Power Analysis For One-Way ANOVA, Independent Samples, SAS
Background
Although Proc Power has a “onewayanova” option, I prefer to approach this
analysis as a multiple regression. If you use the “onewayanova” option, you must
specify the effect size in terms of group means and standard deviations. Psychologists
are accustomed to reporting the effect size in terms of Cohen’s f.
f2
Cohen’s f can be converted into
:  
. Cohen considered an f of .10
1 f 2
to be a small effect, .25 a medium effect, and .40 a large effect. These benchmarks
correspond to 2 values of .0099, .0588, and .1379.
The one-way ANOVA is no different from a multiple regression analysis where
the predictors are k-1 dummy variables, where k is the number of groups in the
analysis. The R2 from such a multiple regression analysis is identical to 2 in ANOVA.
Solving for Sample Size
Suppose that we wish to determine how many subjects we need to have 80%
power for an ANOVA with three groups. Assume that we are using the usual .05
criterion of statistical significance. Here is the SAS code and output:
2
2
proc power;
multreg
model = fixed
nfullpred = 2
nredpred= 0
RsqFull = 0.0099, 0.0588, .1379
RsqRed = 0
ntotal = .
power = 0.8;
run;
Notice that I specified “fixed.” Power depends, in part, on whether the predictor
variables are fixed or random. I nearly all cases psychologists assume that the
predictors in ANOVA are fixed. Since we have three groups, we have 3-1 = 2 predictors
(the treatment df from the ANOVA), . We are testing our model versus a model (the null
hypothesis) that has no parameters other than the intercept (the overall mean), so
nredpred= 0. As values for 2 (RsqFull), I have entered those corresponding to
Cohen’s benchmarks for small, medium, and large. The value of RsqRed is that of the
null hypothesis, 0. Since I want to solve for sample size, I specify “.” for ntotal.
The POWER Procedure
Type III F Test in Multiple Regression
Fixed Scenario Elements
Method
Model
Number of Predictors in Full Model
Number of Predictors in Reduced Model
R-square of Reduced Model
Exact
Fixed X
2
0
0
Nominal Power
Alpha
0.8
0.05
Computed N Total
Index
R-square
Full
Actual
Power
N
Total
1
2
0.0099
0.0588
0.800
0.802
967
158
3
0.1379
0.805
64
Let’s compare our solution with that provided by G*Power.
F tests - ANOVA: Fixed effects, omnibus, one-way
Analysis:
A priori: Compute required sample size
Input:
Effect size f
= 0.1
α err prob
= 0.05
Power (1-β err prob)
= 0.8
Number of groups
= 3
Noncentrality parameter λ
= 9.6900000
Critical F
= 3.0050418
Numerator df
= 2
Denominator df
= 966
Total sample size
= 969
Actual power
= 0.8011010
Output:
SAS told us we need 967 subjects. That is 967/3 = 322.3 per group. Rounding
up to 323 per group and multiplying by 3 give us the 969 reported by G*Power.
Input:
Output:
Effect size f
= 0.25
α err prob
= 0.05
Power (1-β err prob)
= 0.8
Number of groups
= 3
Noncentrality parameter λ
= 9.9375000
Critical F
= 3.0540042
Numerator df
= 2
Denominator df
= 156
Total sample size
= 159
Actual power
= 0.8048873
SAS told us 158 subjects, = 52.67 per group; 3(53) = 159, as reported by
G*Power.
Input:
Effect size f
= 0.4
α err prob
= 0.05
Power (1-β err prob)
= 0.8
Number of groups
= 3
Output:
Noncentrality parameter λ
= 10.5600000
Critical F
= 3.1428085
Numerator df
= 2
Denominator df
= 63
Total sample size
= 66
Actual power
= 0.8180744
SAS told us 64 subjects, = 21.3 per group; 3(22) = 66, as reported by G*Power.
Suppose we intend to have five groups. Changing one line of SAS code,
nfullpredictors = 4, we obtain:
Computed N Total
Input:
Output:
Input:
Output:
Input:
Output:
Index
R-square
Full
Actual
Power
N
Total
1
2
0.0099
0.0588
0.800
0.800
1199
196
3
0.1379
0.803
80
Effect size f
= 0.1
α err prob
= 0.05
Power (1-β err prob)
= 0.8
Number of groups
= 5
Noncentrality parameter λ
= 12.0000000
Critical F
= 2.3793764
Numerator df
= 4
Denominator df
= 1195
Total sample size
= 1200
Actual power
= 0.8006464
Effect size f
= 0.25
α err prob
= 0.05
Power (1-β err prob)
= 0.8
Number of groups
= 5
Noncentrality parameter λ
= 12.5000000
Critical F
= 2.4179625
Numerator df
= 4
Denominator df
= 195
Total sample size
= 200
Actual power
= 0.8097710
Effect size f
= 0.4
α err prob
= 0.05
Power (1-β err prob)
= 0.8
Number of groups
= 5
Noncentrality parameter λ
= 12.8000000
Critical F
= 2.4936960
Numerator df
= 4
Denominator df
= 75
Total sample size
= 80
Actual power
= 0.8030845
As you can see, the SAS analysis is equivalent to that provided by G*Power.
Solving for Power
Suppose that we have four groups with 100 subjects in each group. What is
power?
proc power;
multreg
model = fixed
nfullpredictors = 3
nredpred= 0
RsqFull = 0.0099, 0.0588, .1379
RsqRed = 0
ntotal = 400
power = .;
run;
Computed Power
Index
R-square
Full
Power
1
2
0.0099
0.0588
0.355
0.993
3
0.1379
>.999
F tests - ANOVA: Fixed effects, omnibus, one-way
Analysis:
Post hoc: Compute achieved power
Input:
Effect size f
= 0.1
α err prob
= 0.05
Total sample size
= 400
Number of groups
= 4
Noncentrality parameter λ
= 4.0000000
Critical F
= 2.6274408
Numerator df
= 3
Denominator df
= 396
Power (1-β err prob)
= 0.3552592
Effect size f
= 0.25
α err prob
= 0.05
Total sample size
= 400
Number of groups
= 4
Noncentrality parameter λ
= 25.0000000
Critical F
= 2.6274408
Numerator df
= 3
Denominator df
= 396
Power (1-β err prob)
= 0.9928175
Effect size f
= 0.4
α err prob
= 0.05
Output:
Input:
Output:
Input:
Output:
Total sample size
= 400
Number of groups
= 4
Noncentrality parameter λ
= 64.0000000
Critical F
= 2.6274408
Numerator df
= 3
Denominator df
= 396
Power (1-β err prob)
= 1.0000000
Return to Wuensch’s SAS Lessons Page
Karl L. Wuensch, Dept. of Psychology, East Carolina Univ., Greenville, NC 27858 USA
February, 2010