1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Where did you get your $0.34$? $\endgroup$
    – Henry
    May 24, 2022 at 0:21
  • $\begingroup$ 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 $H_0$ $\endgroup$
    – Glen_b
    May 24, 2022 at 3:37

1 Answer 1

1
$\begingroup$

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.)

enter image description here

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")
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.