Question

# You may need to use the appropriate technology to answer this question. Three different methods for...

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.
Ha: At least two of the population means are different.H0: μ1 = μ2 = μ3
Ha: Not all the population means are equal.    H0: μ1 = μ2 = μ3
Ha: μ1μ2μ3H0: Not all the population means are equal.
Ha: μ1 = μ2 = μ3H0: μ1μ2μ3
Ha: μ1 = μ2 = μ3

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

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.    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.

(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:

Ha: 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.

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