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.

$\overline{x}=23,n=24,\sigma =4,H_0:\mu =22,H_a:\mu \ne 22$

Short answer

The value of z is $1.22$, critical value is$\pm 1.96,P=0.221$and do not reject$H_0$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$$\begin{gathered} \overline{x}=23, \\ n=24, \\ \sigma =4, \\ {H}_0:\mu =22, \\ {H}_a:\mu \ne 22 \end{gathered}$$

Step 2. Consider the test hypothesis.

Consider the given hypothesis,

$\mu$is the population mean.

The test hypothesis,

$$\begin{gathered} H_0:\mu =22vs \\ {H}_a:\mu \ne 22 \end{gathered}$$

Therefore, this is two tailed test.

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

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

$$\begin{gathered} z=\frac{\overline{x}-{\mu }_0}{{\displaystyle \frac{\sigma }{\sqrt{n}}}} \\ =\frac{23-22}{{\displaystyle \frac{4}{\sqrt{24}}}} \\ =1.22 \end{gathered}$$

Therefore, this is left tailed test with$\alpha =0.05$.

Step 3. Take the critical values.

The critical values are given below,

$$\begin{gathered} \pm z_{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\pm z_{\raisebox{1ex}{$0.05$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =\pm z_{0.025} \\ =\pm 1.96 \end{gathered}$$

The rejection region is $z<-z_{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$or $z>z_{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$, i.e., $z<-1.96$or $z>1.96$.

Here, $$\begin{gathered} z=1.22>-z_{0.025} \\ =-1.96 \end{gathered}$$

and $z=1.22<1.96$

Hence, we do not reject $H_0$at $5\%$ level of significance as the value ofz does not fall in the reject region.