# 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.

### Tariq Khan

Normal user

28 Mar 2012, 20:40

### Mark Myatt

Frequent user

29 Mar 2012, 09:21

### 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

Frequent user

14 Apr 2012, 15:57

```
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