# GAM in Nutrition Screening and Survey?

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### Tariq Khan

Normal user

28 Mar 2012, 20:40

Hi,

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?

Thanks.

### Mark Myatt

Consultant Epideomiologist

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?

### Mark Myatt

Consultant Epideomiologist

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 0.1

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 0.1081081

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 0.09594595

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.