Case Study 1:
DEFINE
The process engineer has the following hypothesis tests to perform:
Test 1 is to check whether the population proportion of defectives is less than the target of 5%.
In other words,
Null Hypothesis: Population proportion of defectives = 0.05.
Alternative Hypothesis: Population proportion of defectives < 0.05.
Measure and Analyze Phases
A sample of 200 products is taken, and 3 products have been identified as defective.
Figure 4.1 shows how to select “1 Proportion” in Minitab® to test the population proportion of defectives.
Doing so will open the dialog box shown in Figure 4.2. Enter “3” for “Number of events” and “200” for “Number of trials”. Also, check the box for “Perform hypothesis test” and enter “0.05” for “Hypothesized proportion”. Click on “Options” and the dialog box shown in Figure 4.3 opens. Select “less than” from the drop-down menu for “Alternative” and click on “OK”. This takes you back to the dialog box shown in Figure 4.2.
Click on “OK” and the output shown in Figure 4.4 is the result. Because P-value (0.012) is less than 0.05, reject the null hypothesis. In other words, the population proportion of defectives is lower than 5%, which is good news.
Test 2 is to check whether the length of the product from Machine Q is equal to the target of 10 inches. It is suspected to be less than 10 inches. In other words,
Null Hypothesis: Product length from Machine Q = 10 inches.
Alternative Hypothesis: Product length from Machine Q < 10 inches.
Measure and Analyze Phases
A sample of 20 products is taken from Machine Q and their lengths are measured.
Machine Q
8.9844
7.4928
10.3180
8.7651
9.8638
8.6747
8.2423
9.8338
9.0193
9.9748
9.3649
8.8023
8.2759
8.0240
8.0074
9.5139
7.2129
7.4988
10.7370
9.1634
Because the sample is less than 30 products, “1-Sample t” is the appropriate test. Figure 4.5 shows how to select “1-Sample t”.
Doing so will open the dialog box shown in Figure 4.6. Select “Machine Q” column for “Sample in columns”. Check the box for “Perform hypothesis test” and enter “10” for “Hypothesized mean”. Click on “Options” and the dialog box shown in Figure 4.7 opens.
Select “less than” from the drop-down menu for “Alternative” and click on “OK”. This takes you back to the dialog box shown in Figure 4.6. Click on “OK” and the outputs shown in F4.10 are the results.
95% of the data is less than the H0 (null hypothesis) value of 10. For that same reason, the P-value in the output shown in Figure 4.10 is less than 0.05. Hence, there is sufficient evidence to reject the null hypothesis. In other words, Machine Q is indeed producing shorter products than desired. Therefore, the process engineer recommends fixing Machine Q.
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Case Study 3:
DEFINE
Test 3 is to check whether or not the mean diameters of products are all the same across the three shifts. In other words,
Null Hypothesis: Process mean of Shift 1 = process mean of Shift 2 = process mean of Shift 3.
Alternative Hypothesis: Process means are not all equal.
Measure and Analyze Phases
A sample of 20 products is taken from each of the three shifts, and their diameters are measured.
The process engineer wants to perform an analysis of variance (ANOVA) to check whether the mean diameters of the products from all of these three shifts are the same. However, before performing the ANOVA, it is important to check whether the following conditions for ANOVA are satisfied:
• The data from the three shifts are normally distributed (normality test).
• The data from the three shifts have equal variances (test of equal variances).
For the normality test, select “Probability Plot” as shown in Figure 4.11. Doing so will open the dialog box shown in Figure 4.12. Select “Multiple” and click on “OK”.
This opens the dialog box shown in Figure 4.13. Select “Shift 1 – Shift 3” columns for “Graph variables” and check the box for “Graph variables form groups”. Click on “Multiple graphs” and it opens the dialog box shown in Figure 4.14. Select “On separate graphs” and click on “OK”. This takes you back to the dialog box shown in Figure 4.13. Click on “OK” and the probability plots shown in Figures 4.15–4.17 are the results. Inasmuch as the P-values for the three shifts (0.813, 0.996, 0.519, respectively) are greater than 0.05, the data from each of the three shifts are normally distributed.
For the test of equal variances, it is necessary that the data from the three shifts are stacked into one column. Label two empty columns in CHAPTER_4_TEST_3.MTW as “Size” and “Shift”, as shown in Figure 4.18. Select “Columns” as shown in Figure 4.19. Doing so will open the dialog box shown in Figure 4.20. Select “Shift 1 – Shift 3” columns for “Stack the following columns”, select “Size” for “Column of current worksheet” and check the box for “Use variable names in subscript column”. Click on “OK” and the data are stacked in the “Size” column as shown in Figure 4.22. Type in the shift numbers in the “Shift” column as shown in Figure 4.22. Then, select “Test for Equal Variances” as shown in Figure 4.23. This will open the dialog box shown in Figure 4.24. Select “Size” for “Response” and “Shift” for “Factors”. Click on “OK” and the outputs shown in Figures 4.25 and 4.26 are the results. As shown in these two figures, the P-value (0.862) for Bartlett’s test is greater than 0.05. Hence, the data from the three shifts have equal variances.
To perform Test 3, select “One-Way” as shown in Figure 4.27. Doing so will open the dialog box shown in Figure 4.28. Select “Size” for “Response” and “Shift” for “Factor”. Click on “OK” and the output shown in Figure 4.29 is the result. Because the P-value (0.558) is not less than 0.05, there is insufficient evidence to reject
Improve and Control Phases
Based on the outputs of the three tests, Machine Q is fixed, and the process is improved. The process engineer continues to take samples from all machines and all shifts in order to ensure that the improved process is in control.
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