a. Determine the sample proportion.

b. Decide whether using the one-proportion $z-$test is appropriate.

c. If appropriate, use the one-proportion $z-$test to perform the specified hypothesis test.

$x=8$

$n=40$

$H_0:p=0.3$

$H_2:p<0.3$

$\alpha =0.10$

The given data is

$x=8$

$n=40$

$H_0:p=0.3$

$H_2:p<0.3$

$\alpha =0.10$

The expression for the sample proportion is written as

$\hat{p}=\frac{x}{n}$

The sample proportion is calculated as

$\hat{p}=\frac{8}{40}$

$=0.2$.

The given data is

$x=8$

$n=40$

$H_0:p=0.3$

$H_2:p<0.3$

$\alpha =0.10$

We have to test

$H_0:p=0.3$

$H_0:p<0.3$

Calculate the value of$n{p}_0$

$n{p}_0=\left(40\right)(0.3)$

$=12$

Find the value of$n\left(1-p_0\right)$

$n\left(1-p_0\right)=\left(40\right)(1-0.3)$

$=40(0.7)$

$=28.$

Both $n{p}_0$and $n(1-p_0)$have a value greater than $5$. So, one proportion $z-$test is appropriate to use.

Therefore, it is appropriate to use the one proportion$z-$test.

The given data is

$x=8$

$n=40$

$H_0:p=0.3$

$H_2:p<0.3$

$\alpha =0.10$

Write the expression for $z$

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$

The $z-$value is calculated as

$Z=\frac{0.2-0.3}{\sqrt{\frac{0.3(1-0.3)}{0.40}}}$

$=\frac{-0.1}{0.0725}$

$=-1.3801$

For, $\alpha =0.10$the value of $z$from the table is

$z_{0.05}=-1.282$

The test statistic falls in the rejection region.

So, the hypothesis$H_0$ is rejected.