91Ó°ÊÓ

It is given that $\mathrm{Î\pm }=0.05$

$H_0:p=0.80$

$H_1:p<0.80$

$n=60,x=41$

We know that $\hat{p}=\frac{x}{n}$

Hence, sample proportion is $\hat{p}=\frac{x}{n}=\frac{41}{60}=0.6833$

$0.6833<0.80$, the sample results give some evidence for alternate hypothesis as it agrees to alternate hypothesis$H_1:p<0.80$

It is given that $\mathrm{Î\pm }=0.05$

$H_0:p=0.80$

$H_1:p<0.80$

$n=60,x=41$

The sample proportion is $\hat{p}=\frac{x}{n}=\frac{41}{60}=0.6833$

Test statistics $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}=\frac{0.6833-0.80}{\sqrt{\frac{0.80(1-0.80)}{60}}}=-2.26$

The $P$value is probability of obtaining value of test static, or more extreme value, if null hypothesis is true.

Hence, $P$value is

$P=P(z<-2.26)=0.0119$

It is given that $\mathrm{Î\pm }=0.05$

$H_0:p=0.80$

$H_1:p<0.80$

$n=60,x=41$

Sample proportion is $\hat{p}=\frac{x}{n}=\frac{41}{60}=0.6833$

Test statistic is

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}=\frac{0.6833-0.80}{\sqrt{\frac{0.80(1-0.80)}{60}}}=-2.26$

$P$value using normal probability table is:

$P=P(z<-2.26)=0.0119$

$P<0.05⇒\text{Reject}{H}_0$

Enough convincing proof is that proportion of all students at researcher's high school who owns computer is less than$0.80$