91Ó°ÊÓ

The given data is

$x=16$

$n=20$

$H_0:p=0.7$

$H_a:p≠0.7$

$a=0.05$

Write the expression for the sample proportion as

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

The sample proportion is calculated as

$\hat{p}=\frac{16}{20}$

$=0.8$.

The given data is

$x=16$

$n=20$

$H_0:p=0.7$

$H_{a:}p≠0.7$

$a=0.05$

Test the hypothesis as

$H_0:p=0.7$

$H_0:p≠0.7$

Calculate the value of$n{p}_0$

$n{p}_0=\left(20\right)(0.7)$

$=14$

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

$n\left(1-p_0\right)=\left(20\right)(1-0.7)$

$=20(0.3)$

$=6.$

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

$x=16$

$n=20$

$H_0:p=0.7$

$H_{a:}p≠0.7$

$a=0.05$

Write the expression for $z$as

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

The value is calculated as

$z=\frac{0.8-0.7}{\sqrt{\frac{0.7(1-0.7)}{20}}}$

$=\frac{0.1}{0.1025}$

$=0.9759$

$≈0.98$.

And also, $\mathrm{Î\pm }=0.05$

For, $\mathrm{Î\pm }=0.025,$the critical value from the standard table is

$z_{0.025}=Â\pm 1.96$

The test statistic falls in the acceptance region.

So, the hypothesis$H_0$ is not rejected.