In: Statistics and Probability
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.
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.