Question

A newspaper conducted a statewide survey concerning a proposal to raise taxes to prevent budget cuts to education. The newspaper took a random sample (assume it is an SRS) of 1200 registered voters and found that 580 would vote to raise taxes. Let p represent the proportion of registered voters in the state that would vote to raise taxes.Reference: Ref 20-4How large a sample n would you need to estimate p with margin of error 0.01 with 90% confidence

Answer 1

Answer:

A sample of 6758 would be needed.

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:

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

The newspaper took a random sample (assume it is an SRS) of 1200 registered voters and found that 580 would vote to raise taxes.

This means that $\pi = \frac{580}{1200} = 0.4833$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

How large a sample n would you need to estimate p with margin of error 0.01 with 90% confidence?

This is n for which M = 0.01. So

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

$0.01 = 1.645\sqrt{\frac{0.4833*0.5167}{n}}$

$0.01\sqrt{n} = 1.645\sqrt{0.4833*0.5167}$

$\sqrt{n} = \frac{1.645\sqrt{0.4833*0.5167}}{0.01}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.4833*0.5167}}{0.01})^2$

$n = 6757.5$

Rounding up:

A sample of 6758 would be needed.