Question

Please show your work

11.3.2

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 hypothesis being tested is:

H₀: µ₁ = µ₂ = µ₃ = µ₄ = µ₅

H_a: At least one means is not equal

	Mean	n	Std. Dev
	4.52	22	10.032	Plant 1
	8.83	22	15.353	Plant 2
	4.83	19	4.403	Plant 3
	7.49	19	3.657	Plant 4
	10.38	13	9.555	Plant 5
	6.98	95	9.893	Total
ANOVA table
Source	SS	df	MS	F	p-value
Treatment	450.921	4	112.7302	1.16	.3340
Error	8,749.088	90	97.2121
Total	9,200.009	94

The p-value is 0.3340.

Since the p-value (0.3340) is greater than the significance level (0.01), we fail to reject the null hypothesis.

Therefore, we cannot conclude that there is a difference between some of the suppliers.