Good day we have a serological ELISA test we commonly use for screening herds for a viral infection with an established cutoff for use in individual animals. We wish to investigate the AUC of the test when we mix equal volumes of 1 positive test and 2 negatives (pool of 3) and 1 positive and 4 negatives (pool of 5). I am willing to recalculate the sample size after the initial attempt and add more if necessary. Just wondered what the equation is for a non-pooled AUC and then pooled if there is one? R package if there is one?
Thank you.
Well AUC often means area under a curve. This i recommend integration of some sort. If that's not what you had in mind would you please clarify your.question.
Best wishes,
David Booth
I wonder why you would want to choose sample size based on area under the curve.
Your description is quite brief so I may have easily misunderstood, but:
I assume that you want to compare pools of 3 with pools of 5 and choose which works best. I would think that the smaller pool will always get more accurate test results, and that you may prefer larger pools because it saves costs. Is that correct?
If so, then AUC does not seem a good measure to determine your sample size because it does not consider costs.
Sorry if i was unclear. I wish to identify the optimum new cut-point (balance of sens and spec) for the elisa if we know ahead of time what has 1 positive: 2 negatives and what has 1 positive: 4 negatives.
Does that help at all?
CB
Ok I am going to treating it like this and see if you disagree:
For any study where there are a lot of unknowns I usually recommend two-stage projects where we estimate a SS and then based on the information gained from the preliminary results we calculate a new sample size and then determine how many more we need....
To calculate: TP+FN = [Z2 * Se (1-Se) ] /W2 so I used a bayesian Se (98%) from a previous project for the individual elisa and got 75 participants needed
TN+FP = [Z2 * Sp (1-Sp) ] /W2 and used the specificity of 87% and got 43 participants (W = 0.1) - 10% error
For the Se you would then need (TP+FN/ Prevalence) so I used the 17% from another provincial study and got 75/0.17 = 13 positives needed
For the specificity you would then need (TN+FP/ 1-Prevalence) = 43/0.73 = 52 negative needed
So you would do 13 positive pools and 52 negative pools for each:
13 Positive pools:
52 Negative pools:
So long story short I would run 26 positive test pools and 104 negative pools for a total of 130 tests.
Reasonable?
Recalculate everything after finishing.
I would cite this article:
Article Sample Size Calculation Guide - Part 4: How to Calculate the...
Thanks!
- Badges
- Science topic
- Similar topics
- Mathematics
- Statistics
- Sampling Design
- Sample Size