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.

$x=10$

$n=40$

role="math" localid="1651300220980" $H_0:p=0.3$

$H_a:p<0.3$

role="math" localid="1651300430510" $\alpha =0.05$

The given data is

$x=10$

$n=40$

$H_0:p=0.3$

$H_a:p<0.3$

$\alpha =0.05$

The expression for the given sample is written as

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

The sample proportion is calculated as

$\hat{p}=\frac{10}{40}$

$=0.25.$

The given data is

$x=10$

$n=40$

$H_0:p=0.3$

$H_a:p<0.3$

$\alpha =0.05$

Let's test the hypothesis

$H_0:p=0.3$

$H_0:p<0.3$

Calculate the value of$n{p}_0$

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

$=12$

Calculate 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 one proportion$z-$ test.

The given data is

$x=10$

$n=40$

$H_0:p=0.3$

$H_a:p<0.3$

$\alpha =0.05$

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

$z=\frac{0.25-0.3}{\sqrt{\frac{0.3(1-0.3)}{0.40}}}$

$=\frac{-0.05}{0.0725}$

$=-0.6901.$

And also, $\alpha =0.05$

From the standard table

$z_{0.05}=-1.645$

Negative value is taken because the test is left handed.

The test statistic falls in the acceptance region.

So, the hypothesis $H_0$is not rejected. Therefore, the hypothesis $H_0$is not rejected.