Question

Exercise 6.13 presents the results of a poll evaluating support for the health care public option in 2009, reporting that 52% of Independents in the sample opposed the public option. If we wanted to estimate this number to within 1% with 90% confidence, what would be an appropriate sample size?

Answer 1

Answer:

A sample size of 6755 or higher would be appropriate.

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 M is given by:

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

90% confidence level

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

52% of Independents in the sample opposed the public option.

This means that $p = 0.52$

If we wanted to estimate this number to within 1% with 90% confidence, what would be an appropriate sample size?

Sample size of size n or higher when $M = 0.01$. So

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

$0.01 = 1.645\sqrt{\frac{0.52*0.48}{n}}$

$0.01\sqrt{n} = 0.8218$

$\sqrt{n} = \frac{0.8218}{0.01}$

$\sqrt{n} = 82.18$

$\sqrt{n}^{2} = (82.18)^{2}$

$n = 6754.2$

A sample size of 6755 or higher would be appropriate.