Power analysis examples

* ======================================================================= 
*  File:  	power analysis.SPS .
*  Date:  	09-Oct-2003 .
*  Author:  	Bruce Weaver, weaverb@mcmaster.ca .
*  Notes:	Power analysis examples.
* ======================================================================= .

* The following examples are from a paper by D'Amico, Neilands, and
* Zambarano in Behavior Research Methods, Instruments, & Computers,
* 2001, 33(4), 479-484.

* These examples use the MATRIX DATA command to input the data.
* For a brief explanation of how this works, go to
*  http://www.utexas.edu/cc/faqs/stat/spss/spss33.html .


* -------------------------------------------------- .
* Example 1:  ANCOVA with 3 groups and 2 covariates .
* -------------------------------------------------- .

* The data for this example are as follows.

*               Patient   # of     Parent   .
*               Anxiety  Siblings  Anxiety  .
* AGE GROUP      M    SD    M  SD    M   SD .
*  6-12         7.5  1.9    3  1     4  2.3 .
* 13-19         6.8  2.5    2  2     5  1.4 .
* 20-45         7.1  2.1    4  1     6  1.8 .

* Patient Anxiety is the dependent variable.
* Age Group is the independent variable.
* # of siblings and Parent Anxiety are covariates.

* The following syntax reads in these data in matrix format.

matrix data 
 variables = agegroup rowtype_ pat_anx sibnumbr prnt_anx
 /factor = agegroup
 /format = lower nodiagonal.
begin data.
1 mean 7.5 3.0 4.0
1 n 50 50 50
2 mean 6.8 2.0 5.0
2 n 50 50 50
3 mean 7.1 4.0 6.0
3 n 50 50 50
. sd 2.17 1.33 1.83
. corr 0.3
. corr 0.3 0.3
end data.

* NOTE:  	Only 3 SDs are read in, one for the DV, and one for each covariate.
*		These are MEAN standard deviations, i.e., mean of the SDs for the
*		3 age groups.  For Patient Anxiety, for example, the mean SD is
*		(1.9 + 2.5 + 2.1)/3 = 2.17.  Using the mean SD makes sense if
*		the assumption of homogeneous variances is tenable.  You can enter
*		the SDs for each group (like you do the means), but you have to
*		also enter the mean SDs as we did in order for the procedure to
*		work properly.  (I tried it with only the individual group SDs,
*		and got an error message indicating that needed SDs were missing.)

manova
 pat_anx by agegroup(1,3) with prnt_anx sibnumbr
 /method = unique
 /error = within+residual
 /matrix = in(*)
 /power t (.05) F (.05)
 /print signif (mult averf)
 /noprint param(estim).

* For the main effect of Age Group, power = 0.742.
* This is a bit lower than the usual figure one shoots for, 0.80.
* So let's repeat the exercise, but with increased sample sizes.
* The following sets each group size = 60 instead of 50.

matrix data 
 variables = agegroup rowtype_ pat_anx sibnumbr prnt_anx
 /factor = agegroup
 /format = lower nodiagonal.
begin data.
1 mean 7.5 3.0 4.0
1 n 60 60 60
2 mean 6.8 2.0 5.0
2 n 60 60 60
3 mean 7.1 4.0 6.0
3 n 60 60 60
. sd 2.17 1.33 1.83
. corr 0.3
. corr 0.3 0.3
end data.

manova
 pat_anx by agegroup(1,3) with prnt_anx sibnumbr
 /method = unique
 /error = within+residual
 /matrix = in(*)
 /power t (.05) F (.05)
 /print signif (mult averf)
 /noprint param(estim).

* With n=60 per group, Power for the main effect of Age Group = 0.823 .

* ---------------------------------------- .
* Power for One-way ANOVA on the same data .
* ---------------------------------------- .

* Suppose you dropped the two covariates from the analysis,
* and used one-way ANOVA to analyze these data.  Power would 
* be calculated as shown below (for n=60 per group).

matrix data 
 variables = agegroup rowtype_ pat_anx 
 /factor = agegroup .
begin data.
1 mean 7.5 
1 n 60 
2 mean 6.8 
2 n 60 
3 mean 7.1 
3 n 60 
. sd 2.17 
. corr 1
end data.

* NOTE THE FOLLOWING CHANGES TO THE MATRIX DATA COMMAND (compared to previous example).

*   [1] The "/format = lower nodiagonal" line has been removed;
*   [2] The correlation matrix consists of a single line "corr 1".

* We only need one line to represent the correlation matrix, because
* the correlation is of the DV with itself.

manova
 pat_anx by agegroup(1,3)
 /method = unique
 /error = within+residual
 /matrix = in(*)
 /power t (.05) F (.05)
 /print signif (mult averf)
 /noprint param(estim).

* Without the 2 covariates, power drops from .823 to .329.


* ---------------------------------------------------------- .
* Example 2:  MANOVA with 3 groups and 2 dependent variables .
* ---------------------------------------------------------- .

* Research Question:  Are there differences among 3 ethnic groups
* on ratings of risks and benefits of alcohol consumption?

* Independent variable: ethnic group (3 levels).
* Dependent variables:	1) ratings of risks of alcohol consumption
*				2) ratings of benefits of alcohol consumption.

* Read in data in matrix format.

matrix data
 variables = group rowtype_ risk benefit
 /factor = group
 /format = lower nodiagonal.
begin data.
1 mean 4.8 3.9
1 n 20 20
2 mean 5.3 4.0
2 n 20 20
3 mean 5.8 3.5
3 n 20 20 
. sd 1.27 1.4
. corr .3
end data.

manova
 risk benefit by group(1,3)
 /method = unique
 /error = within+residual
 /matrix = in(*)
 /power t (.05) F (.05)
 /print signif (mult averf)
 /noprint param(estim).

* There are 3 Power estimates to examine here:

* 	1) Power = 0.57 for the univariate test for RISK
* 	2) Power = 0.17 for the univariate test for BENEFIT
* 	3) Power = 0.65 to 0.67 for the various multivariate tests .

* You would probably want to add more subjects here, no matter what.
* But how many more would depend on which of the DVs you were
* primarily interested in.  If RISK is of primary concern, and
* you were not too bothered about BENEFIT, you would not need to
* add as many subjects to get the Power for RISK up to acceptable
* levels.  But if you wanted to have 80% power for BENEFIT, you'd
* have to add a lot more.

* Let's try again with n = 30 per group instead of 20.

matrix data
 variables = group rowtype_ risk benefit
 /factor = group
 /format = lower nodiagonal.
begin data.
1 mean 4.8 3.9
1 n 30 30
2 mean 5.3 4.0
2 n 30 30
3 mean 5.8 3.5
3 n 30 30 
. sd 1.27 1.4
. corr .3
end data.

manova
 risk benefit by group(1,3)
 /method = unique
 /error = within+residual
 /matrix = in(*)
 /power t (.05) F (.05)
 /print signif (mult averf)
 /noprint param(estim).

* We now have roughly 85% power for the multivariate test,
* power = 0.77 for RISK, and power = 0.23 for BENEFIT .

* ---------------------------------------------------------- .
* Example 3:  Mixed design (between-within) ANOVA            .
* ---------------------------------------------------------- .

* Dependent variable: 		Ratings of depression .
* Between-subjects factor:  	Group (3 levels).
* Within-subjects factor:	Time (DV measured on 3 occasions).

* The effect of most interest is the Group x Time interaction.
* How many subjects do we need to have 80% power for this term?.

* Read in data in matrix format .

matrix data
 variables = group rowtype_ depress1 depress2 depress3
 /factor = group
 /format = lower nodiagonal.
begin data.
1 mean 11 10 9
1 n 33 33 33
2 mean 11 10 10
2 n 33 33 33
3 mean 11 11 10
3 n 34 34 34
. sd 2.3 2.0 1.8
. corr 0.3
. corr 0.3 0.3
end data.

manova
 depress1 depress2 depress3 by group(1,3)
 /wsfactors depress(3)
 /method = unique
 /error = within+residual
 /matrix = in(*)
 /power t (.05) F (.05)
 /print signif (mult averf)
 /noprint param(estim).

* Power = 0.56 for the interaction term, so we need to increase n.
* Try again with n = 55 per group .

matrix data
 variables = group rowtype_ depress1 depress2 depress3
 /factor = group
 /format = lower nodiagonal.
begin data.
1 mean 11 10 9
1 n 55 55 55
2 mean 11 10 10
2 n 55 55 55
3 mean 11 11 10
3 n 55 55 55
. sd 2.3 2.0 1.8
. corr 0.3
. corr 0.3 0.3
end data.

manova
 depress1 depress2 depress3 by group(1,3)
 /wsfactors depress(3)
 /method = unique
 /error = within+residual
 /matrix = in(*)
 /power t (.05) F (.05)
 /print signif (mult averf)
 /noprint param(estim).

* With n=55 per group, power = .813 for the interaction term.

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