Question

Please Help

A supermarket chain wants to know if its "buy one, get one free" campaign increases customer traffic enough to justify the cost of the program. For each of 5 stores it selects two days to run the test. For one of those days the program will be in effect. At 1% significance level, test the claim that the program increases traffic. Use t-distribution.

For parts (a), (b), (c), round your answers to 2 decimal places.

(a) ¯dd¯ =

You MUST answer part (a) before filling in the two right columns in the table.

(b) Fill in the table below.

Store	With Program	Without Program	d=d= With−-Without	d−¯dd-d¯	(d−¯d)2(d-d¯)2
1	140	136
2	233	235
3	110	108
4	332	328
5	151	144
Total

(c) The standard deviation for the differences is sd=sd=

At a 1% significance level, test the company's claim. Use the t-test for matched pairs and the formula involving the mean of the differences μμ,

t=¯d−μ(sd√n)t=d¯-μ(sdn)

(d) State the null and alternative hypotheses, and identify which one is the claim.

H0H0: Select an answer μ p s x̄ σ  ? = < ≠ ≥ ≤ >  

H1H1: Select an answer p x̄ σ s μ  ? > ≠ < ≤ ≥ =  

Which one is the claim:

• H0H0
• H1H1

For parts (e), (f) use the correct sign for the t-value and test statistic, either positive or negative, and round your answers to 3 decimal places.

(e) What is the critical t-value?

(f) What is the test statistic?

(g) Is the null hypothesis rejected?

• Yes,reject the null hypothesis.
• No, do not reject the null hypothesis.

(h) Is the claim supported?

• At 1% significance level, there is not sufficient sample evidence to support the claim that the program increases traffic.
• At 1% significance level, there is sufficient sample evidence to support the claim that the program increases traffic.

Answer 1

a)

∑d = -5

n = 5

Mean , x̅d = Ʃd/n = 15/5 = 3

b)

Store	With	Without	Difference, d	d - d̅	(d - d̅)²
1	140	136	4	1	1
2	233	235	-2	-5	25
3	110	108	2	-1	1
4	332	328	4	1	1
5	151	144	7	4	16
Total	-	-	-	-	89

c)

Standard deviation, s = √(Ʃ(d - d̅)²/(n-1)) = √(89/(5-1)) = 3.3166

d)

Null and Alternative hypothesis:

Ho : µd ≤ 0

H1 : µd > 0

Claim: H1

e)

df = n-1 = 4

Critical value, t-crit = ABS(T.INV(0.01, 4)) = 3.747

Reject Ho if t > 3.747

f)

Test statistic:

t = (x̅d)/(sd/√n) = (3)/(3.3166/√5) = 2.023

g)

No, do not reject the null hypothesis.

h)

At 1% significance level, there is not sufficient sample evidence to support the claim that the program increases traffic.