Question

Customer participation in-store loyalty card programs. Refer to the Pew Internet & American Life Project Survey (January 2016) study of 250 store customers and their participation in a store loyalty card program, Exercise 7.69 (p. 425). Recall that a store owner claimed that more than 80% of all customers would participate in a loyalty card program. You conducted a test of H0: p = .8 versus Ha: p 7 .8 using a = 01. What is the probability that the test results will support the claim if the true percentage of customers who would participate in a loyalty card program is 79%?

Short answer

Power is $0.0038$

Step-by-step solution

The test hypothesis is given below

Null hypothesis:

$H_0:p=0.8$

This means that the genuine proportion of all consumers enrolling in a shop loyalty card program is less than 0.8.

Null hypothesis:

$H_0:p>0.8$

This means that the genuine proportion of all consumers enrolling in a shop loyalty card program is more than 0.8.

Rejection region

Rejection region for $\alpha =0.01$

The test is right-tailed, and the level of significance is $\alpha =0.01$

The critical value for the right-tailed test $\alpha =0.01$is $z_{0.01}=2.33$. Hence, the region is $z_c>2.33$

Therefore,

$$\begin{gathered} z=\frac{\hat{p}-p}{{\displaystyle \frac{\sqrt{0.8(1-0.8)}}{250}}} \\ 0.0589=\hat{p}-0.8 \\ \hat{p}=0.8589 \end{gathered}$$

Thus, the rejection region is $\hat{p}>0.8589$

The power explains

$p=79\%$

Power =$P(\overline{p}>0.858\overline{)9}p=0.79)$

$$\begin{gathered} =P\left(z>\frac{0.8589-0.79}{\sqrt{{\displaystyle \frac{0.79(1-0.79)}{250}}}}\right) \\ =P\left(z>\frac{0.0689}{0.0258}\right) \\ =P(z>2.67) \\ =0.5-0.4962 \\ =0.0038 \end{gathered}$$

Thus, if the genuine proportion of consumers enrolling in a loyalty card program is 79%, the probability that the test findings will back up the claim is 0.0038.