Null hypothesis
Null hypothesis Ho: µ = µo (Mean traffic without the program is equal to µo=145)
Alternative hypothesis HA: µ ≠ µo (Mean traffic without the program is greater µo=145)
Test Statistics
One sample T-test because the sample size is less than 30 that is n =10
Degress of freedom, df = n-1
Tcalculated =
where, = ; sample mean of traffic without program
S1 = ; sample variance
X2 number of traffic without the program in every store
Critical Value and Decision Criteria
Reject Ho if > Ttabulated
Ttabulated = From t-tables at α = 0.05 and df = 9
Calculations and Results
∑ X1 = 1471, = 68084.9
Implying = = 147.1 and S1 = = 86.977
Therefore Tcal = = 0.0764
Conclusion
Since Tcal = 0.0764 is less than T0.025, 9 = 2.262, therefore, we fail to reject null hypothesis an conclude that the mean traffic without the program is the same as the mean of the previous store traffic of 145 customers.
2. Null hypothesis Ho: µ = µo (Mean traffic with the program is equal to µo=145)
Alternative hypothesis HA: µ > µo (Mean traffic with the program is greater µo=145)
Test Statistics
One sample T-test because the sample size is less than 30 that is n =10
Degress of freedom, df = m-1
Tcalculated =
where, = ; sample mean of traffic without program
S2 = ; sample variance
X2 number of traffic with the program in every store
Critical Value and Decision Criteria
Reject Ho if > Ttabulated
Ttabulated = From t-tables at α = 0.05 and df = 9
Calculations and Results
∑ X2 = 1501, = 67072.9
Implying = = 150.1 and S2 = = 86.328
Therefore Tcal = = 0.1868
Conclusion
Since Tcal = 0.1868 is less than T0.05, 9 = 1.833, therefore, we fail to reject null hypothesis an conclude that the mean traffic without the program is not greater than the mean of the previous store traffic of 145 customers.
Null hypothesis Ho: µ1 = µ2 (Mean traffic with the program is equal to mean traffic without the program)
Alternative hypothesis HA: µ1 < µ2 (Mean traffic without the program is less than the mean traffic without the program)
Test Statistics
Pooled T-test with n = m=10
Degress of freedom, df = n +m -2
Tcalculated = -
where,
= ; sample mean of traffic without program
= ; sample mean of traffic with program
S1 = ; sample variance without program
S2 = ; sample variance with program
X1 number of traffic without the program in every store
X2 number of traffic with the program in every store
Critical Value and Decision Criteria
Reject Ho if > T18, 0.05
Ttabulated = From t-tables at α = 0.05 and df = 10+10-2 = 18
Calculations and Results
∑ X1 = 1471 and = 68084.9
Implying = = 147.1 and S1 = = 86.977
Also, ∑ X2 = 1501, = 67072.9
Implying = = 150.1 and S2 = = 86.328
Therefore, = - = 0.0774
Conclusion
Since = 0.0774 is less than T0.05, 9 = 1.734, therefore, we fail to reject null hypothesis and conclude that the program did not significantly increase the traffic in the stores at 95% confidence level.
Null hypothesis Ho: µ1 = µ2 (Mean traffic with the program is equal to mean traffic without the program)
Alternative hypothesis HA: µ1 < µ2 (Mean traffic without the program is less than the mean traffic without the program)
Test Statistics
Pooled T-test with n =10
Degress of freedom, df = n - 1
Tcalculated = -
where,
X1 number of traffic without the program in every store
X2 number of traffic with the program in every store
d = X2 – X1; differences between the traffic without and with program
= ; sample mean of differences in traffic
Sd = ; sample variance of differences in traffic
Critical Value and Decision Criteria
Reject Ho if > T9, 0.05
Ttabulated = From t-tables at α = 0.05 and df = 0-2 = 9
Calculations and Results
∑d = -30 and = 184
Implying = = -3 and Sd = = 20.444
Therefore, = - = 0.4640
Conclusion
Since = 0.4640 is less than T0.05, 9 = 1.833, therefore, we fail to reject null hypothesis and conclude that the program did not significantly increase the traffic in the stores at 95% confidence level. As a conclusion for question 3 and 4, the results differ with those of question 1 and 2 but, the conclusions are similar for both sections. However, method 1 and 2 are the best as they give values which are far from the critical value hence clear in making conclusions.
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