repeated measures ('within subjects', related samples) versus 'between subjects' (nonrelated) ANOVA

[frank.d...@gmail.com]

Hi there,

Just an idea that occurred to me while doing some statistical work: could it be that any repeated measures x-way ANOVA is equivalent to a (x+1)-way ANOVA for nonrelated samples - just by factoring in the entities in whom the repetition occurs, as an independent factor? With "is equivalent to" I mean "gives the same ANOVA tables".
I would be grateful if you share your thoughts on this.

Kind regards,

Frank

[Ray Koopman]

On Apr 21, 12:21 pm, frank.degee...@gmail.com wrote:
>
> Just an idea that occurred to me while doing some statistical work: could it be that any repeated measures x-way ANOVA is equivalent to a (x+1)-way ANOVA for nonrelated samples - just by factoring in the entities in whom the repetition occurs, as an independent factor? With "is equivalent to" I mean "gives the same ANOVA tables".

The 2^x - 1 SS's and df's involving only the x between-subjects
factors will be the same in both analyses, but the SS and df for the
Subjects main effect and the 2^x - 1 interactions with Subjects in the
repeated-measures analysis will all be pooled into the within-cell SS
and df in the independent-groups analysis.

[Jeff Miller]

On Monday, April 22, 2013 9:11:28 AM UTC+12, Ray Koopman wrote:
> On Apr 21, 12:21 pm, frank.degee...@gmail.com wrote:
>
> >
>
> > could it be that any repeated measures x-way ANOVA is equivalent to a (x+1)-way ANOVA for nonrelated samples

> The 2^x - 1 SS's and df's involving only the x between-subjects
> factors will be the same in both analyses, but the SS and df for the
> Subjects main effect and the 2^x - 1 interactions with Subjects in the
> repeated-measures analysis will all be pooled into the within-cell SS
> and df in the independent-groups analysis.

As I interpret the OP's question, the (x+1) way ANOVA includes a
subjects factor and its interactions with the other factors,
so these will be separated in the analysis--not
pooled into the within-cell SS and df.

On that interpretation, the answer to the OP's question is that
the two analyses will give the same SS & df breakdowns, but:
(a) the "independent groups ANOVA" can't really be carried
through to the end. Specifically, F's can't be computed,
because is no estimate of the within-cell error term
(i.e., no df's are left over for this error term, because
the subjects factor and its interactions will have used
them all).
(b) even if you had an estimate of within-cell error, the "independent
groups ANOVA" would be inappropriate because it would use this
same error term in the denominator of all F's rather than using
the appropriate treatment X S interaction error term that would be
used by the appropriate repeated-measures analysis.

[Rich Ulrich]

I've always been willing to say, Yes, they are basically the same
thing. It is useful to know, especially if you have to analyze RM
data where a few data points are missing -- Use ANOVA.

But you do have to be careful to pull out and use the
right terms as error terms for tests.


--
Rich Ulrich

[Ray Koopman]

On Apr 22, 6:01 pm, Jeff Miller <milleratot...@yahoo.com> wrote:
> On Monday, April 22, 2013 9:11:28 AM UTC+12, Ray Koopman wrote:
>> On Apr 21, 12:21 pm, frank.degee...@gmail.com wrote:
>>
>>> could it be that any repeated measures x-way ANOVA is

>>> equivalent to a (x+1)-way ANOVA for nonrelated samples.

>>
>> The 2^x - 1 SS's and df's involving only the x between-subjects
>> factors will be the same in both analyses, but the SS and df for
>> the Subjects main effect and the 2^x - 1 interactions with Subjects
>> in the repeated-measures analysis will all be pooled into the
>> within-cell SS and df in the independent-groups analysis.
>
> As I interpret the OP's question, the (x+1) way ANOVA includes
> a subjects factor and its interactions with the other factors,
> so these will be separated in the analysis--not
> pooled into the within-cell SS and df.

You're right. The question I answered was
::: could it be that any repeated measures (x+1)-way ANOVA is
::: equivalent to an x-way ANOVA for nonrelated samples.

[frank.d...@gmail.com]

On Tuesday, April 23, 2013 3:01:10 AM UTC+2, Jeff Miller wrote:
> On Monday, April 22, 2013 9:11:28 AM UTC+12, Ray Koopman wrote: > On Apr 21, 12:21 pm, frank.degee...@gmail.com wrote: > > > > > > could it be that any repeated measures x-way ANOVA is equivalent to a (x+1)-way ANOVA for nonrelated samples > The 2^x - 1 SS's and df's involving only the x between-subjects > factors will be the same in both analyses, but the SS and df for the > Subjects main effect and the 2^x - 1 interactions with Subjects in the > repeated-measures analysis will all be pooled into the within-cell SS > and df in the independent-groups analysis. As I interpret the OP's question, the (x+1) way ANOVA includes a subjects factor and its interactions with the other factors, so these will be separated in the analysis--not pooled into the within-cell SS and df. On that interpretation, the answer to the OP's question is that the two analyses will give the same SS & df breakdowns, but: (a) the "independent groups ANOVA" can't really be carried through to the end. Specifically, F's can't be computed, because is no estimate of the within-cell error term (i.e., no df's are left over for this error term, because the subjects factor and its interactions will have used them all). (b) even if you had an estimate of within-cell error, the "independent groups ANOVA" would be inappropriate because it would use this same error term in the denominator of all F's rather than using the appropriate treatment X S interaction error term that would be used by the appropriate repeated-measures analysis.

Thanks,

Frank

[frank.d...@gmail.com]

On Tuesday, April 23, 2013 4:19:54 AM UTC+2, Rich Ulrich wrote:
> On Mon, 22 Apr 2013 18:01:10 -0700 (PDT), Jeff Miller <miller...@yahoo.com> wrote: >On Monday, April 22, 2013 9:11:28 AM UTC+12, Ray Koopman wrote: >> On Apr 21, 12:21 pm, frank.degee...@gmail.com wrote: >> >> > >> >> > could it be that any repeated measures x-way ANOVA is equivalent to a (x+1)-way ANOVA for nonrelated samples > >> The 2^x - 1 SS's and df's involving only the x between-subjects >> factors will be the same in both analyses, but the SS and df for the >> Subjects main effect and the 2^x - 1 interactions with Subjects in the >> repeated-measures analysis will all be pooled into the within-cell SS >> and df in the independent-groups analysis. > >As I interpret the OP's question, the (x+1) way ANOVA includes a >subjects factor and its interactions with the other factors, >so these will be separated in the analysis--not >pooled into the within-cell SS and df. > >On that interpretation, the answer to the OP's question is that >the two analyses will give the same SS & df breakdowns, but: >(a) the "independent groups ANOVA" can't really be carried >through to the end. Specifically, F's can't be computed, >because is no estimate of the within-cell error term >(i.e., no df's are left over for this error term, because >the subjects factor and its interactions will have used >them all). >(b) even if you had an estimate of within-cell error, the "independent >groups ANOVA" would be inappropriate because it would use this >same error term in the denominator of all F's rather than using >the appropriate treatment X S interaction error term that would be >used by the appropriate repeated-measures analysis. I've always been willing to say, Yes, they are basically the same thing. It is useful to know, especially if you have to analyze RM data where a few data points are missing -- Use ANOVA. But you do have to be careful to pull out and use the right terms as error terms for tests. -- Rich Ulrich

Thanks,

Frank