Question

In the given exercise, we have provided a sample mean, sample size, and population standard. In each case, use the one-mean z-test to perform the required hypothesis test at the 5% significance level.

$\stackrel{—}{x}=20,n=24,\sigma =4,H_0:\mu =22,H_{\mathrm{a}}:\mu \ne 22$

Short answer

Ans: We have to reject $H_{0}at5\%$the level of significance as the value of z falls in the rejection region.

Step-by-step solution

Step 1. Given information.

given,

$\stackrel{—}{x}=20,n=24,\sigma =4,H_0:\mu =22,H_{\mathrm{a}}:\mu \ne 22$

Step 2. Let's perform a hypothesis test.

$\begin{array}{r}{H}_0:\mu =22\\ H_a:\mu \ne 22\end{array}$

Where $\mu$is the population mean.

Since the alternative hypothesis contains the symbol localid="1651574793371" $\ne$.

And the test is a two-tailed test.

Now,

The level of significance is, $\alpha =0.05$.

Step 3. Now, we find the hypothesis test about the mean, μ.

we have,

$\stackrel{—}{x}=20,n=24,\sigma =4$

Test statistic

$\begin{array}{r}z=\frac{\overline{x}-{\mu }_0}{\frac{\sigma }{\sqrt{n}}}\\ =\frac{20-22}{\frac{4}{\sqrt{24}}}\\ =-2.45\end{array}$

Step 4. Since this is a two-tailed test with α=0.05,  

By using the normal distribution table, the critical values are:

$\begin{array}{r}\pm z_{\alpha /2}=\pm z_{0.05/2}\\ =\pm z_{0.025}\\ =\pm 1.96\end{array}$

Here the rejection regions are $z<-z_{\alpha /2}orz>z_{\alpha /2}$

i.e., $z<-1.96orz>1.96$

Here $z=-2.45<-z_{0.025}=-1.96$

Since the level of significance as the value of z falls in the rejection region.

So, we have to reject localid="1651211012259" $H_{0}at5\%$.