# 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.

• Where did you get your $0.34$? May 24, 2022 at 0:21
• Your formula for W is wrong. Presumably you intend W to be the average of the X's. Now consider - what is the distribution of W under $H_0$ May 24, 2022 at 3:37

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")