Question

Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 329 with 49 successes at a confidence level of 99.5%.

Answer 1

Answer:

The magin of error is 0.0552 = 5.52 percentage points.

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:

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

For this problem, we have that:

$n = 329, \pi = \frac{49}{329} = 0.1489$

99.5% confidence level

So $\alpha = 0.005$, z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$, so $z = 2.81$.

M.E:

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

$M = 2.81\sqrt{\frac{0.1489*(0.8511)}{329}} = 0.0552$

The magin of error is 0.0552 = 5.52 percentage points.