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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

Expert verified

Power is 0.0038

Step by step solution

01

The test hypothesis is given below

Null hypothesis:

H0: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:

H0:p>0.8

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

02

Rejection region

Rejection region for =0.01

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

The critical value for the right-tailed test =0.01is z0.01=2.33. Hence, the region is zc>2.33

Therefore,

z=p^-p0.8(1-0.8)2500.0589=p^-0.8p^=0.8589

Thus, the rejection region is p^>0.8589

03

The power explains

p=79%

Power =P(p>0.8589p=0.79)

=Pz>0.8589-0.790.79(1-0.79)250=Pz>0.06890.0258=P(z>2.67)=0.5-0.4962=0.0038

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.

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