Question

Levi-Strauss Co manufactures clothing. The quality control department measures weekly values of different suppliers for the percentage difference of waste between the layout on the computer and the actual waste when the clothing is made (called run-up). The data is in table #11.3.3, and there are some negative values because sometimes the supplier is able to layout the pattern better than the computer ("Waste run up," 2013). Do the data show that there is a difference between some of the suppliers? Test at the 1% level. Table #11.3.3: Run-ups for Different Plants Making Levi Strauss Clothing Plant 1 Plant 2 Plant 3 Plant 4 Plant 5 1.2 16.4 12.1 11.5 24 10.1 -6 9.7 10.2 -3.7 -2 -11.6 7.4 3.8 8.2 1.5 -1.3 -2.1 8.3 9.2 -3 4 10.1 6.6 -9.3 -0.7 17 4.7 10.2 8 3.2 3.8 4.6 8.8 15.8 2.7 4.3 3.9 2.7 22.3 -3.2 10.4 3.6 5.1 3.1 -1.7 4.2 9.6 11.2 16.8 2.4 8.5 9.8 5.9 11.3 0.3 6.3 6.5 13 12.3 3.5 9 5.7 6.8 16.9 -0.8 7.1 5.1 14.5 19.4 4.3 3.4 5.2 2.8 19.7 -0.8 7.3 13 3 -3.9 7.1 42.7 7.6 0.9 3.4 1.4 70.2 1.5 0.7 3 8.5 2.4 6 1.3 2.9

Answer 1

The one way ANOVA analysis is used to test the null hypothesis that all the means are equal. The hypothesis is defined as,

Null Hypothesis: All the means are equal,  

Alternative Hypothesis: At least one means differ significantly.

The test is performed in excel by using the following steps,

Step 1: Write the data values in excel. The screenshot is shown below

Step 2: DATA > Data Analysis > ANOVA: Single Factor > OK.  The screenshot is shown below,

Step 3: Select Input Range: All the data values column. The screenshot is shown below,

The result is obtained.  The screenshot is shown below,

Conclusion: From the ANOVA table,

P-value = 0.34547

Since the P-value = 0.34547 > 0.05 which means there is no difference in means hence we failed to reject the null hypothesis.

Hence we can conclude that there is no difference in the mean percentage difference of waste between the layout on the computer and the actual waste among the 5 supplier groups.