W\leq c$. For the following, we will test the null hypothesis $H_0$ at a significance level $\alpha=0.05$.
What I need help with: How do I show that $c=0.4247$ gives a test with a significance level of $\alpha=0.05$?
My initial thought was to check if the area under the graph for the normal distribution with mean 0 and variance 1 was 0.05 to the right of the critical value, but I got 0.34 instead (thought the critical value looked a bit off when I drew the graph...). I don't know what I've overlooked. My best guess is that my "critical value" isn't actually the critical value.
 A: The traditional approach would be to do a right-sided
z test at the 5% level of $H_0: \mu = 0$ vs.
$H_a: \mu = 0.5.$
The test statistic is
$$Z = \frac{\bar X - \mu_0}{\sigma/\sqrt{n}} = \frac{0.716 - 0}{1/\sqrt{15}} = 2.773.$$
Then, $Z \sim\mathsf{Norm}(0,1),$
which has 5% of its probability above $1.645.$
Thus, the critical value for test statistic $Z$ is
$c^\prime = 1.645,$ from a CDF table of the
standard normal distribution. Because $Z > c^\prime = 1.645$
you reject $H_0$ in favor of the alternative $H_a.$
The figure below shows the standard normal distribution. The solid vertical line shows
the observed value of the test statistic $Z$ and
the vertical dotted line shows the critical value.
(See end note for R code.)

However, you are asked to use test statistic
$W = \bar X = 0.716.$ The statistic $W$ is normally
distributed, but not standard normal.
Accordingly, it seems you are told to use critical value $c = 0.4247 \approx 1.645/\sqrt{15}.$
Because this is a self-study problem, I will leave it to
you to explain why the 5% critical value in terms of
test statistic $W$ should be $c=0.4247.$
Note: R code for figure.
hdr = "Standard Normal Density"
curve(dnorm(x), -3, 3, 
      lwd=2, ylab="Density", xlab="z", main=hdr)
abline(h=0, col="green2")
abline(v=0, col="green2")
abline(v=2.773, lwd=2)
abline(v=1.645, lwd=2, col="orange", lty="dotted")

