Question

A random sample of elementary school children in New York state is to be selected to estimate the proportion p who have received a medical examination during the past year. An interval estimate of the proportion p with a margin of error of 0.06 0.06 and 99% 99 % confidence is required.

Answer 1

Answer:

The minimum sample size required is 461.

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

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

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

An interval estimate of the proportion p with a margin of error of 0.06. What is the minimum sample size required?

The minimum sample size required is n, which is found when M = 0.06.

We don't have an estimate for the true proportion, which means that we use $\pi = 0.5$. So

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

$0.06 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.06\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.06}$

$(\sqrt{n})^2 = (\frac{2.575*0.5}{0.06})^2$

$n = 460.5$

Rounding up

The minimum sample size required is 461.