Question

In a study to find the proportion of children with asthma in a certain population, we desire a 95% confidence interval with standard error no more than 0.05. Without any prior knowledge of the proportion, what should be our minimum sample size

Answer 1

The minimum sample size should be of 381.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is given by:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

Without any prior knowledge of the proportion, what should be our minimum sample size

We dont know the population proportion, so we use
`\pi = 0.5`, which is when the largest sample size is needed. We have to find n for which
`M = 0.05`. So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.05 = 1.95\sqrt{(0.5*0.5)/(n)}`

`0.05√(n) = 1.95*0.5`

`√(n) = (1.95*0.5)/(0.05)`

`(√(n))^2 = ((1.95*0.5)/(0.05))^2`

`n = 380.25`

Rounding up

The minimum sample size should be of 381.