Gary, it sounds as if you are doing a sample size calculation after the data collection. It seems as if you are not stating what differences other than head circumference are present in these groups. If both groups have few differences other than head shape, then you are correct to find variation differences. This sounds like a t-test power analysis would work; however the standard deviation for each type of measurement (scanner versus camera) can be incorporated into the sample size calculation by using the maximum variance. You would need sufficient power to calculate a difference between two groups assuming the maximum variation. This website should be useful:

http://www.stat.ubc.ca/~rollin/stats/ssize/n2.html

Joan,

Thank you for the response. You are correct that I didn't mention how the subjects were categorized into the two groups. In this case we are comparing the effects of two different surgical repairs for an uncommon birth defect (sagittal craniosynostosis). I understand your response as saying that we might simply add up the two sources of variation: 1 - the variation of the measurement (head size) in the population. and 2 - the variation in matched measurements with the two imaging systems (camera and scanner). Is that correct?

Use the larger for the calculation. i would check the websites for t-test power calculations.

Do I read you correctly that since head size variation is greater than the variation between imaging systems one may ignore the imaging system error?

The imaging system error is irrelevant. You will have measurements, and they vary. The variation is a combination of technical and biological variance (as always!). No test cares about the source of the variance. The test sees the values, and they "contain" all these sources of variability.

As Joan said: use the larger for the calculation.

It's dificult to answer without details on the experiment for which you need the sample size... What do you want to compare, and which method will be used, on which ground? If you compare the two on the same patient, for instance, or before/after surgery analysis (paired samples), between subjects patient is irrelevant even if it is higher, but methods variance will be the key point...

Using the higher is relevant if the other sources are much smaller, otherwise you will overestimate your power/underestilmate your sample size, but you need a model for the variance of what you will test, and for that detail the experimental protocol...

To complete Emmanuels point "If you compare the two on the same patient, for instance, or before/after surgery analysis (paired samples), between subjects patient is irrelevant even if it is higher, but methods variance will be the key point...":

Also for this design Joans advice is correct: use the larger for the calculation.

Gary - 

Two additional comments: (1) confidence intervals are generally more practically interpretable than p-values, and (2) where you have "matched camera and scanner data," you could do a simple regression to compare them for future consideration. 

Cheers - Jim 

Using the largest deviance is not the optimal solution. It is just the most conservative and since no one has given a rationale for it, you should not accept that proposal. Consider this extreme case:

group 1: 30, 31, 30, 31, 30, 31, 30, 31, 30, 31, 30, 31, 30, 31, 30, 31 (low variance)

group 2: 15, 30, 45, 16, 31, 46,  15, 30, 45,16, 31, 46, 15, 30, 45, 16 (high variance)

Surely, you know a lot more about the true mean of group 1 than if you just used the high variance from group 2. In R you would model different variances for each group using the "weights" argument, but still compare means, like this:

lm(head_size ~ group, data=your_data, weights=varIdent(form=~1|group))

So much for modelling. If you want to do power calculations, it depends on the variance and the estimated mean, so it depends on which device you're going to use. You have lower power with a high-variance device.  There's a range of functions and guides for power analysis on linear models inr R (like the one above) if you google it.

Here is an ad-hoc method coded in R that I use in this situation:

# R-code for estimating number of samples for comparison of two means using independent samples,

# without assuming equal cariance

# Example: Calculate sample size for comparing Mu1 and Mu2, sigma1_0 = 1, sigma2_0 = 1.5 (initial guesses for sd's)

# difference to be detected = DElta = Mu2 - Mu1 = 1,

# then use conservative effect size = Diff/sigma2_0 = e0 = 1/1.5 = 0.66

library(pwr)

pwr.t.test(d=0.66,power=.7,sig.level=.05,type="two.sample",alternative="two.sided")

# this will yield n2 = 30

# in this example, ratio of large var/small var = (1.5)^2 = 2.25

# then we can take 30/2.25 or 14 samples in group with smaller variance

# and 30 samples from group with larger variance

# Note that this is an ad-hoc procedure I cooked up using the fact that

# for 1-sample problem, n = (z^2)(sigma^2)/(d^2)

# n2/n1 = sigma2^2/signa1^2 = 2.25 for this example, hence n1 = n2/2.25

Why not use directly the power.t.test function of R ?

power.t.test(d=0.66,power=.7,sig.level=.05,type="two.sample",alternative="two.sided")

Two-sample t test power calculation

n = 29.32845

delta = 0.66

sd = 1

sig.level = 0.05

power = 0.7

alternative = two.sided

NOTE: n is number in *each* group

Emanuel, you are absolutely right, that works and gives a conservative n for both groups. One might want to take a smaller n for the group with smaller bariance, as I show in my earlier answer.

Oops! Variance not barianve

thanks to the many contributors to this thread. Perhaps I can re-state the problem. We have two independent patient groups. One group was seen in St. Louis the second group in North Carolina. We have 3D images of all the patient's heads and will obtain head shape measurements from these images. The images were generated using two different sorts of cameras.  

The group in North Carolina was able to obtain paired images on some patients using the two sorts of cameras. While there was no obvious bias there was, of course some discrepancy in the measurements of the paired data. The mean difference between the cameras was 2% of the values measured. 

The standard deviation in head shape about the mean is roughly 4%. 

I hope that clarifies things for those still finding my initial project description ambiguous.

I still have a few more questions:

Two Group 1 (St Louis),   Group 2 (NC) 

Group 1 has 3D images (using Camera 1), Group 2 has 3D images (using camera 2)

Group 2 has a few paired images using both cameras,  relative bias = 2%

standard deviation = 2 (for both groups, either camera?)

A.K.

Thanks for your continued interest. My addenda to your description are in bold italics.

Two Group 1 (St Louis), Group 2 (NC) [Yes - exactly right]

Group 1 has 3D images (using Camera 1), Group 2 has 3D images (using a laser scanner)

Group 2 has a few paired images using a camera and laser scanner. There is no relative bias but the standard deviation between the scores = 2%

standard deviation of head shape in the general population = 4 (for both groups, either camera or laser scanner.)

So from the paired data (of Group 2), we can conclude that measurements from the two 'cameras' follow (more or less) the same distribution; we assume approximate normality when calculating number of samples for comparing two means. In your case, the two population variances are equal, so Emanuel's solution using R works:

library(pwr)

pwr.t.test(d=0.25,power=.7,sig.level=.05,type="two.sample",alternative="two.sided")

# this will yield n1 = n2 = 199

I had incorrectly assumed that the two variances are equal in my solution. 

Hope this is helpful.

Drs. Curis and Singh,

Thank you for the instructive and constructive advice.

Sincerely,

Gary Skolnick

You are welcome!