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Anonymous 2885

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?

Mark Myatt

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.

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