- Thread starter Learner1
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Formula (1):

n= t² x p(1-p)

m²

Description:

n = required sample size

t = confidence level at 95% (standard value of 1.96)

p = estimated prevalence of malnutrition in the project area

m = margin of error at 5% (standard value of 0.05)

Example

In the Al Haouz project in Morocco, it has been estimated that roughly 30% (0.3) of the children in the project area suffer from chronic malnutrition. This figure has been taken from national statistics on malnutrition in rural areas. Use of the standard values listed above provides the following calculation.

Calculation:

n= 1.96² x .3(1-.3)

.05²

n = 3.8416 x .21

.0025

n = .8068

.0025

n = 322.72 ~ 323

Resource : http://www.ifad.org/gender/tools/hfs/anthropometry/ant_3.htm

Definitely, in the above example, the Confidence Interval (CI) is 95%, and yours 80% which is below the statistical expectation, then look for the Z value, but I am not sure the exact value.

Formula (2)

n=pq(z)2/((e)2)

Simply this equation expect half of the population (50%) and by substitution 95% CI:

n=((0.5)(0.5) 〖x(1.96)〗^2)/〖0.05〗^2

n= 384

Again, this is under 95% CI, so look for 80% in Z, and make your substitution.

Try to learn other methods.

( I am sorry, the wed doesn't support word equation, or I may not be aware of how to handel it well).

Thanks a ton.

if my required sucess rate is 80%, so should i calculate the sample size as follows:

n= 1.96^2 x 0.8 (1-0.8)/0.05^2

n=245

OR

Should the calculation be:

Required sucess rate 80% (1.282) and

since i do not know the estimated prevalence, i may use an assumption of 50%,

then

n= 1.282^2 x 0.5 (1-0.5)/0.05^2

n=164

Thanks for your support.

Please help me calculate the power also?

This is a great forum. I am learning a lot by reading different posts.

Thanks to all members who are answering each and every question.

I understand that in the sample size calculation for a proportion, the sample size (n) decreases with increase in the estimated proportion (p).

If I assume that if p =95% with a t(CI) = 95% and m =0.05 then the n = 73;

If I enroll 73 patients, and the success rate is 70%, How do I interpret the result?

Please help.

Thanks to all members who are answering each and every question.

I understand that in the sample size calculation for a proportion, the sample size (n) decreases with increase in the estimated proportion (p).

If I assume that if p =95% with a t(CI) = 95% and m =0.05 then the n = 73;

If I enroll 73 patients, and the success rate is 70%, How do I interpret the result?

Please help.

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