91Ó°ÊÓ

The rejection zone in left-tailed test is at the extreme left of the distribution. The null hypothesis in this case is that the claimed value is less than or equal to the population mean value

Because the problem is the percentage of adults in the United States who own a cell phone, this is an assessment of a single population proportion.

We'll have to put our theories to the truth.

$H_0:p=0.92vs{H}_a:p<0.92$

$\mathrm{Î\pm }=5\%=0.05$is used.

As this is a left-tailed test, the < tells.

The problem's graph is as follows:

A total of $n=200$American adults were polled, with $x=174$claiming to own a cell phone. The stochastic process $P$represents the percentage of Americans who own a cell phone. The test's distribution is normal, i.e.

$P^{\mathrm{′}}:\mathcal{N}\left(p,\sqrt{\frac{p(1−p)}{n}}\right)=\mathcal{N}\left(0.92,\sqrt{\frac{0.92×0.08}{200}}\right)$

We use the assumption of normality of the mean to get the

$\text{P-value}=P\left(p^{\mathrm{′}}<0.87\right)≈0.0046$

where its data from getting is given as in the issue

$$\begin{gathered} p=\frac{x}{n} \\ p=\frac{174}{200} \\ p=0.87 \end{gathered}$$

We've come to the conclusion that

$\mathrm{Î\pm }=0.05>0.0046=\mathrm{P}\text{-value.}$

As a result, $H_0:p=0.92$is rejected.

The sample data indicate sufficient evidence that the percent of people in the United States who own a mobile phone is far less than$92\%$at the $5\%$level of significance.

"The proportion of American residents who own a smartphone is$92\%$,$H_0$

says the alternate hypothesis.

As a result, the Type I error is: Reject the null hypothesis that the percentage of adults in the United States who own a cell phone is less than $92\%$when it really is $92\%$

A type II error is characteristic as failing to reject the null hypothesis when it is wrong.

As a result, the Type II mistake is: Not to reject the null hypothesis that the percentage of adults in the United States who own a cell phone is$92\%$when it is not.