Normal user

30 Mar 2015, 12:02

Comment déterminer la taille de l'échantillon dans une enquête SLEAC d'évaluation d'intervention de nutrition avec une méthodologie LQAS?
How do you determine the sample size in a SLEAC survey of a nutrition intervention with an LQAS methodology?

Consultamt Epidemiologist

Frequent user

31 Mar 2015, 08:41

SLEAC uses a "standard" sample size of n = 40 SAM cases. This provides good performance for a simple three-class classifier when the distance between the lower and upper classification thresholds (i.e. the width of the moderate class) is at 20% and there are about N = 400 SAM cases in the surveys area. By "good performance we mean:
(1) Errors tend to occur when coverage is close to class boundaries.
(2) Gross errors (i.e. true low as high / true high as low) almost never occur.
The 20% width of the moderate class was selected from experience ... during development and testing we found that users tended to use classes such as 20%/50% and 50%/70%.
If you need finer classifications somewhere in the range of 20% to 80% then you will need a larger sample size. At higher and lower proportions you can still use a sample size similar to n = 40 and get finer classification with good performance. For example, the prevalence of transmitted HIV drug resistance is routinely classified using 5% and 15% lower and upper class thresholds using sample sizes of N <= 57.
The N = 400 was decided using computer based simulations of million sof SLAEAc surveys with assumptions:
(1) The largest service delivery unit for which coverage will be classified is a health district and these are seldom larger than 100,000 total population.
(2) Approximately 20% of the population are aged between 6 and 59 months.
(3) SAM prevalence may be about 2%.
That is:
```
N = 100000 * 0.2 * 0.02 = 400
```

If your population of SAM cases is smaller than N = 400 then you can adjust the SLEAC sample size downward using:
```
new_n = (old_n * population) / (old.n + population - 1)
```

For example if the populations is 105,000, the proportion 6-59 months is 17%, and SAM prevalence is 1.5% then the required sample size would be:
```
N = 105000 * 0.17 * 0.015 = 268
n = (40 * 268) / (40 + 268 - 1)
n = 34
```

This correction gives a sample size with the same errors as the "standard" n = 40 sample size in a population of N = 400 SAM cases.
I hope this helps.