Question

A researcher would like to estimate p, the proportion of U.S. adults who support recognizing civil unions between gay or lesbian couples.
If the researcher would like to be 95% sure that the obtained sample proportion would be within 1.5% of p (the proportion in the entire population of U.S. adults), what sample size should be used?
(a) 17,778
(b) 4,445
(c) 1,112
(d) 67
(e) 45
Due to a limited budget, the researcher obtained opinions from a random sample of only 2,222 U.S. adults. With this sample size, the researcher can be 95% confident that the obtained sample proportion will differ from the true proportion (p) by no more than (answers are rounded):
(a) .04%
(b) .75%
(c) 2.1%
(d) 3%
(e) There is no way to figure this out without knowing the actual sample proportion that was obtained.

Answer 1

Answer:

Question 1:

(b) 4,445

Question 2:

(c) 2.1%

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 of:

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

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 $Z = 1.96$.

Question 1:

We have no previous estimate for the population proportion, so we use $\pi = 0.5$.

The sample size is n for which M = 0.015. So

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

$0.015 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.015\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.015}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.015})^2$

$n = 4268$

Samples above this value should be used, and the smaller sample above this value is of 4445, so the answer is given by option b.

Question 2:

Now we find M for which $n = 2222$.

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

$M = 1.96\sqrt{\frac{0.5*0.5}{2222}}$

$M = 0.021$

So 2.1%, and the correct answer is given by option c.