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 is 10, 760 SSTR is 4,520

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
Error
Total

b.Use a= 0.05 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 -value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 12

What is your conclusion?

Answer 1

Part A:

The filled table is given below:

SSE = SST - SSTR = 10760 - 4520 = 6240

Degree of Freedom (error) = n -1 = 30 - 1 = 29 (total members per sample - 1)

Degree of Freedom (treatment) = k -1 = 3 - 1 = 2 (total groups - 1)

MSE = SSE/df

MSTR = SSTR/df

F = MSTR/MSE

The F table is given below:

So, the P value is between Less than 0.01

Exact P value from Excel is 0.00037

Part B:

Here, the hypothesis are:

H₀: U₁ = U₂ = U₃ (Null Hypothesis)

H_a: All the means are not equal (Alternate Hypothesis)

U is the mean...

As the P value is less than 0.05, it can be stated that there occurs a significant difference in the means of all the three groups. In this case the null hypothesis is not accepted that means the means of all the three groups are not equal.

End of the Solution...