Question

Exercise 13.07 Algorithmic as Question 3 of 14 Check My Work eBook 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 10,890; SSTR -4,590. a. Set up the ANOVA table for this problem (to 2 decimals, if necessary). p-value (to 4 decimals) F Source of Variation Sum of Squares Degrees of Freedom Mean Square Treatments Error Total Ob.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). The p-value is Select your answer- What is your conclusion? Select your answer- Check My Work

Answer 1

(a) There are 3 treatments, therefore k = 3 and DF1 = 3 - 1 = 2

Total Observation = 30. DF2 = N - k = 30 - 3 = 27

Given SSTR = 4590

MSTR = SSTR / DF1 = 4590 / 2 = 2295

SS error = SST - SSTR = 10890 - 4590 = 6300

MS error = SS error / DF2 = 6300 / 27 = 233.33

Therefore F = MSTR / MS error = 2295 / 233.33 = 9.84

The Complete ANOVA table is below

Source	Sum of Squares	DF	Mean Square	F	p value
SSTR	4590	2	2295.00	9.84	0.0006
Error	6300	27	233.33
Total	10890	29

(b) The Test Statistic = F = 9.84

The p value is p < 0.001

Conclusion: There is sufficient evidence to conclude that there is a significant difference in the means for the three assembly methods.