91Ó°ÊÓ

The given data is

$x=35$

$n=50$

$H_0:p=0.6$

$H_a:p>0.6$

$\mathrm{Î\pm }=0.05$

The expression for the sample proportion is

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

The sample proportion is calculated as

$\hat{p}=\frac{35}{50}$

$=0.7$.

The given data is

$x=35$

$n=50$

$H_0:p=0.6$

$H_a:p>0.6$

$\mathrm{Î\pm }=0.05$

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

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

$=50(0.4)$

$=20$

Both $n{p}_0$and $n(1-p_0)$have 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

$x=35$

$n=50$

$H_0:p=0.6$

$H_a:p>0.6$

$\mathrm{Î\pm }=0.05$

Write the expression for $z$

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{P_0\left(1-P_0\right)}{n}}}$

Then the value is calculated as

$z=\frac{0.7-0.6}{\sqrt{\frac{0.6(1-0.6)}{50}}}$

$=\frac{0.1}{0.0693}$

$=1.4434$

And also, $\mathrm{Î\pm }=0.05$

From the standard table

$z_{0.05}=1.645$

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

The test statistic falls in the acceptance region.

So, the hypothesis $H_0$is not rejected.