# Re: [R] Discrepancy between R and SPSS in 2-way, repeated measures ANOVA

From: John Maindonald <john.maindonald_at_anu.edu.au>
Date: Sat 10 Sep 2005 - 22:17:01 EST

There are 20 distinct individuals, right? expno breaks the 20 individuals into five groups of 4, right? Is this a blocking factor?

If expno is treated as a blocking factor, the following is what you get:

> xy <- expand.grid(expno=letters[1:5],cond=letters[1:4],

```+                                    time=factor(paste(1:2)))
> xy\$subj <- factor(paste(xy\$expno, xy\$cond, sep=":"))
> xy\$cond <- factor(xy\$cond)
> xy\$expno <- factor(xy\$expno)
```

> xy\$y <- rnorm(40)
> summary(aov(y~cond*time+Error(expno/cond), data=xy))

Error: expno

Df Sum Sq Mean Sq F value Pr(>F) Residuals 4 3.59 0.90

Error: expno:cond

```           Df Sum Sq Mean Sq F value Pr(>F)
cond       3   1.06    0.35    0.36   0.78
```
Residuals 12 11.86 0.99

Error: Within

```           Df Sum Sq Mean Sq F value Pr(>F)
time       1   2.27    2.27    1.38   0.26
```
cond:time 3 3.27 1.09 0.67 0.59 Residuals 16 26.19 1.64

If on the other hand this is analyzed as for a complete randomized design, the following is the output:

> summary(aov(y~cond*time+Error(subj), data=xy))

Error: subj

```           Df Sum Sq Mean Sq F value Pr(>F)
cond       3   1.06    0.35    0.37   0.78
```
Residuals 16 15.46 0.97

Error: Within

```           Df Sum Sq Mean Sq F value Pr(>F)
time       1   2.27    2.27    1.38   0.26
```
cond:time 3 3.27 1.09 0.67 0.59 Residuals 16 26.19 1.64

On 10 Sep 2005, at 8:00 PM, Larry A Sonna wrote:

> From: "Larry A Sonna" <larry_sonna@hotmail.com>
> Date: 10 September 2005 12:10:06 AM
> To: <r-help@stat.math.ethz.ch>
> Subject: [R] Discrepancy between R and SPSS in 2-way, repeated
> measures ANOVA
>
>
> Dear R community,
>
> I am trying to resolve a discrepancy between the way SPSS and R
> handle 2-way, repeated measures ANOVA.
>
> An experiment was performed in which samples were drawn before and
> after treatment of four groups of subjects (control and disease
> states 1, 2 and 3). Each group contained five subjects. An
> experimental measurement was performed on each sample to yield a
> "signal". The before and after treatment signals for each subject
> were treated as repeated measures. We desire to obtain P values
> for disease state ("CONDITION"), and the interaction between signal
> over time and disease state ("CONDITION*TIME").
>
> Using SPSS, the following output was obtained:
> DF SumSq (Type 3) Mean Sq F
> value P=
>
> COND 3 42861 14287
> 3.645 0.0355
>
> TIME 1 473
> 473 0.175 0.681
>
> COND*TIME 3 975 325
> 0.120 0.947
>
> Error 16 43219 2701
>
>
>
> By contrast, using the following R command:
>
> summary(aov(SIGNAL~(COND+TIME+COND*TIME)+Error(EXPNO/COND),
> Type="III"))
>
> the output was as follows:
>
> Df Sum Sq Mean Sq F value Pr(>F)
>
> COND 3 26516 8839 3.2517 0.03651 *
>
> TIME 1 473 473 0.1739 0.67986
>
> COND:TIME 3 975 325 0.1195 0.94785
>
> Residuals 28 76107 2718
>
>
>
> I don't understand why the two results are discrepant. In
> particular, I'm not sure why R is yielding 28 DF for the residuals
> whereas SPSS only yields 16. Can anyone help?
>
>

John Maindonald email: john.maindonald@anu.edu.au phone : +61 2 (6125)3473 fax : +61 2(6125)5549 Centre for Bioinformation Science, Room 1194, John Dedman Mathematical Sciences Building (Building 27) Australian National University, Canberra ACT 0200.

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