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.

$\overline{x}=21,n=32,\sigma =4,H_0:\mu =22,H_a:\mu <22$

Short answer

The value of z is $-1.41$, critical value is$-1.645$,$P=0.002$and reject$H_0$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$$\begin{gathered} \overline{x}=21, \\ n=32, \\ \sigma =4, \\ {H}_0:\mu =22,H_a:\mu <22 \end{gathered}$$

Step 2. Consider the test hypothesis.

Consider the given hypothesis,

$\mu$is the population mean.

The test hypothesis,

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

Therefore, the test is left 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 $\mu$,

$$\begin{gathered} z=\frac{\overline{x}-{\mu }_0}{{\displaystyle \frac{\sigma }{\sqrt{n}}}} \\ =\frac{21-22}{{\displaystyle \frac{4}{\sqrt{32}}}} \\ =-1.41 \end{gathered}$$

Therefore, this is left tailed test with $\alpha =0.05$, the critical value is $$\begin{gathered} -z_a=-z_{0.05} \\ =-1.645 \end{gathered}$$

The rejection region is $z<-z_{0.05}$.

Here, data-custom-editor="chemistry" $\mathrm{z}=-1.41>-1.645$.

Therefore, we do not reject $H_0$at $5\%$ level of significance as the value of the test statistics z does not fall in the rejection region.