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# % of Stunting prevelence or Mean WHZ score

This question was posted the Assessment and Surveillance forum area and has 3 replies.

### Mark Myatt

Frequent user

6 May 2015, 08:52

OK. Here is some "hand-waving" to get this started ... I assume: (1) You will compare your outcomes in "Arm 1" with your outcomes in "Arm 2" at an end point. (2) Your outcome is some measure of stuntedness and when you say "mean WHZ score" you mean "mean HAZ score". (3) The end-point is "graduation" at two years of age. You will collect data (e.g. height and age) at or near "graduation". You can use either prevalence HAZ < -2 of mean HAZ for your outcome. Using means may be more efficient. The difficulty with using means is that we often want to work with prevalences as they can be easier to understand. It is not too difficult convert between means and prevalences when the quantitative indicator (e.g. HAZ) is quasi-normal. If we assume that the mean HAZ is -1 with SD = 1.1 we would expected about: ``` > pnorm(q = -2, mean = -1, sd = 1.1) * 100  18.16511 ``` That is 18.2% with HAZ < -2. If we manage to shift the mean HAZ upwards to -0.7 with SD = 1.1 we would expect about: ``` > pnorm(q = -2, mean = -0.7, sd = 1.1) * 100  11.86389 ``` That is 11.9% with HAZ < -2. That is equivalent to a reduction of about 6.3% in the prevalence of stunting. I used R above but the calculation can be done in (e.g.) Excel using: ``` =100 * NORMDIST(-2, -1, 1.1, 1) <- yields 18.165107 =100 * NORMDIST(-2, -0.7, 1.1, 1) <- yields 11.863892 ``` You can play with these calculations (using available survey data to get expected means and SDs) to finds the difference between means that corresponds to a desired difference in prevalence. I will work continue by using what I have done above. Ignoring design effects (intra-cluster correlations), a simple sample size calculation to detect this difference with 90% power at the 5% significance level would be: ``` n = ((1.28 + 1.96)^2 * (1.1^2 + 1.1^2)) / ((-1) - (-0.7))^2 n = 282 ``` children in each arm. You have a complex design. This means that you will likely need a larger sample size (e.g. double what you calculate using standard formulae would probably be OK) and you will need to analyse the data using appropriate methods (these are available in most statistical packages). You will need to weight within-community results by population size. The sample size will also need to be increased to account for losses to follow-up (LTFU). About 20% LTFU is probably OK. The sample size would then be: ``` n = 2 * (282 * 1.2) n = 677 ``` in each arm. What you have above suggests that you will have no problems reaching this sample size. In that case you could collect data from about: ``` m = (677 / 5000) * 191 m = 26 ``` communities in each arm which could be selected at random. I would increase this to at least m = 30. That will reduce costs (i.e. you will only need to sample is m = 60 communities rather than m = 382 communities). There are other designs. A longitudinal design (e.g) will need fewer children but you will need to measure them more often (e.g. quarterly over two years (that is 8 measurements) rather than once at graduation. This may be more expensive and data management and analysis will be complicated. Sequential trials are robust and very efficient and will typically cut the required sample sizes in half. Such trials are not commonly used in the emergency nutrition field. If you decide to go for a sequential design then you should consult an experienced statistician. What I have here is "rough and ready" and is intended only to illustrate how you could proceed. These types of trial can be very difficult to get right. You may want to consider consulting a statistician with experience with clustered trials. I hope this is of some use.

### Mark Myatt

Frequent user

25 Jan 2016, 11:33

This is a "quasi-experiment" in the sense that there is no randomisation between control and intervention groups with the controls just "found". Such studies need to be interpreted with care as bias and confounding can easily creep in. In the case (e.g.) of a secular trend of reducing stunting you may see a difference in favour of the intervention group even when an intervention effect was absent.

You have identified a potential bias with age. Ideally you would want to compare children at the same ages. If (e.g.) you had a pair with the control ages 33 months at at recruitment then the comparison would be with the intervention child at (or very close to) 33 months. In many cases stunting occurs before about 30 months and the child remains stunting for some time after that (i.e. until the pre-adolescent growth spurt). You may, therefore, want to compare measurements at 30 months (intervention) and >= 30 months (control). Note that age can be subject to considerable measurement error and this will likely effect the control group more than the intervention group.

Data-analysis should be straightforward. Something like a paired t-test on HAZ should do the job.

I'd be tempted to choose the simplest design. You may find the quasi-experimental design a bit complicated to run.

I hope this is of some help.