Question

A company’s executives have selected four areas that they believe are suitable locations to relocate one of its manufacturing plants. However, they want to determine if the average wages are significantly different in any of the locations. A survey of hourly wages of similar workers in each of the four areas is performed. Summary statistics and the ANOVA table are given below.

Summary

Groups	Count	Sum	Average	Variance
Area 1	5	56	11.2	1.7
Area 2	5	80	16	2.5
Area 3	5	69	13.8	1.7
Area 4	5	87	17.4	2.8

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ANOVA

Source of Variation SS df MS F Between Groups 110 3 36.6667 16.8582 Within Groups 34.8 16 2.175 Total 144.8 19

a. Use a 0.05 significance level to test the claim that there is no difference in wages between the four areas.

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b. If there is a difference in means, perform the Bonferroni method with EER = 0.05. Summarize your results with the underline approachand also state which means can be concluded larger than which. Hint: if needed, the proper t score to be used with the formula is 3.03. Continue on extra paperif needed.

Answer 1

anova table
	SS	df	MS	F	p-value	F-critical
Between:	110.000	3	36.667	16.858	0.0000	3.239
Within:	34.800	16	2.175
Total:	144.800	19
					α =	0.05

1)Ho: µ1=µ2=µ3=µ4
2) H1: not all means are equal
3) test stat = 16.858

4) p value=0.000

5)

   Decision:   p-value<α , reject null hypothesis
      
conclusion :    there is enough evidence of significant mean difference among three treatments  

b)

Level of significance=	0.05/6 = 0.0083
no. of treatments,k=	4
DF error =N-k=	16
MSE=	2.1750
t-critical value,t(α/2,df)=	3.0083

critical value=tα/2,df √(MSE(1/ni+1/nj)) = 2.81

population mean difference	critical value			result
µ1-µ2	4.800	2.80599			means are different
µ1-µ3	2.600	2.81			means are not different
µ1-µ4	6.200	2.81			means are different
µ2-µ3	2.200	2.81			means are not different
µ2-µ4	1.400	2.81			means are not different
µ3-µ4	3.600	2.81			means are different