How this calculation works
Use p and the absolute margin as proportions, not whole percentages. A proportion of 50% gives the largest sample size under this formula. The estimate assumes a large population and simple random sampling.
n = ceiling(z² × p × (1 − p) ÷ margin²)A worked example
At 95% confidence, expected proportion 50% and a 5 percentage point margin, at least 385 responses are needed.
Good to know
Normal-approximation planning formula without finite-population correction, design effects or nonresponse allowance. It does not make a convenience sample representative.
Halving the margin of error needs about four times the sample
The margin is squared in the denominator. Under the same confidence and proportion assumptions, moving from a five-percentage-point margin to 2.5 points multiplies the unrounded sample-size requirement by four.
An expected proportion of 50% gives the largest p(1 − p) term. When there is no sound prior estimate, that value provides a conservative normal-approximation planning scenario rather than a guess that exactly half the population will agree.
Responses and invitations are not the same count
The result estimates completed independent responses under a simple random-sampling model. If some invited people do not respond, the number of invitations would need separate planning and a justified response-rate assumption.
The tool omits finite-population correction and design effects. Cluster samples, weighting, subgroup analysis and very rare outcomes can need a different approach. A large convenience sample is not automatically representative.
A few questions, answered
Does a 5% margin mean five percentage points?
Here it means an absolute margin of five percentage points around a proportion.
Why does the answer round up?
A sample needs a whole number of responses and must not fall below the calculated requirement.