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

Answer:

They should sample the results of 1990 patient results.

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

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 $1 - \frac{0.01}{2} = 0.995$, 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

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.025 = 2.575\sqrt{\frac{0.25*0.75}{n}}$

$0.025\sqrt{n} = 2.575\sqrt{0.25*0.75}$

$\sqrt{n} = \frac{2.575\sqrt{0.25*0.75}}{0.025}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.25*0.75}}{0.025})^2$

$n = 1989.2$

Rounding up:

They should sample the results of 1990 patient results.