Question

An advertising firm wants to confirm estimates that 24% of households will buy something on the Internet next year. What sample size will produce a margin of error of 9% with 90% confidence level

Answer 1

Answer:

A sample size of 61 will produce a margin of error of 9% with 90% confidence level

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 of the confidence interval is:

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

For this problem, we have that:

$p = 0.24$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.05$, so $Z = 1.645$.

What sample size will produce a margin of error of 9% with 90% confidence level

This is n when M = 0.09. So

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

$0.09 = 1.645\sqrt{\frac{0.24*0.76}{n}}$

$0.09\sqrt{n} = 1.645\sqrt{0.24*0.76}$

$\sqrt{n} = \frac{1.645\sqrt{0.24*0.76}}{0.09}$

$(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.24*0.76}}{0.09})^{2}$

$n = 61$

A sample size of 61 will produce a margin of error of 9% with 90% confidence level