Question

Use the given data to find the minimum sample size required to estimate the population proportion.

a. Margin of error: 0.004: confidence level: 95% p and q unknown
b. Margin of error: 0.009, confidence level: 99%; p and q unknown
c. Margin of error: 0.01; confidence level: 95%; from a prior study, p is estimated by the decimal equivalent of 52%.

Answer 1

Answer:

a. The minimum sample size required is of 60,025.

b. The minimum sample size required is of 20,465.

c. The minimum sample size required is 9,589.

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:

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

Question a:

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 $Z = 1.96$.

p and q unknown means that we use $\pi = 0.5$, which is when the largest sample size is needed.

Margin of error: 0.004

We have to find n for which M = 0.004. So

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

$0.004 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.004\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.004}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.004})^2$

$n = 60025$

The minimum sample size required is of 60,025.

b. Margin of error: 0.009, confidence level: 99%; p and q unknown

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

Margin of error: 0.009

We have to find n for which M = 0.009. So

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

$0.009 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.009\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.009}$

$(\sqrt{n})^2 = (\frac{2.575*0.5}{0.009})^2$

$n = 20464.9$

Rounding up:

The minimum sample size required is of 20,465.

c. Margin of error: 0.01; confidence level: 95%; from a prior study, p is estimated by the decimal equivalent of 52%.

Here we have $\pi = 0.52$.

The minimum sample size is n for which M = 0.01. So

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

$0.01 = 1.96\sqrt{\frac{0.52*0.48}{n}}$

$0.01\sqrt{n} = 1.96\sqrt{0.52*0.48}$

$\sqrt{n} = \frac{1.96\sqrt{0.52*0.48}}{0.01}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.52*0.48}}{0.01})^2$

$n = 9588.6$

Rounding up:

The minimum sample size required is 9,589.