# Coverage prevalence survey for blanket SFP for chronic malnutrition

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

### Anonymous 2232

Normal user

18 Mar 2013, 06:51

### Mark Myatt

Frequent user

18 Mar 2013, 17:41

### Mark Myatt

Frequent user

7 Jun 2013, 08:31

```
probability = proportion
```

We can reverse this so that we have:
```
proportion = probability
```

This is a perfectly legitimate reversal.
The problem now becomes one of estimating probability. This required us to specify an appropriate probability model for our variable of interest. The appropriate model for H/A is the normal distribution. We then estimate the probability of a child will have (e.g.) a HAZ < -2. If we can do that we have prevalence of HAZ < -2 since:
```
probability = proportion = prevalence
```

The normal distribution is fully described by two parameters. These are the mean and the standard deviation (SD). With small sample sizes we prefer to use robust (i.e. resistant to outliers) estimators for the mean and SD. These are:
```
estimate of population mean = sample median
estimate of population SD = sample IQR / 1.34898
```

where:
```
IQR = upper quartile - lower quartile
```

Once we have these we can use the cumulative normal probability function (available in most stats packages and spreadsheets) to calculate the probability (prevalence) we are interested in.
Here is an example ... assuming (from our sample), we calculate:
```
median = -1.6
IQR / 1.34898 = 1.4
```

In Microsoft Excel (e.g.) we would use:
```
=NORMDIST(-2;-1.6;1.4;TRUE)
```

which gives:
```
0.3875484811
```

which is an estimated prevalence of 38.8%.
In R (e.g.) we would use:
```
pnorm(-2, mean = -1.6, sd = 1.4, lower.tail = TRUE)
```

which gives:
```
[1] 0.3875485
```

which is (also) an estimated prevalence of 38.8%.
If you need to calculate a 95% CI (we usually need to do this) then we first calculate a 95% CI on the median and use that. An approximate formula for a 95% CI on the median which is safe for long-tailed (i.e. somewhat non-normal situations) is:
```
median +/- 1.58 * (IQR / sqrt(n))
```

This formula is from:
```
Velleman PF, Hoaglin DC, Applications, Basics, and Computing
of Exploratory Data Analysis, Duxbury Press, Boston,
Massachusetts, USA, 1981
```

with the example data and a sample size of n = 200 we have:
```
-1.6 + 1.58 * (1.9 / 14.14) = [-1.4; -1.8]
```

In Microsoft Excel we have:
```
=NORMDIST(-2,-1.4,1.4,TRUE)
=NORMDIST(-2,-1.8,1.4,TRUE)
```

which give:
```
0.4432015032
0.3341175709
```

We have an estimate of 38.8% (95% CI = 33.4%; 44.3%).
You should check my arithmetic here.
BTW : This procedure is (somewhat confusingly) referred to as PROBIT and is is used in RAM and S3M type surveys.
I hope this helps.### Mark Myatt

Frequent user

7 Jun 2013, 09:15

Please visit BFA-02 - Development of multisectoral coordination tools and mechanisms for nutrition in Burkina Faso for ToR and application information.