Please visit COD-02 - Conduct a FRAT study in the Democratic Republic of Congo for ToR and application information.
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START WITH:
guess at prevalence = 10%
required precision = 3%
overall sample size (simple random sample) = 384
guess at design effect = 2.0
overall sample size (cluster sample) = 384 * 2.0 = 768
number of clusters = 25
within-cluster sample size = 31
BUT :
average village population = 90
proportion aged 6 - 59 months = 17%
average village population aged 6 - 59 months = 90 * 0.17 = 15
SO :
New design effect = 1.25 (we have more small clusters)
New sample size = 384 * 1.25 = 480
within-cluster sample size = 15
number of clusters = 480 / 15 = 32
The proportion aged 6 - 59 months can also affect the overall sample size.
Sample size formulae tend to assume that we sample from an infinitely large population. This assumption is reasonable when the sampling fraction (i.e. the ratio of the sample size to the population size) is small. The error (e.g. the width of the 95% CI) is essentially the same regardless of population size as long as the sampling fraction is less than about 5%.
In the example above we have n = 480. If the size of the population aged 6 - 59 months were about:
N = 480 * 20 = 9600
or larger then we need not worry.
If the sampling fraction was greater than about 5% then we would apply a "finite population correction" (FPC) in order to account for the added precision gained by sampling a large proportion of the population. The FPC can be calculated as:
FPC = sqrt((Population - Sample Size) / (Population - 1))
If we assume a population of 4,800 and a sample size of 480 then we have a sampling fraction of 10%. Since this is above about 5% we should calculate an FPC:
FPC = sqrt((4800 - 480) / (4800 - 1)) = 0.95
The required sample size is now:
n = 480 * 0.95 = 456
Continuing with our example ... we might collect this as 30 clusters of 15 (n = 450 is close enough to n = 456 to make no difference) and save ourselves a little work and a little money.
We do not usually apply an FPC in SMART surveys as (1) savings are usually small and (2) the SMART software does not adjust results to account for the sampling fraction.
I hope this is of some use.Answered:
11 years ago
new.n = (old.n * population) / (old.n + (population - 1))
Continuing with the example we get :
new.n = (480 * 4800) / (480 + (4800 - 1)) = 436
which we might collect as 29 clusters or 15.
Sorry for any confusion.
The moral here (for me) is to check what I write before posting.
I fear my mind is going.Answered:
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failureCount <- 0
for(test in 1:100000)
{
randomHeights <- round(runif(n = 500, min = 650, max = 1100), 0)
finalDigits <- substr(randomHeights, nchar(randomHeights), nchar(randomHeights))
table(finalDigits)
p <- chisq.test(table(finalDigits))$p.value
if(p < 0.05)
{
failureCount <- failureCount + 1
}
}
failedProportion = failureCount / 100000
failedProportion
Simulates 100,000 surveys with no digit preference (just random variation). The result is that the null is rejected (as expected) in 5% (actually 4.968% in the simulation that I ran) of the surveys. This means that 1 in 20 good surveys are rejected.
As the sample size increases even small deviations from expected distributions can be statistically significant. This will usually only be a problem with very large sample sizes. The opposite is also true. With small sample sizes we might not detect clear digit preference.
Here are some examples ... clear digit preference for last digit = 0 or 5:
last digit count
---------- -----
0 10
1 5
2 5
3 5
4 5
5 10
6 5
7 5
8 5
9 5
---------- -----
60
-----
chi-square = 6.6667, df = 9, p-value = 0.6718
but we fail to reject the null of no digit preference.
Here is the same pattern but with 10 times the sample size:
last digit count
---------- -----
0 100
1 50
2 50
3 50
4 50
5 100
6 50
7 50
8 50
9 50
---------- -----
600
-----
chi-square = 66.6667, df = 9, p-value = 0.0000
we reject the null of no digit preference.
This makes it very difficult to use simple significance tests for monitoring when (e.g.) a team might bring in data from two clusters per day (i.e. n = 60 or less).
Since plausibility tests can "detect" problems that are not there (about 5% of the time) and can (with small samples)fail to detect real problems we should be careful when we use them. It is probably better to "eyeball" (visually inspect) the data to see if we have (e.g.) too many ".0" or ".5" final digits for height.
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11 years agoHi... there are not many studies on adolescent nutrition in my country, and no studies carried out in the location in which I am intending to carry out my research.
Is it advisable to do a cross-sectional study before starting the intervention study to be able to know the gap and topics that will be relevant during the intervention and also to compare the outcome at the end? thank you
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11 years agoHi , i am Dr.Baidar Bakht Habib
i have one question , minimum for how many HHs samples we need to include in the SMART assessment in the planing to give us Nutrition GAM rate results .as standard or we do not have standard for sampling.
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8 years ago