Question

You may need to use the appropriate technology to answer this 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 = 10,710; SSTR = 4,520.

(a)

Set up the ANOVA table for this problem. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.)

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

(b)

Use α = 0.05 to test for any significant difference in the means for the three assembly methods.

State the null and alternative hypotheses.

H₀: At least two of the population means are equal.
H_a: At least two of the population means are different.H₀: μ₁ = μ₂ = μ₃
H_a: Not all the population means are equal.    H₀: μ₁ = μ₂ = μ₃
H_a: μ₁ ≠ μ₂ ≠ μ₃H₀: Not all the population means are equal.
H_a: μ₁ = μ₂ = μ₃H₀: μ₁ ≠ μ₂ ≠ μ₃
H_a: μ₁ = μ₂ = μ₃

Find the value of the test statistic. (Round your answer to two decimal places.)

Find the p-value. (Round your answer to four decimal places.)

p-value =

State your conclusion.

Do not reject H₀. There is not sufficient evidence to conclude that the means of the three assembly methods are not equal.Reject H₀. There is not sufficient evidence to conclude that the means of the three assembly methods are not equal.    Do not reject H₀. There is sufficient evidence to conclude that the means of the three assembly methods are not equal.Reject H₀. There is sufficient evidence to conclude that the means of the three assembly methods are not equal.

Answer 1

(a)

ANOVA Table is set up asfollows:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p - value
Treatments	4520	k - 1 = 3 - 1 = 2	4520/2=2260	2260/229.26=9.86	0.0006
Error	6190	29- 2 = 27	6190/27=229.26
Total	10710	N- 1 = 30 - 1 = 29

(b)

Correct option:

H_a: Not all the population means are equal.    H₀: μ₁ = μ₂ = μ₃

(c)

Test Statistic is given by:

F = 2260/229.26=9.86

(d)

By Technology,

p - value = 0.0006

(e)

Correct option:

.Reject H₀. There is sufficient evidence to conclude that the means of the three assembly methods are not equal.