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 = 10,780; SSTR = 4,570.

1. Set up the ANOVA table for this problem (to 2 decimals, if necessary). Round p-value to four decimal places.
   

   Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
   Treatments
   Error
   Total

Answer 1

Define the null and the alternate hypothesis as:

The means of the three groups is same.

The means of atleast one of the three groups is different.

The given study involves a total of persons assigned to different methods (or treatments). Each treatment involves data of 10 persons.

The ANOVA table is computed using the formulas:

Source of Variation	Sum of Squares	Degrees of freedom	Mean Square	F	p-value
Treatments	SSTR
Error
Total	SST

The following information is provided in the problem:

• 
• 
• 
• 

Substituting the given values in the above table:

Source of Variation	Sum of Squares	Degrees of freedom	Mean Square	F	p-value
Treatments
Error
Total

Therefore, the final ANOVA Table is:

Source of Variation	Sum of Squares	Degrees of freedom	Mean Square	F	p-value
Treatments
Error
Total

The p-value is calculated using the formula:

The value of the F-distribution will be calculated using the command "=F.DIST.RT()" in MS-Excel as shown below:

Therefore,

Since the p-value is less than 0.05, hence, there exists sufficient evidence to reject the null hypotthesis at 5% level of significance. Therefore, the means of atleast one of the three groups is not different from the rest.