Question

Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST =10,800; SSTR =4,500. a. Set up the ANOVA table for this problem (to 2 decimals, if necessary). Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value (to 4 decimals) Treatments 9.87 0.0006 4560 6240 Ⓡ 2 27 ♡ 2280 * 231.11 > Error Total 10800 29 b.Use a = .05 to test for any significant difference in the means for the three assembly methods. Calculate the value of the test statistic (to 2 decimals). 9.87 The p-value is between .01 and .025 ) What is your conclusion? Conclude not all means of the three assembly methods are equal

Answer 1

Ans:

a)SS(error)=10800-4500=6300

Source	SS	df	MS	F	p-value
treatments	4500	2	2250.00	9.64	0.0007
error	6300	27	233.33
Total	10800	29

b)

Test statistic:

F=9.64

p-value=0.0007

p-value is less than 0.01

Reject the null hypothesis and conclude that not all the means of the three assembly methods are equal.