Abstract: A supermarket chain wants to know if their “buy one, get one free” cam
ID: 3228460 • Letter: A
Question
Abstract: A supermarket chain wants to know if their “buy one, get one free” campaign increases customer traffic enough to justify the cost of the program. For each of 10 stores they select two days at random to run the test. For one of those days (selected by a coin flip), the program will be in effect. They want to determine whether the program increases the mean traffic. The results in number of customer visits to the 10 stores are in the data set. Answer each of the questions below using hypothesis testing. Follow the seven-step procedure for testing a hypothesis in your text book as a guide for answering the questions. Use a .05 significance level. 1. Previous data suggests mean store traffic is 145. Is the mean traffic without the program different from 145? 2. Is the mean traffic with the program greater than 145? 3. Did the program increase store traffic? Use a pooled t-test. 4. Did the program increase store traffic? Use a paired difference t-test. Turn in your findings as described below. Generally the report will be graded for clarity (how easy it is to understand you), completeness (no significant gaps in the information provided) and correctness (the values and descriptions are correct). The report will also be graded on adherence to the report standard. The report will be structured as follows Section 1: For each question, provide • The null hypothesis • The alternate hypothesis • The test statistic chosen (including which test) • The critical value and decision rule • The P-value • Your findings
Supermarket study Store # With Program Without Program
1 140 136
2 233 235
3 110 108
4 42 35
5 332 328
6 135 135
7 151 144
8 33 39
9 178 170
10 147 141
Explanation / Answer
1) Previous data suggests mean store traffic is 145. Is the mean traffic without the program different from 145?
In the given dataset we use the data for without program for answering this question.
step 1: state the hypothesis
H0:The mean traffic without program is 145 .i.e H0: =145
against
H1:The mean traffic without program is different from 145 .i.e. H1:145.
step 2: Test statistics
The one sample t test for mean is use because sample size is small (<30) and population standard deviation is unknown.
sample mean(x) for without program data is 147.1 , std dev(s) is 86.328,n=10
the test statistics is
t =(x-)/(s/n)
=(147.1-145)/(86.328/10)
=0.0769
since,t distribution with (10-1)=9 degrees of freedom.
step 3: critical value and decision rule
here =0.05,/2=0.025
criitical value for two sided t(0.025,9) is ±2.262
decision rule : The calsulated test statistics is 0.0769 is less than the 2.262 (critical value),hence we failed to reject null hypothesis.
step 4: p-value
P(t>|0.0769|)=P(t<-0.0769)+P(t>0.0769) =0.9404
p-value is greater than (0.05),hence failed to reject null hypothesis.
step 5: Conclusion
We can conclude that the data is not providing enough evidence to conclude that the mean traffic without the program is significantly different from 145.
2) Is the mean traffic with the program greater than 145?
In the given dataset we use the data for with program for answering this question.
step 1: state the hypothesis
H0:The mean traffic with program is 145 .i.e H0: =145
against
H1:The mean trffic with the program is greater than 145 .i.e. H1:>145.
step 2: Test statistics
The one sample t test for mean is use because sample size is small (<30) and population standard deviation is unknown.
sample mean(x) for with program data is 150.1 , std dev(s) is 86.977,n=10
the test statistics is
t =(x-)/(s/n)
=(150.1-145)/(86.977/10)
=0.1854
since,t distribution with (10-1)=9 degrees of freedom.
step 3: critical value and decision rule
here =0.05,
criitical value for one sided t(0.05,9) is 1.8331
decision rule : The calsulated test statistics is 0.1854 is less than the 1.8331 (critical value),hence we failed to reject null hypothesis.
step 4: p-value
P(t>0.1854)=0.4285
p-value is greater than (0.05),hence failed to reject null hypothesis.
step 5: Conclusion
We can conclude that the data is not providing enough evidence to conclude that the mean traffic with the program is significantly greater than145.
3) Did the program increase store traffic? Use a pooled t-test.
Here we use the two sample data of with and without
step 1: state the hypothesis
H0:The program did not increase the store traffic.i.e H0:with - without=0
against
H1:The program increased the store traffic.i.e. H1:with - without>0
step 2: Test statistics
The two sample pooled t test for independent samples is use because sample size is small (<30) and population standard deviation is unknown.
xwith=150.1 , s1= 86.977 ,xwithout=147.1 , s2= 86.328 ,n1=n2=10
sp=pooled standard deviation=[9*(86.977)2+9*(86.328)2]/(10+10-2)=7508.767=86.6531
the test statistics is
t =(xwith- xwithout)/sp(1/n1+1/n2)
=(150.1-147.1)/(86.6531/1/10+1/10)
=3/38.75246
=0.07741
since,two sample t distribution with (10+10-2)=18 degrees of freedom.
step 3: critical value and decision rule
here =0.05,
criitical value for one sided t(0.05,18) is 1.7341
decision rule : The calsulated test statistics is 0.07741 is less than the 1.7341 (critical value),hence we failed to reject null hypothesis.
step 4: p-value
P(t>0.07741)=0.4696
p-value is greater than (0.05),hence failed to reject null hypothesis.
step 5: Conclusion
We can conclude that the data is not providing enough evidence to conclude that the program increased the traffic significantly.
4. Did the program increase store traffic? Use a paired difference t-test.
Here we use the two sample data of with and without.
step 1: state the hypothesis
H0:The program did not increase the store traffic.i.e H0:d=0
against
H1:The program increased the store traffic.i.e. H1:d>0
step 2: Test statistics
The two sample pooled t test for independent samples is use because sample size is small (<30) and population standard deviation is unknown.
here we use excel for paired t test and the output is
the test statistics is
t =2.098136
since,two sample t distribution with (10-1)=9 degrees of freedom.
step 3: critical value and decision rule
here =0.05,
criitical value for one sided t(0.05,9) is 1.8331
decision rule : The calsulated test statistics is 2.098136 is greater than the 1.8331 (critical value),hence we reject the null hypothesis.
step 4: p-value
P(t>2.098136)=0.0327
p-value is smaller than (0.05),hence we are rejecting null hypothesis.
step 5: Conclusion
We can conclude that the data is providing enough evidence to conclude that the program increased the traffic significantly.
In Questions 3 we were interested to test using a pooled t-test for independent samples which we did and we found out that there is no significant improvement. In question 4 however the paired t-test was used and found out that there exists a significant difference. The reason is, the paired t-test considers that the samples are dependent thus it also takes care of the correlation between the variables, whereas the independent sample ttest does not. Now if we look at the data we can see that the samples are dependent as the sample consists of same stores. Thus this is a dependent sample and so the test in question 4 is most appropriate.
Hence the final conclusion is the program increased the traffic significantly.