Question

In: Statistics and Probability

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, 36 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 12 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 = 12,960; SSTR = 4,590.

(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: 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.)

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

p-value =

State your conclusion.

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.    

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 not sufficient evidence to conclude that the means of the three assembly methods are not equal.

Solutions

Expert Solution

Solution:

Given:

k = Number of methods for assembling a product = 3

N = total observations = 36

SST = 12,960;

SSTR = 4,590.

Part a) Set up the ANOVA table for this problem

SSE= SST - SSTR

SSE = 12960 - 4590

SSE = 8370

dftreatments = k - 1

dftreatments = 3- 1

dftreatments = 2

dftotal = N - 1

dftotal = 36 - 1

dftotal = 35

and

dferror = dftotal - dftreatments   

dferror = 35 - 2

dferror = 33

Mean Squares:

MSTR = SSTR /  dftreatments   

MSTR =  4590 / 2

MSTR = 2295

MSE = SSE / dferror

MSE = 8370 / 33

MSE = 253.64

F test statistic:

F = MSTR / MSE

F = 2295 / 253.64

F = 9.05

P-value:

To find P-value use Excel command:

=F.DIST.RT(F test statistic , dftreatments , dferror )

=F.DIST.RT(9.05,2,33)

=0.0007

Thus P-value = 0.0007

Thus we get:

Source SS DF MS F P-value
Treatments 4590 2 2295 9.05 0.0007
Error 8370 33 253.64
Total 12960 35

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

Find the value of the test statistic.

F = 9.05

Find the p-value.

P-value = 0.0007

State your conclusion.

Since P-value = 0.0007 < 0.05 level of significance , we reject H0.

Thus :

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


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