Question 13 (4 points) ~ SavedThe Fast N' Hot food chain wants to test if their "Buy One, Get One Free" programincreases customer traffic enough to support the cost of the program. For each of 15stores, one day is selected at random to record customer traffic with the program ineffect, and one day is selected at random to record customer traffic with programnot in effect. The results of the experiment are documented in DATA. For each store,compute difference = traffic with program minus traffic without program. At a =0.05, test the hypothesis that the mean difference is at most 0 (at best the programmakes no difference, or worse it decreases traffic) OPTIONAL ANSWERS. The pvalue of 0.221 indicates that the data provide insignificant evidenceagainst HO. HO is not rejected at a = 0.05. You decide to conclude the study andnot to recommend the program. - The p-value of 0.002 provides overwhelming evidence against the null hypothesis. Therefore, the null hypothesis is rejected at a significance level of 0.05. Based on this evidence, you decide that the program results in increased customer traffic overall and recommend the program be implemented. = 0.05. You decide the evidence is not strong enough to recommend furtherevaluation of the program. None of the answers match my calculation.The pvalue of 0.002 provides overwhelming evidence against HO. HO is rejecteat x = 0.05. You decide that the program results in increased customer traffic,overall, and recommend the program be implemented. - The p-value of 0.002 provides overwhelming evidence against the null hypothesis. Therefore, the null hypothesis is rejected at a significance level of 0.05. Based on this evidence, you decide that the program results in increased customer traffic overall and recommend the program be implemented. This statement aligns with the interpretation of the p-value being very low (0.002), indicating strong evidence against the null hypothesis. The decision to reject the null hypothesis in favor of the alternative hypothesis suggests a significant impact of the program on customer traffic, leading to the recommendation for implementation. The pvalue of 0.033 provides strong evidence against HO. HO is rejected at a =0.05. You decide to recommend further evaluation of the program. The pvalue rejects HO: Mean difference > 0. The pvalue of 0.084 provides weak evidence against HO. HO is not rejected at a= 0.05. You decide the evidence is not strong enough to recommend furtherevaluation of the program. None of the answers match my calculation. The pvalue of 0.002 provides overwhelming evidence against HO. HO is rejectedat x = 0.05. You decide that the program results in increased customer traffic.overall, and recommend the program be implemented. Customer Traffic With Program Without Program 134 140 246 233 116 110 42 42 335 332 145 135 164 151 31 33 183 178 149 147 176 162 257 243 155 149 61 48 343 346