Question

We have been provided a sample mean, sample size, and population standard deviation. In the given case, use the one-mean z-test to perform the required hypothesis test at the $5\%$ significance level.

$\overline{x}=24,n=15,\sigma =4,H_0:\mu =22,H_a:\mu >22$

Short answer

The value of z is $1.94$, critical value is $1.645,P=0.026$and reject $H_0$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$$\begin{gathered} \overline{x}=24, \\ n=15, \\ \sigma =4, \\ {H}_0:\mu =22,H_a:\mu >22 \end{gathered}$$

Step 2. Consider the test hypothesis.

Consider the given hypothesis,

$\mu$is the population mean.

The test hypothesis is given below,

$$\begin{gathered} H_0:\mu =22Vs \\ {H}_a:\mu >22 \end{gathered}$$

Therefore, the test is right tailed test.

And the level of significance is$\alpha =0.05$.

Step 3. Use the test statistics.

We want to find the hypothesis test about the mean $\mu$,

$$\begin{gathered} z=\frac{\overline{x}-{\mu }_0}{{\displaystyle \frac{\sigma }{\sqrt{n}}}} \\ =\frac{24-22}{{\displaystyle \frac{4}{\sqrt{15}}}} \\ =1.94 \end{gathered}$$

Therefore, this is right tailed test with $\alpha =0.05$, the critical value is given below,

role="math" localid="1651164260948" $$\begin{gathered} z_a=z_{0.05} \\ =1.645 \end{gathered}$$

The rejection region is $z>z_{0.05}$.

Here, $$\begin{gathered} z=1.94>z_{0.05} \\ =1.645 \end{gathered}$$

Hence, we reject $H_0$at $5\%$ level of significance as the value of the test satisfice falls in the rejection region.