Question

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 x̄ s p μ σ  ? ≤ ≥ = ≠ > <  

H1H1: Select an answer σ x̄ s p 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?

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

(h) Is the claim supported?

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

Answer 1

a)

mean of difference ,    D̅ =ΣDi / n =   3.0000

b)

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
140	136	4	1.000
233	235	-2	25.000
110	108	2	1.000
332	328	4	1.000
151	144	7	16.000

c)

Level of Significance ,    α =    0.01                  
                          
sample size ,    n =    5                  
                          
mean of sample 1,    x̅1=   193.200                  
                          
mean of sample 2,    x̅2=   190.200                  
                          
mean of difference ,    D̅ =ΣDi / n =   3.0000                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    3.3166                  
                          

d)

Ho :   µd=   0                  
Ha :   µd >   0                  
                          

e)

Degree of freedom, DF=   n - 1 =    4      
t-critical value , t* =        3.747 [excel function: =t.inv(α,df) ]   

f)

std error , SE = Sd / √n =     3.3166    / √    5    =    1.4832        
                            
  
                          
t-statistic = (D̅ - µd)/SE = (   3   -   0   ) /    1.4832   =   2.02

(g) Is the null hypothesis rejected?

• 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.

Please let me know in case of any doubt.

Thanks in advance!

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