91Ó°ÊÓ

A two-tailed test is required when the degree of uncertainty cannot be identified on the higher or lower side.

Because the problem is the percentage of pupils in the class who will desire to go on a field trip to the local zoo, this is a check of a singular sample size.

We must put your ideas to the reality.

$H_0:p=0.85vs{H}_a:p≠0.85$

We have, $\mathrm{Î\pm }=1\mathrm{\%}=0.01$

As this is a two-tailed test, the phrase "is the same or different from" indicate this is a two-tailed test.

The teacher polls $n=50$students, $x=39$

say they'd want to visit the zoo.

The random variable $P$represents the percentage of children in the school who desire to travel to the nearby zoo on a school trip. The test's distribution is normal, i.e

$P^{\mathrm{′}}:\mathcal{N}\left(p,\sqrt{\frac{p(1−p)}{n}}\right)=\mathcal{N}\left(0.85,\frac{0.85×0.15}{\sqrt{50}}\right).$

Now we determine the $p$-value for a mean using the data is normally distributed:

$$\begin{gathered} P-\mathrm{value}=P\left(p^{\mathrm{′}}<0.78\text{or}{p}^{\mathrm{′}}>0.78\right) \\ =2\mathrm{P}\left({\mathrm{p}}^{\mathrm{′}}<0.78\right)≈0.166 \end{gathered}$$

where its data from getting is given as in the issue

$$\begin{gathered} p=\frac{x}{n} \\ p=\frac{39}{50} \\ p=0.78 \end{gathered}$$

The conclusion is

$\mathrm{Î\pm }=0.01<0.166=\mathrm{P}\text{-value.}$

As a result, we do not rule out $H_0:p=0.85$In other words, the sample proportion $p$has a $0.166$chance of being $0.79or0.92$or less.

The sample data do not reveal sufficient evidence that the percentage of kids who will desire to go on a field trip to a local zoo is different from$85\%$at$1\%$level of significance.

The following is the diagram for the problem