Question

A teacher believes that $85\%$of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from $85\%$. The teacher samples $50$students and $39$reply that they would want to go to the zoo. For the hypothesis test, use a $1\%$level of significance.

First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion.

Short answer

a. The test is two-tailed test .

b. The value of p is $0.166$

c. We conclude that there is insufficient information to establish that the percent of pupils in the class who will desire to attend on a field trip to the local zoo is the same or different from $85\%$at the $5\%$significance level.

d. Graph is,

Step-by-step solution

Introduction

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

Explanation Part a

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.

$H_0:p=0.85vs{H}_a:p\ne 0.85$

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

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

Explanation Part b

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

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

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

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

where its data from getting is given as in the issue

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

Explanation Part c

The conclusion is

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

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.

Explanation Part d

The following is the diagram for the problem