Levi-Strauss Co manufactures clothing. The quality control department measures w
ID: 3371511 • Letter: L
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 the following table, and there are some negative values because sometimes the supplier is able to layout the pattern better than the computer ("Waste run up," 2013).
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
Do the data show that there is a difference between some of the suppliers? Test at the 1% level
**********************************************************************
Let x1 = percentage difference of waste between the layout on the computer and the actual waste when the clothing is made (called run-up) from plant 1
Let x2 = percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 2
Let x3 = percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 3
Let x4 = percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 4
Let x5 = percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 5
Let u1 = mean percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 1
Let u2 = mean percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 2
Let u3 = mean percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 3
Let u4 = mean percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 4
Let u5 = mean percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 5
(i) Which of the following statements correctly defines the null hypothesis HO?
A. All five mean percentage differences are equal
B. Two of the mean percentage differences are not equal
C. At least four of the mean percentage differences are equal
D. At least two of the mean percentage differences are not equal
Let u1 = mean percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 1
Let u2 = mean percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 2
Let u3 = mean percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 3
Let u4 = mean percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 4
Let u5 = mean percentage difference of waste between the layout on the computer and the actual waste of run-up from plant 5
(ii) Which of the following statements correctly defines the alternate hypothesis HA?
A. All five mean percentage differences are equal
B. Two of the mean percentage differences are not equal
C. At least four of the mean percentage differences are equal
D. At least two of the mean percentage differences are not equal
(iii) Enter the level of significance ? used for this test:
(iv) Calculate sample mean and sample standard deviation for Plant 1 sample
(v) Calculate sample mean and sample standard deviation for Plant 2 sample
(vi) Calculate sample mean and sample standard deviation for Plant 3 sample
(vii) Calculate sample mean and sample standard deviation for Plant 4 sample
(viii) Calculate sample mean and sample standard deviation for Plant 5 sample
(ix) Using technology, determine F ratio test statistic and corresponding p-value.
(x) Comparing p-value and ? value, which is the correct decision to make for this hypothesis test?
A. Reject Ho
B. Fail to reject Ho
C. Accept Ho
D. Accept HA
(xi) Select the statement that most correctly interprets the result of this test:
A. The result is not statistically significant at .01 level of significance. Sufficient evidence exists to support the claim that there is a difference between some of the suppliers.
B. The result is statistically significant at .01 level of significance. There is not enough evidence to support the claim that there is a difference between some of the suppliers.
C. The result is statistically significant at .01 level of significance. Sufficient evidence exists to support the claim that there is a difference between some of the suppliers.
D. The result is not statistically significant at .01 level of significance. There is not enough evidence to support the claim that there is a difference between some of the suppliers.
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
Explanation / Answer
i)
option A)
ii)
D. At least two of the mean percentage differences are not equal
iii)
alpha = 0.01
iv) - viii)
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance sd
Plant 1 22 99.5 4.522727273 100.6418398 10.03204066
Plant 2 22 194.3 8.831818182 235.7289394 15.35346669
Plant 3 19 91.8 4.831578947 19.38783626 4.403162075
Plant 4 19 142.3 7.489473684 13.37432749 3.657092764
Plant 5 13 134.9 10.37692308 91.29858974 9.555029552
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 450.9206647 4 112.7301662 1.159631096 0.334012276 3.534991543
Within Groups 8749.088388 90 97.2120932
Total 9200.009053 94
ix) F = 1.159631096
p-value = 0.334012276
x) since p-value > 0.01
we fail to reject the null hypothesis
xi)
D. The result is not statistically significant at .01 level of significance. There is not enough evidence to support the claim that there is a difference between some of the suppliers.
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