Question

6. +-116.7 points ASWSBE13 13.E.019 My Notes Ask Your Teacher You may need to use the appropriate appendix table or technology to answer this question To test for any significant difference in the number of hours between breakdowns for four machines, the following data were obtained Machine Machine Machine Machine 2 3 4 6.4 8.9 9.8 12.6 12.0 7.8 5.4 7.3 8.3 10.9 10.1 9.4 10.1 8.8 8.3 7.5 9.7 10.7 11.2 10.9 10.2 9.3 10.2 Use the Bonferroni adjustment to test for a significant difference between all pairs of means. Assume that a maximum overall experimentwise error rate of 0.05 is desired Find the value of LSD. (Round your comparisonwise error rate to three decimal places. Round your answer to two decimal places LSD Find the pairwise absolute difference between sample means for each pair of machines 1 2 4 2 3 4 4

Answer 1

One-way ANOVA: machine 1, machine 2, machine 3, machine 4

Method

Null hypothesis         All means are equal
Alternative hypothesis At least one mean is different
Significance level      α = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor Levels Values
Factor       4 machine 1, machine 2, machine 3, machine 4

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value
Factor   3   51.24 17.080    17.01    0.000
Error   20   20.08   1.004
Total   23   71.32

Model Summary

      S    R-sq R-sq(adj) R-sq(pred)
1.00200 71.85%     67.62%      59.46%

Means

Factor     N    Mean StDev       95% CI
machine 1 6   7.100 1.043 ( 6.247, 7.953)
machine 2 6   9.300 1.018 ( 8.447, 10.153)
machine 3 6   9.600 0.955 ( 8.747, 10.453)
machine 4 6 11.200 0.990 (10.347, 12.053)

Pooled StDev = 1.00200

Tukey Pairwise Comparisons

Grouping Information Using the Tukey Method and 95% Confidence

Factor     N    Mean Grouping
machine 4 6 11.200 A
machine 3 6   9.600 A B
machine 2 6   9.300    B
machine 1 6   7.100      C

Means that do not share a letter are significantly different.

Tukey Simultaneous Tests for Differences of Means

                       Difference       SE of                            Adjusted
Difference of Levels     of Means Difference       95% CI      T-Value   P-Value
machine 2 - machine 1       2.200       0.579 ( 0.580, 3.820)     3.80     0.006
machine 3 - machine 1       2.500       0.579 ( 0.880, 4.120)     4.32     0.002
machine 4 - machine 1       4.100       0.579 ( 2.480, 5.720)     7.09     0.000
machine 3 - machine 2       0.300       0.579 (-1.320, 1.920)     0.52     0.954
machine 4 - machine 2       1.900       0.579 ( 0.280, 3.520)     3.28     0.018
machine 4 - machine 3       1.600       0.579 (-0.020, 3.220)     2.77     0.054

Individual confidence level = 98.89%

Tukey Simultaneous 95% CIs

Fisher Pairwise Comparisons

Grouping Information Using the Fisher LSD Method and 95% Confidence

Factor     N    Mean Grouping
machine 4 6 11.200 A
machine 3 6   9.600    B
machine 2 6   9.300    B
machine 1 6   7.100      C

Means that do not share a letter are significantly different.

Fisher Individual Tests for Differences of Means

                      Difference       SE of                            Adjusted
Difference of Levels     of Means     Difference       95% CI      T-Value   P-Value
machine 2 - machine 1       2.200       0.579 ( 0.993, 3.407)     3.80     0.001
machine 3 - machine 1       2.500       0.579 ( 1.293, 3.707)     4.32     0.000
machine 4 - machine 1       4.100       0.579 ( 2.893, 5.307)     7.09     0.000
machine 3 - machine 2       0.300       0.579 (-0.907, 1.507)     0.52     0.610
machine 4 - machine 2       1.900       0.579 ( 0.693, 3.107)     3.28     0.004
machine 4 - machine 3       1.600       0.579 ( 0.393, 2.807)     2.77     0.012

treatments differ significantly
option 1),2),3),5),6),