Recent content by Doctor_G

Though, on second thought, I dunno. Your method might be good enough. My main concern was that by treating each of the reps as a separate independent observation, you're artificially boosting your n and reducing your SD. It's the difference between weighing nine different test subjects and...

I would respectfully differ with that prognosis. I think, just by looking at the data, that the three reps in each set are a lot closer to each other than they are to the reps from different sets. If the data looked like this: 0____0___0_________0_0__0____0_____0__0 then yeah, I think a fair...

Yes, there was a typo. Set 2.1 should be 8476.44, not, as I wrote, 8746.44. I would go back and fix the original, but I don't see any way to do that. Well, are we talking precision or accuracy here? But, to try to answer your questions, I didn't come with any a priori assumptions of either...

I'm afraid I don't follow you here. When you say 'spread', are you referring to the SD of the meta-average, or are you referring to the range? When you refer to the uncertainty of the measurements, do you mean their respective SEMs? Also, how did we go from 17 to 10?

My answers are 'Wha...?' and 'Sure, I guess?' respectively. Also, if 'my guess is the probability to pick set 2 as an unbiased sample must be rather small' is a complicated way of saying 'Set 2 seems like the odd man out', yeah, but not really. The meta-average is about 12,000. Set 3 is about...

Yes.

For simplicity's sake, I haven't shown the dose response. I figure the process should be the same for each dose. Again, for simplicity's sake. But, as you wish. Just looking at the control dose (i.e. no drug), my measurements are: Set 1: 1. 12446.52 2. 12757.86 3. 11786.68 Set 2: 1. 8476.44...

It's more of the first one. I'm not sure why I would want the observed SD reduced. I mean, from a presentation standpoint, that does make my data look nicer, but from the standpoint of intellectual honesty, I don't follow. If I have a set of data, like so...

Thanks. That helps set me straight. But it exposes what I feel is an intrinsic shortcoming of the approach. My original formula quadratically sums the variance of the three set averages and the variances of their uncertainty. But I don't really care about the second one. I want to factor in how...

This is an issue I've seen asked and answered before on this forum some years ago. However, the answer doesn't quite make sense to me, so I want to see if I can get either a more satisfactory answer or a better explanation of the original. Suppose that I have some cells that produce a...