Question

# Three different methods for assembling a product were proposed by an industrial engineer. To investigate the...

Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 42 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 14 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 = 13,960; SSTR = 4,510.

(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: μ1μ2μ3
Ha: μ1 = μ2 = μ3

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

H0: At least two of the population means are equal.
Ha: At least two of the population means are different.

H0: μ1 = μ2 = μ3
Ha: μ1μ2μ3

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

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

t stat =

p-value =

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.

Reject H0. There is 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.

(a)

The ANOVA Table isset up for the problem as follows:

 Source of Variation Sum of Squares Degrees of Freedom Mean Square F p - value Treatments 4510 k - 1 = 3 - 1 = 2 4510/2 = 2255.00 2255.00/242.31 = 9.31 0.0005 Error 13960 - 4510 = 9450 N - k = 42 - 3 = 39 9450/39 = 242.31 Total 13960 N - 1 = 42 - 1 = 41

(b)

Correct option:

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

(c)

F stat = 2255.00/242.31 = 9.31

(d)

Degrees of Freedom for numerator = 2

Degrees of freedom for denominator = 39

By Technology, p - value = 0.0005

(e)

Correct option:

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