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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 |
Sum |
Degrees |
Mean |
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.
H_{0}: At least two of the population means are
equal.
H_{a}: At least two of the population means are
different.H_{0}: μ_{1} =
μ_{2} = μ_{3}
H_{a}: Not all the population means are
equal. H_{0}:
μ_{1} = μ_{2} =
μ_{3}
H_{a}: μ_{1} ≠
μ_{2} ≠
μ_{3}H_{0}: Not all the
population means are equal.
H_{a}: μ_{1} =
μ_{2} =
μ_{3}H_{0}:
μ_{1} ≠ μ_{2} ≠
μ_{3}
H_{a}: μ_{1} =
μ_{2} = μ_{3}
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 H_{0}. There is not sufficient evidence to conclude that the means of the three assembly methods are not equal.Reject H_{0}. There is not sufficient evidence to conclude that the means of the three assembly methods are not equal. Do not reject H_{0}. There is sufficient evidence to conclude that the means of the three assembly methods are not equal.Reject H_{0}. 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:
H_{a}: Not all the population means are equal. H_{0}: μ_{1} = μ_{2} = μ_{3}
(c)
Test Statistic is given by:
F = 2260/229.26=9.86
(d)
By Technology,
p - value = 0.0006
(e)
Correct option:
.Reject H_{0}. There is sufficient evidence to conclude that the means of the three assembly methods are not equal.
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