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 in DATA. For each store, compute difference = traffic with program minus traffic without program. At ∝ =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 mean difference >0 (the program increases traffic).

Customer Traffic With /Program Without Program

144 /140

236 /233

108 /110

43 /42

337 /332

134/ 135

148 /151

30/ 33

181/ 178

146 /147

159/ 162

248/ 243

150 /149

54 /48

349/ 346

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

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

C. The pvalue of 0.002 provides overwhelming evidence against H0: H0 is rejected at ∝ =0.05. you decide that the program results in increased customer traffic, overall and recommend the program be implemented.

D. The pvalue of 0.021 indicates that the data provide insignificant evidence against H0. H0 is not rejected at ∝ =0.05. You decide to conclude the study and not recommend the program.

E. The pvalue rejects H0: mean difference > 0. F. None of the answers are correct

Answer 1

Sol:

Its paired t test

Customer Traffic With Program	Customer Traffic Without Program	Difference
144	140	4
236	233	3
108	110	-2
43	42	1
337	332	5
134	135	-1
148	151	-3
30	33	-3
181	178	3
146	147	-1
159	162	-3
248	243	5
150	149	1
54	48	6
349	346	3
	sample mean d bar=	1.2
	sample sd,sd=	3.189268
	sample size=n=	15
	t=1.2/(3.189268/sqrt(15))
	1.457256027
	p=T.DIST.RT(1.457256027,14)
	0.083554504

t=1.46

p=0.0836

p>0.05

Fail to reject null hypothesis

Accept null hypothesis.

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