Question

How many American adults must be interviewed to estimate the proportion of all American adults who actively try to avoid drinking regular soda or pop within 0.01 margin of error with 95% confidence using the large-sample confidence interval

Answer 1

Answer:

9604 American adults must be interviewed.

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 z-score that has a p-value 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 p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

How many adults must be interviewed for the margin of error to be 0.01?

This is n, which is found for M = 0.01.

We have no estimate for the proportion, so we use $\pi = 0.5$, which is when the largest sample size will be needed.

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

$0.01 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.01\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.01}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.01})^2$

$n = 9604$

9604 American adults must be interviewed.