Download One-Factor Analysis of ANOVA - California State University, East Bay

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One-Factor Analysis of ANOVA
Reading Assignment: O/L Sections 8.1–8.4
Model: Yij = i + eij, i = 1, ..., g = 4, j = 1, ..., n = 10; eij iid NORM(0, ).
Note: Simulated data here: 1 = 110, 2 = 110, 3 = 120, 4 = 100, and  = 10.
Data: Entered into c1 - c4 of MTB worksheet ("Unstacked" format).
Data Display
Row
1
2
3
4
5
6
7
8
9
10
D1
103
112
113
103
118
123
114
105
124
103
D2
111
126
117
101
99
131
108
114
103
128
D3
107
113
137
127
134
140
103
128
125
99
Pl
82
112
113
89
92
107
116
106
105
91
One-way ANOVA: Response versus Group
Copyright © 2007 by Bruce E. Trumbo. All rights reserved. Department of Statistics and Biostatistics. California State University East Bay (Hayward). This is a draft. Corrections/Permissions: [email protected]
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Source
Group
Error
Total
Level
1
2
3
4
DF
3
36
39
N
10
10
10
10
SS
2043
5001
7044
Mean
111.80
113.80
121.30
101.30
MS
681
139
F
4.90
StDev
8.15
11.54
14.74
11.80
P
0.006
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(-------*------)
(-------*------)
(------*-------)
(------*-------)
------+---------+---------+---------+--100
110
120
130
Pooled StDev = 11.79
Copyright © 2007 by Bruce E. Trumbo. All rights reserved. Department of Statistics and Biostatistics. California State University East Bay (Hayward). This is a draft. Corrections/Permissions: [email protected]
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Problems:
1) Compute the ANOVA table for the original simulation by hand from the data table and
compare your results with those from Minitab.
The possible differences are due to rounding differently.
2) Make a column of the 40 residuals using Minitab. Verify several values.
Residuals can be calculated and stored when running ANOVA in Minitab. See the handout for
One-way Anova for detail. And check a few of the residuals by the definition of residual:
ˆij  yij  y j .
3) Do several simulations with the same parameters as shown, except  = 15. Comment on the
results.
Please refer to p. 185 of the textbook O/L for how to generate random data. Here is the brief
version:
Minitab: Calc>> random data>> normal and type the required number of observations as the
number of rows and the required mean and standard deviation (15). Repeat this for each of the four
groups. Then run Anova and see if we can detect the difference among groups.
4) How big would n (for each group) have to be in order for the F-test to have 95% power?
Copyright © 2007 by Bruce E. Trumbo. All rights reserved. Department of Statistics and Biostatistics. California State University East Bay (Hayward). This is a draft. Corrections/Permissions: [email protected]
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Minitab: Stat>>Power and sample size>> one-way anova and type in the required number of
levels (groups) and other information and leave sample size blank.
Power and Sample Size
One-way ANOVA
Alpha = 0.05
SS
Means
200
Assumed standard deviation = 15
Number of Levels = 4
Sample Target
Maximum
Size Power Actual Power Difference
21
0.95
0.956730
20
When the maximum difference in group means is 20 and  = 15, then sample size should be 21 for
each group.
5) Do a simulation with the population means and variance of the original example.
Also use n1 = n2 = n3 = 10, but use n4 = 15. Explain the entries in the DF column.
Explain why the CIs for the group means have different lengths.
The df of total changes (45-1=44) since the total sample size is increased to be 45. Therefore, the df
of error term changes too: df of total – df of group = 44 - 3=41. The CIs for the D3 groups have the
same width while the CI for the Placebo group is much narrower since the standard deviation of
sample mean is /sqrt(n). The larger the sample size is, the narrower the CI is.
Copyright © 2007 by Bruce E. Trumbo. All rights reserved. Department of Statistics and Biostatistics. California State University East Bay (Hayward). This is a draft. Corrections/Permissions: [email protected]