Hello, I calculate minimal sample size to compare means of
two groups (different treatment) of datapoints by a t test.
For continuously distributed datapoints when the number of datapoints in each
group is more than 30 I can assume normal distribution of a sample mean and
estimate minimal sample size by using the formular:
sample size of each group = 2 * [sigma^2] * [(Zcrit+Zpwr)^2] / [D^2]
sigma = standard deviation (~ equal in each group)
D = difference between the means of each group
Zcrit = 1,96 for significance 0,05
Zpwr = 1,28 for desired power 0,9
My question is: does this formular still hold in case I have a technical error when
estimating the value of each datapoint? Lets assume this technical error is normally distributed with standard deviation sigma2, which is of course smaller than sigma.
How small should the observation error (sigma2) be for the given formular to remain ~ valid? Is there any method for sample size calculation in case the observation error is not small enough? Any program (may be SPSS samplepower) can perform sample size calculations for such situations? Thank you!
The observation error will show itself as a larger interspecimen variation which automatically leads to a larger standard deviation, and therefore a smaller power. Therefore, measurement errors are already hidden and incorporated in the standard deviation, which you use for calculating the power. If you have a small sigma, you might rest assured that the measurement error is already small; otherwise, you were not likely to get a small standard deviation.Originally Posted by bondsergey
bondsergey (10-26-2012)
It may also help to thoroughly describe what you are calling an observation error and whether it randomly occurs in all groups or not, and whether you think it increases or decreases variability.
Thanks I agree
Sometimes an observation error can lead to a constant increase or decrease in all the observed values, without affecting the variation (and sigma). In such a case (when there is a measurement error but that is not incorporated in the sigma), however, the existing measurement error is unlikely affecting the test power, as it affects both groups similarly without affecting their contrast, which is the factor relevant to the test power. So I think the standard deviation is the key to check if the power can be low or high.
It is a measurement error, which does occur randomly in both groups, in my case it occurs due to chance: in one moment there can be 40 particular cells in one ml of blood, in another moment there can be 45 cells, normal distribution. I can improve evaluation of mean by increasing the number of replicates, but it is too expensive. The standard deviation (sigma) is 22 within each of the two groups, the difference between means of each group is 25 (mean1-mean2=43-18=25). The standard deviation sigma2 for measurement of each datapoint is ~4 (evaluated by taking 4 replicate measurements of one of the datapoints). I do not think it increases the variability within each group, but this is just my feeling. Do you think I should not worry? If I should what are my actions? May be I should estimate a possible change in variability (sigma) by computer simulations and use new estimated sigma (if different) for sample size evaluation? Would it be proper way?
Low on time, but the big thing is that you understand what is going on so that you can list it as a limitation and that it occurs randomly in both groups.
bondsergey (10-26-2012)
And note that if you still have an acceptable test power (> 0.8), the measurement error is not disrupting your comparison.
This is all given that this error is never so large that it discredits the validity of the data. It must be marginal and not disruptive to testing and proving hypotheses.
What do you mean by saying "the error should be marginal"? Marginal error is 1.96*sigma / [(n)^1/2]? It goes to zero if n goes to infinity. How to check if the error is "not disruptive to testing and proving hypotheses"?
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