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Jul 8, 2010 7:39 PM
(1010 views)

i did a multiway ANOVA using 4 crossed factors . JMP shows a significant difference but i noticed that within a factor the standard deviation is exactly the same for all the variables

i guess there must be something wrong but i don't know what i did wrong?

would someone have any idea of what could cause that? or maybe some one knows how the standard deviation is calculated in the multiway ANOVA?

Thanks in advance

4 REPLIES

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Jul 9, 2010 12:56 AM
(910 views)

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Jul 9, 2010 7:34 AM
(910 views)

all those factors are crossed, given for each orientation (N,S,E,W), 3 sample were taken, at respectively 0, 1 and 2 m from the trunk. and for each of these distances the cocons were counted in the litter, between 0-5 cm and between 5-10cm.

this design was repeated three time per station.

so at the end i have 3 samples*3 depths* 3 distances* 4 orientations (= 108 samples) for each of the 4 stations.

and i think you are right because for example the standard error is the same for the station and the orientation which both contain 4 levels.

but is it the standard error that allows the ANOVA to say whether or not there is a difference between the factors?

thanks again

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Jul 9, 2010 3:07 PM
(910 views)

Assuming this is a fixed-effects model (and I'm assuming here that it is), it sounds as if you've got 432 observations - probably in a data table with 432 rows - in a four-level factorial design, i.e. 4 Stations x 3 Depths x 4 Orientations x 3 Distances, with three reps of each one. If those three replicates are simply different random samples (e.g. there's nothing systematic to distinguish the first from the second, or the second from the third) you shouldn't include Replicate Number in the experimental design, so for a factorial up to and including all two-factor interactions only (I've simplified it to make the example a bit easier), the breakdown of the degrees of freedom should look like this:

Station 3

Depth 2

Orientation 3

Distance 2

Station*Depth 6

Station*Orientation 9

Station*Distance 6

Depth*Orientation 6

Depth*Distance 4

Orientation*Distance 6

Model (i.e. the sum of the above) has 47 degrees of freedom

Corrected Total has 431 degrees of freedom (i.e. no. of observations minus 1)

Error has 384 degrees of freedom (i.e. the difference between the two above)

Incidentally, don't forget to ensure that all of the factors you put into the ANOVA have their modelling types set to "Nominal" before you run the analysis. This will only actually matter if one or more of them has been recorded as a number (because character variables will be set to Nominal by default), but if any of them

Moving on to the standard errors of the means, you should find that each one has been calculated as the square root of (the residual mean square divided by the number of observations with that factor level) - and

SE(any Station mean) = sqrt(S² / (432/4))

SE(any Depth mean) = sqrt(S² / (432/3))

SE(any Orientation mean) = sqrt(S² / (432/4))

SE(any Distance mean) = sqrt(S² / (432/3))

I don't have your data here so I can't verify the above equations directly on that, but I've checked them with a dummy data set of my own, and they work.

In the above, "S²" is the Error Mean Square in the Analysis of Variance table, which is the variance of the "random noise" left over when everything else has been fitted.

Does that help?

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Jul 9, 2010 7:34 PM
(910 views)

yes it helps a lot, it made everything clear in my mind. i really appreciate your help

thanks again

thanks again