Question

Use the given data to find the minimum sample size required to estimate a population proportion or percentage. Margin of error: seven percentage points, confidence level 95%, from a prior study,^p is estimated by the decimal equivalent of 42%
n=_____ (round to the nearest integer.)

Answer 1

Answer:

$n = 191$

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 interval $1-\alpha$, 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:

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

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 $Z = 1.96$.

In this problem, we have that:

$M = 0.07, \pi = 0.42$

We have to find n

$0.07 = 1.96*\sqrt{\frac{0.42*0.58}{n}}$

$0.07\sqrt{n} = 1.96*\sqrt{0.42*0.58}$

$0.07\sqrt{n} = 0.9674$

$\sqrt{n} = \frac{0.9674}{0.07}$

$\sqrt{n} = 13.82$

$\sqrt{n}^{2} = (13.82)^{2}$

$n = 190.9$

So, rounded to the nearest integer

$n = 191$