Question

Assume that you wish to estimate a population proportion, p. For the given margin of error and confidence level, determine the sample size required:
A researcher wants to determine what proportion of adults in one town regularly buy organic food. Obtain a sample size that will ensure a margin of error of at most 0.04 for a 90% confidence interval. In previous years, the proportion has been 0.16 .

Answer 1

Answer:

The sample size required is 228.

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:

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

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

In previous years, the proportion has been 0.16.

This means that $\pi = 0.16$

Obtain a sample size that will ensure a margin of error of at most 0.04 for a 90% confidence interval.

This is n for which M = 0.04. So

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

$0.04 = 1.645\sqrt{\frac{0.16*0.84}{n}}$

$0.04\sqrt{n} = 1.645\sqrt{0.16*0.84}$

$\sqrt{n} = \frac{1.645\sqrt{0.16*0.84}}{0.04}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.16*0.84}}{0.04})^2$

$n = 227.3$

Rounding up:

The sample size required is 228.