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Jul 8, 2010 7:39 PM
(1415 views)

hello,

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

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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If within any one factor all the levels are equally replicated then the standard errors (and I'm assuming it's the standard errors of the various factor means you're referring to here) *would* all be the same - because they'll all be calculated from the residual mean square of the whole ANOVA (which is the overall estimate of the variance S²) and the number of experimental units in each of the factor levels. If you'd like to post details of the experimental design, i.e. how many factors and the various replication levels, it'll be easy enough to check.

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thanks for your help. i'm testing the effect of different factors on the distribution of cocoons in the ground. the factors are: stations ( 4 different stations) depth ( litter, 0-5cm and 5-10cm), orientation ( N,S,E,W) and the distance from the trunk (0, 1 and 2m).

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

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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Hello again - the ANOVA won't tell you whether there's a difference between the factors (e.g. it won't tell you whether there's a difference between stations and depths - that's a meaningless question), but it *will* tell you whether there's a difference between the levels of any factor you're interested in.

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*has* been recorded as a numerical code it could make the difference between the analysis being correct and incorrect. For example, if Distance (from the trunk) has been recorded in your data set as 0, 1 and 2, the Analyze | Fit Model procedure will by default assume that you want to analyse Distance as a continuous variable (with one degree of freedom) as opposed to a factor with three levels (i.e. two degrees of freedom). This *could* actually make sense for Distance, which is actually recorded on pretty much an interval scale anyway, but it would make no sense at all for the Station factor if the stations were recorded as 1, 2, 3 and 4.

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*that's* why the standard error should be the same for every level of any one factor in a factorial experiment. In this example:

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?

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

thanks again

thanks again