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grif
03/06/2003, 10:29 PM
Dr Ron - did each replicate have a different (controlled) level copper? The table seems to imply that each replicate had a different level of copper in it.

grif
03/06/2003, 10:30 PM
level of copper, or copper level.

Sheesh, I can't type tonight.

rshimek
03/07/2003, 11:38 AM
Hi Grif,

There were no replicates for the copper. The copper run was simply a sequence of beakers with copper concentrations varied by a factor 10. Copper is a known poison, so this sequence simply shows what level it has its effects at.

This is called a "positive control" as it shows that the test will "positively" detect mortality caused by a poison.

:D

grif
03/07/2003, 05:37 PM
So n=1 for each treatment (water type), at each level of "poison".

I'm wondering about the averaging of the replicates within a water type then, given they are not dealing with a homogeneous sample (differing copper levels).

I guess I'm wondering if ANOVA is the best test because of that. I'm wondering if independent t-tests might be more appropriate, but I'm concerned you'd have run out of degree's of freedom.

Results don't appear they'd lose significance, just wondering if the ANOVA is most appropriate. I guess I need to pull Keppel's text back out to be sure.

rshimek
03/07/2003, 06:32 PM
Hi,

I did an ANOVA to examine the variances. Once those were determined as being very different I used pairwise t-tests of the means for samples with differing variances.

simonh
03/07/2003, 07:28 PM
Originally posted by grif
So n=1 for each treatment (water type), at each level of "poison".


From my reading:

Poison was only added to the colum labelled 'Cupric Dichloride Larve' used as a positive control with n=1 for each dosage. For all other water types n=10 with no poison added to any of them.

If you totally ignore the ANOVA due to unequal variances and just run multiple comparisions i.e. 6 un-equal variance t-tests (as Ron did) and use a Bonferoni adjustment to control the familywise alpha at 0.05 i.e. only declare individual results significant if p-value less than 0.05/6 = 0.008 it does change the 'significant' results.

HTH.

grif
03/07/2003, 09:03 PM
Okay, I can see where the table could be read like that.

Are the results still significant, just at a higher p-value, or is significance lost? Just a seat of the pants look at the result (I'm on the road and away from my texts and s/w to refer to or I'd run some analysis) seem to indicate at the least there are some comparisons that would remain significant almost no matter what.

You'd also have to be careful to only test a priori comparisons against 1 reference point (seawater?) and not try and compare each mix to each other mix. If you did, the familywise adjustment would get out of hand (well beyond the 6 you mention, 6 factorial I believe).