Question

A sociologist wishes to estimate the percentage of the United States population living in poverty. What size sample should be obtained if she wishes the estimate the population proportion to be within 2 percentage points with 90% confidence, a) If she uses 1999 estimate of 11.8% obtained from the Current Population Survey. b) She does not use any estimate.

Answer 1

Answer:

a) We need a sample size of at least 705.

b) We need a sample size of at least 1692.

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 level

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

a) If she uses 1999 estimate of 11.8% obtained from the Current Population Survey.

We need a sample of size at least n

n is found when $M = 0.02, \pi = 0.118$.

So

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

$0.02 = 1.645\sqrt{\frac{0.118*0.882}{n}}$

$0.02\sqrt{n} = 1.645\sqrt{0.118*0.882}$

$\sqrt{n} = \frac{1.645\sqrt{0.118*0.882}}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.118*0.882}}{0.02})^{2}$

$n = 704.08$

Rounding up

We need a sample size of at least 705.

b) She does not use any estimate.

Same thing as above, we just use $\pi = 0.5$ when do not use any estimate.

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

$0.02 = 1.645\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 1.645\sqrt{0.5*0.5}$

$\sqrt{n} = \frac{1.645\sqrt{0.5*0.5}}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.5*0.5}}{0.02})^{2}$

$n = 1691.2$

Rounding up

We need a sample size of at least 1692.