Customers who participate in a store’s free loyalty card program save money on their purchases but allow the store to keep track of the customer’s shopping habits and potentially sell these data to third parties. A Pew Internet & American Life Project Survey (January 2016) revealed that 225 of a random sample of 250 U.S. adults would agree to participate in a store loyalty card program, despite the potential for information sharing. Let$\mathit{p}$ represent the true proportion of all customers who would participate in a store loyalty card program.

a. Compute a point estimate of$\mathit{p}$

b. Consider a store owner who claims that more than 80% of all customers would participate in a loyalty card program. Set up the null and alternative hypotheses for testing whether the true proportion of all customers who would participate in a store loyalty card program exceeds .8

c. Compute the test statistic for part b.

d. Find the rejection region for the test if $\mathit{\alpha }\mathbf{=}\mathbf{.}\mathbf{01}$.

e. Find the p-value for the test.

f. Make the appropriate conclusion using the rejection region.

g. Make the appropriate conclusion using the p-value.

As per Pew Internet & American Life Project Survey (January 2016), out of 250 U.S adults surveyed, 225 agree to participate in a store loyalty card program, despite the potential for information sharing.

That is

The size of the sample$n=250$

The sample proportion is

$$\begin{gathered} \hat{p}=\frac{225}{250} \\ =0.9 \end{gathered}$$

The point for the true proportion of all customers who would participate in a store loyalty card program is$\hat{p}=0.9$

The null and alternative hypotheses are given as

$H_0:p=0.80$

That is, the true proportion of all customers who would participate in a store loyalty card program does not exceed .8

And

$H_a:p>0.80$

That is, the true proportion of all customers who would participate in a store loyalty card program exceeds .8

The test statistic for testing store owner’s claim is

$$\begin{gathered} Z=\frac{\hat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}} \\ =\frac{0.90-0.80}{\sqrt{\frac{0.80\left(1-0.80\right)}{250}}} \\ =\frac{0.10}{\sqrt{0.00064}} \\ =3.95 \end{gathered}$$

We have first to find the critical value for the critical region

Here

$\alpha :$The level of significance

$\alpha =.01$

Using the standard normal table, the critical value at the 1% significance level is 2.326

Hence the rejection region is

Reject the null hypothesis if

$Z_0>2.326$

Where

$Z_0$is the value of the test statistic.

We have $Z=3.95$, and the test is right-tailed (as an alternative hypothesis is right-tailed)

Therefore p-value in this scenario is

$$\begin{gathered} p-value=P\left(Z>Z_0\right) \\ =P\left(Z>3.95\right) \\ =0.000039.....\left[usingstandardnormaltable\right] \end{gathered}$$

We can see that

$Z=3.95>2.326$

Hence, we reject the null hypothesis.

Conclusion:

At a 1% significance level, we have sufficient evidence to conclude that the true proportion of all customers who would participate in a store loyalty card program exceeds .8

We can see that

$p-value=0.000039<0.01$

That is, the obtained p-value is less than the significance level.

Hence, we reject the null hypothesis.

Conclusion:

Based on the p-value, we have sufficient evidence to conclude that the true proportion of all customers who would participate in a store loyalty card program exceeds .8