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GAM in Nutrition Screening and Survey?

This question was posted the Assessment and Surveillance forum area and has 6 replies. You can also reply via email – be sure to leave the subject unchanged.

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Tariq Khan

Normal user

28 Mar 2012, 20:40

I just encountered a question. Can the results of a exhaustive nutrition screening (based only on MUAC) be presented in terms of GAM or GAM can only be presented in Surveys?

Mark Myatt

Frequent user

29 Mar 2012, 09:21

Of course ... MUAC is a far better indicator than WHZ. The problem is that a lot of people want prevalence by WHZ even though key programs (e.g. CMAM) tend not to use WHZ. The case definitions in common use are:

SAM : MUAC < 115 mm or OEDEMA
MAM : 115 <= MUAC < 125
GAM : MUAC < 125 or OEDEMA

An "at risk" category:

125 <= MUAC < 135

is commonly used.

Tamsin Walters

en-net moderator

Forum moderator

29 Mar 2012, 19:43

From Blessing Mureverwi:

You can only present GAM if the sampling was indeed exhaustive,and
therefore every child was reached.However,if this was conducted in a
central place,rather than from household to
household,representativeness may be challenged.To compute GAM in this
case you need to be able to estimate the sampling error and therefore
come up with confidence intervals.This has become common practice.Did
your survey fulfill the above?

Tariq Khan

Normal user

8 Apr 2012, 18:01

In fact the sampling was supposed to be exhaustive say, of 7400 children 6800 children were screened and this was done in a central place, where all caretakers were asked to bring their children.

Mark Myatt

Frequent user

14 Apr 2012, 15:57

There is a potential for bias if the kids that were not measured were more (or less) likely than the kids that were measured to be SAM or MAM cases. It is good practice to test case-finding exhaustivity using a capture recapture study. You do not have this.

What you do have is a very large sampling fraction:

	SF = 6800 / 7400 = 91.9%

If you found (e.g.) 680 GAM cases then you would estimate GAM in the usual way (i.e. 10%).

You could then put forward some "what-if" scenarios ... with (e.g.) the same prevalence as the sample, a much higher prevalence (20%) than the sample, and a much lower prevalence (5%) than the sample in the missing 600 children and estimate with these "what-if" numbers.

The the same prevalence we would have 680 + 0.1 * 600 = 740. Our estimate would be:

    data:  740 and 7400 
    95 percent confidence interval:
     0.09325574 0.10706118 
    sample estimates:
    probability of success 

With the much higher prevalence scenario we have 680 + 0.2 * 600 = 800 cases. Our estimate would now be:
    data:  800 and 7400 
    95 percent confidence interval:
     0.1011211 0.1154055 
    sample estimates:
    probability of success 

With the much lower prevalence scenario we have 680 + 0.05 * 600 = 710 cases. Our estimate would now be:
    data:  710 and 7400 
    95 percent confidence interval:
     0.08932819 0.10288391 
    sample estimates:
    probability of success 

We can present the average for the estimate:
    estimate = (0.1 + 0.1081081 + 0.09594595) / 3 = 0.1013514 = 10.14%

and the smallest and largest confidence limits found in the what-if scenarios giving 10.14% (95% CI = 8.93% - 11.54%).

When you report this estimate you should also report on what you did and why you chose the two what-if scenarios.

Is this any help?

Anonymous 730

Nutrition and Food Security Officer

Normal user

27 Apr 2012, 16:30

Hie Mark,
This looks very logical.Many thanks.

Mark Myatt

Frequent user

27 Apr 2012, 16:56

You are most welcome.

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