Question

A reproductive clinic had a success rate of 25% of patients with live births in 2008. They want to update their success rate (the proportion of live births) for 2020. If they want their estimate to be within 2.5%, how many patient's results should they sample if they plan to use a 99% confidence interval? Round your answer up to the nearest integer.

Answer 1

They should sample the results of 1990 patient results.

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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is of:

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

A reproductive clinic had a success rate of 25% of patients with live births in 2008.

This means that
`\pi = 0.25`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a p-value of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

If they want their estimate to be within 2.5%, how many patient's results should they sample if they plan to use a 99% confidence interval?

They should sample n patients.

n is found for
`M = 0.025`. So

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

`0.025 = 2.575\sqrt{(0.25*0.75)/(n)}`

`0.025√(n) = 2.575√(0.25*0.75)`

`√(n) = (2.575√(0.25*0.75))/(0.025)`

`(√(n))^2 = ((2.575√(0.25*0.75))/(0.025))^2`

`n = 1989.2`

Rounding up:

They should sample the results of 1990 patient results.