Question

A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence? Explain/show how you obtain your answer

Answer 1

Answer:

We need a sample of size at least 13.

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:

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

90% confidence interval: (0.438, 0.642).

The proportion estimate is the halfway point of these two bounds. So

$\pi = \frac{0.438 + 0.642}{2} = 0.54$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?

We need a sample of size at least n.

n is found when M = 0.08. So

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

$0.08 = 1.96\sqrt{\frac{0.54*0.46}{n}}$

$0.08\sqrt{n} = 1.96\sqrt{0.54*0.46}$

$\sqrt{n} = \frac{1.96\sqrt{0.54*0.46}}{0.08}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.54*0.46}}{0.08})^{2}$

$n = 12.21$

Rounding up

We need a sample of size at least 13.