Please do these calculations by hand and not in minitab Problem 3-23 from Design
ID: 3208149 • Letter: P
Question
Please do these calculations by hand and not in minitab
Problem 3-23 from Design and Analysis of Experiments (Montgomery, 8th Edition): The effective life of insulating fluids at an accelerated load of 35 kVis being studied. Test data have been obtained for four types of fluids. The results were as follows: (a) Is there any indication that the fluids differ? Set up appropriate hypotheses and assumptions for investigating this issue. (b) By hand calculation, test the appropriate hypotheses using alpha = 0.05. What are your conclusions? (c) By hand calculation, find the P-value for the F statistic calculated in part (c) (d) Confirm your answers above using Minitab. Do the simultaneous confidence intervals identified differ significantly from what you calculated in part (d)? (e) Analyze the residuals and draw conclusions about model adequacy.Explanation / Answer
Part a
There is no indication that the fluids differ because average value of four types of fluid are approximately near to each other.
Here, we have to use the one way analysis of variance for checking the claim or hypothesis whether there is any significant difference in the average life of fluid at 35kV load. The null and alternative hypotheses are given as below:
Null hypothesis: H0: There is no any significant difference in the average life of four types of fluid at 35kV load.
Alternative hypothesis: Ha: There is a significant difference in the average life of four types of fluid at 35kV load.
Part b
Here, we have to calculate the ANOVA table.
The calculation table is given as below:
X1
X2
X3
X4
X1^2
X2^2
X3^2
X4^2
17.6
16.9
21.4
19.3
339.68
285.61
457.96
372.49
18.9
15.3
23.6
21.1
398.79
234.09
556.96
445.21
16.3
18.6
19.4
16.9
275.47
345.96
376.36
285.61
17.4
17.1
18.5
17.5
304.5
292.41
342.25
306.25
20.1
19.5
20.5
18.3
367.83
380.25
420.25
334.89
21.6
20.3
22.3
19.8
427.68
412.09
497.29
392.04
Total
111.9
107.7
125.7
112.9
2113.95
1950.41
2651.07
2136.49
ANOVAs: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
Fluid Type 1
6
111.9
18.65
3.811
Fluid Type 2
6
107.7
17.95
3.439
Fluid Type 3
6
125.7
20.95
3.531
Fluid Type 4
6
112.9
18.81667
2.417667
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
30.165
4-1=3
30.165/3=10.055
10.055/3.2996=3.047278
0.052463
3.098391
Within Groups
65.99333
23-3=20
65.99/20=3.299667
Total
96.15833
24-1=23
F test statistic is less than F critical, so we do not reject the null hypothesis that there is no any significant difference in the average life of four types of fluid at 35kV load.
Part c
We are given
F = 10.055/3.2996 = 3.0473
DF for between = 3
DF for within = 20
So, P-value = 0.052 by using F-table
Part d
The Minitab output is given as below:
Welcome to Minitab, press F1 for help.
One-way ANOVA: X1, X2, X3, X4
Analysis of Variance
Source DF SS MS F P
Factor 3 30.17 10.06 3.05 0.052
Error 20 65.99 3.30
Total 23 96.16
Individual 95% CIs For Mean
Based on Pooled StDev
Level N Mean StDev --------+---------+---------+--------
X1 6 18.650 1.952 (------*-------)
X2 6 17.950 1.854 (-------*------)
X3 6 20.950 1.879 (-------*------)
X4 6 18.817 1.555 (-------*-------)
--------+---------+---------+--------
Pooled StDev = 1.816 18.0 20.0 22.0
From this output, it is observed that ANOVA is correct.
X1
X2
X3
X4
X1^2
X2^2
X3^2
X4^2
17.6
16.9
21.4
19.3
339.68
285.61
457.96
372.49
18.9
15.3
23.6
21.1
398.79
234.09
556.96
445.21
16.3
18.6
19.4
16.9
275.47
345.96
376.36
285.61
17.4
17.1
18.5
17.5
304.5
292.41
342.25
306.25
20.1
19.5
20.5
18.3
367.83
380.25
420.25
334.89
21.6
20.3
22.3
19.8
427.68
412.09
497.29
392.04
Total
111.9
107.7
125.7
112.9
2113.95
1950.41
2651.07
2136.49