Example of 2-Sample Poisson ratemain topic interpreting results session command see also
Example of 2-Sample Poisson rateCompany A and Company B manufacture televisions and have counted the number of units with defective screens for the past ten years. Company A counts the number of defective units each quarter (3 month period), but Company B counts the number of defective units in every 6 month period. You manage an electronics retail shop, and you want to stock televisions with the fewest defects. To decide which company's products to stock, you conduct a 2-Sample Poisson Rate test to determine which company has the lowest monthly defect rate.
1 Open the worksheet TVDEFECT.MTW.
2 Choose Stat > Basic Statistics > 2-Sample Poisson Rate.
3 Choose Each sample is in its own column.
4 In Sample 1, enter 'Defective A '.
5 In Sample 2, enter 'Defective B '.
6 Click Options. In Lengths of observation, enter '3 6 ' (three and six).
7 Click OK in each dialog box.
Session window output
Test and CI for Two-Sample Poisson Rates: Defective A, Defective B
Total “Length” of Rate of Mean Variable Occurrences N Observation Occurrence Occurrence Defective A 713 40 3 5.94167 17.825 Defective B 515 20 6 4.29167 25.750
Difference = rate(Defective A) - rate(Defective B) Estimate for difference: 1.65 95% CI for difference: (1.07764, 2.22236) Test for difference = 0 (vs ≠ 0): Z = 5.65 P-Value = 0.000 Exact Test: P-Value = 0.000
Difference = μ (Defective A) - μ (Defective B) Estimate for difference: -7.925 95% CI for difference: (-10.5053, -5.34474) Test for difference = 0 (vs ≠ 0): Z = -6.02 P-Value = 0.000 Exact Test: P-Value = 0.000 |
The "length" of observation corresponds to the number of months during which each company counts defects. Because these time periods differ, direct comparison of the mean number of defects does not answer the research question. Therefore, Minitab uses the "length" entries to calculate the average number of defects per month for each company. The hypothesis test determines whether the difference between the two monthly rates is statistically significant.
The p-value for this hypothesis test is zero. Therefore, you should reject the null hypothesis that the defect rates are equal. Furthermore, because the confidence interval for (Defect Rate A - Defect Rate B) contains only positive numbers, you can conclude with 95% confidence that televisions from Company A have a higher defective rate. As the manager of an electronics retail store, you choose to stock televisions from Company B.