91Ó°ÊÓ

The given data is

$x=40$

$n=50$

$H_0:p=0.6$

$H_a:p>0.6$

$\mathrm{Î\pm }=0.01$

The expression for the sample data is given as

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

The sample proportion is calculated as

$\hat{p}=\frac{40}{50}$

$=0.8$.

The given data is

$x=40$

$n=50$

$H_0:p=0.6$

$H_a:p>0.6$

$\mathrm{Î\pm }=0.01$

Let's test the hypothesis

$H_0:p=0.6$

$H_0:p>0.6$

Calculate the value of$n{p}_0$

$n{p}_0=\left(50\right)(0.6)$

$=30$

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

$n\left(1-p_0\right)=\left(50\right)(1-0.6)$

$=50(0.4)$

$=20$

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

$x=40$

$n=50$

$H_0:p=0.6$

$H_a:p>0.6$

$\mathrm{Î\pm }=0.01$

The expression for $z$is

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

The $z$value is calculated as

$z=\frac{0.8-0.6}{\sqrt{\frac{0.6(1-0.6)}{50}}}$

$=\frac{0.2}{0.0693}$

$=2.8868$

And also,

$\mathrm{Î\pm }=0.01$

From the standard table

$z_{0.05}=2.326$

A positive value is taken because the test is Right-handed.

The test statistic falls in the rejection region.

So, the hypothesis $H_0$is rejected.