Eagle Outfitters is a chain of stores specializing in outdoor apparel and camping gear. They are considering a promotion that involves mailing discount coupons to all their credit card customers. This promotion will be considered a success if more than of those receiving the coupons use them. Before going national with the promotion, coupons were sent to a sample of credit card customers. Click on the datafile logo to reference the data. a. Develop hypotheses that can be used to test whether the population proportion of those who will use the coupons is sufficient to go national. b. The file Eagle contains the sample data. Develop a point estimate of the population proportion. c. Use αα= .05 to conduct your hypothesis test. Should Eagle go national with the promotion?

Step-by-step explanation:

The question is incomplete. The complete question is:

Eagle Outfitters is a chain of stores specializing in outdoor apparel and camping gear. It is considering a promotion that involves mailing discount coupons to all its credit card customers. This promotion will be considered a success if more than 10% of those receiving the coupons use them. Before going national with the promotion, coupons were sent to a sample of 100 credit card customers. Out of the 100 customers, 13 customers said that they used the discount coupons to make a purchase at a Eagle Outfitters store. Use a 0.05 level of significance. (a) Develop the null and alternative hypotheses that can be used to test whether the population proportion of those who will use the coupons is sufficient to go national. (b) Compute the sample proportion. (c) Compute the test statistic. (d) Compute the critical value. (e) Based on the critical value, do we reject H0 or do we not reject H0? (f) Based on the result of the hypothesis test, should Eagle Outfitter go national with the promotion?

Solution:

a) We would set up the hypothesis test.

For the null hypothesis,

H0: p ≥ 0.1

For the alternative hypothesis,

Ha: p < 0.1

This is a left tailed test

Considering the population proportion, probability of success, p = 0.1

q = probability of failure = 1 - p

q = 1 - 0.1 = 0.9

b) Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 13

n = number of samples = 100

p = 13/100 = 0.13

c) We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.13 - 0.1)/√(0.1 × 0.9)/100 = 1

The calculated test statistic is 1 for the right tail and - 1 for the left tail.

d) Since α = 0.05, the critical value is determined from the normal distribution table.

For the left, α/2 = 0.51/2 = 0.025

The z score for an area to the left of 0.025 is - 1.96

For the right, α/2 = 1 - 0.025 = 0.975

The z score for an area to the right of 0.975 is 1.96

e) In order to reject the null hypothesis, the test statistic must be smaller than - 1.96 or greater than 1.96

Since - 1 > - 1.96 and 1 < 1.96, we would fail to reject the null hypothesis.

Therefore, based on the result of the hypothesis test, the Eagle Outfitter should go national with the promotion

answer: sine

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