Question

The Fast N' Hot food chain wants to test if their "Buy One, Get One Free" program increases customer traffic enough to support the cost of the program. For each of 15 stores, one day is selected at random to record customer traffic with the program in effect, and one day is selected at random to record customer traffic with program not in effect. The results of the experiment are documented below. At proportional to = 0.05, test the hypothesis that the mean difference is at most 0 (at best the program makes no difference, or worse it decreases traffic) against the alternative that the mean difference > 0 ( the program increases traffic).

Customer Traffic
With Program	Without Program
149	140
225	233
112	110
52	42
336	332
146	135
143	151
35	33
185	178
147	147
167	162
249	243
161	149
44	48
347	346

What is the best option:

a. The pvalue of 0.221 indicates that the data provide insignificant evidence against H0. H0 is not rejected at a = 0.05. You decide to conclude the study and not to recommend the program.

b. The pvlaue of 0.084 provides weak evidence against H0. H0 is not rejected at a = 0.05. You decide the evidence is not strong enough to recommend further evaluation of the program.

c. The pvalue of 0.033 provides strong evidence against H0. H0 is rejected at a = 0.05. You decide to recommend further evaluation of the program.

d. The pvalue rejects H0: Mean difference > 0.

e. None of the answers are correct.

f. The pvalue of 0.002 provides overwhelming evidence against H0. H0 is rejected at a = 0.05. You decide that the program results increased customer traffic, overall, and recommend the program be implemented.

Answer 1

= (9 + (-8) + 2 + 10 + 4 + 11 + (-8) + 2 + 7 + 0 + 5 + 6 + 12 + (-4) + 1)/15 = 3.27

s_d = sqrt(((9 - 3.27)^2 + (-8 - 3.27)^2 + (2 - 3.27)^2 + (10 - 3.27)^2 + (4 - 3.27)^2 + (11 - 3.27)^2 + (-8 - 3.27)^2 + (2 - 3.27)^2 + (7 - 3.27)^2 + (0 - 3.27)^2 + (5 - 3.27)^2 + (6 - 3.27)^2 + (12 - 3.27)^2 + (-4 - 3.27)^2 + (1 - 3.27)^2)/14) = 6.35

The test statistic t = ( - D)/(s_d/)

                             = (3.27 - 0)/(6.35/)

                             = 1.99

P-value = P(T > 1.99)

             = 1 - P(T < 1.99)

             = 1 - 0.9668

             = 0.0332 = 0.033

Since the P-value is less than the significance level (0.033 < 0.05), so we should reject the null hypothesis.

Option - C is correct.