Question

Della wants to make a one-sample interval to estimate what proportion of her community members favor a
tax increase for more local school funding. She wants her margin of error to be no more than -2% at the 95%
confidence level
What is the smallest sample size required to obtain the desired margin of error?

Answer 1

Answer:

2401

Step-by-step explanation:

answer on Khan Academy

Answer 2

Using the z-distribution, it is found that the smallest sample size required to obtain the desired margin of error is of 2401.

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $\alpha$, 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 $\frac{1+\alpha}{2}$.

The margin of error is of:

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

95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so $z = 1.96$.

In this problem, there is no prior estimate, hence $\pi = 0.5$ is used.

The smallest sample size needed is n for which M = 0.02, hence:

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

$0.02 = 1.96\sqrt{\frac{0.5(0.5)}{n}}$

$0.02\sqrt{n} = 1.96(0.5)$

$\sqrt{n} = \frac{1.96(0.5)}{0.02}$

$(\sqrt{n})^2 = \left(\frac{1.96(0.5)}{0.02}\right)^2$

$n = 2401$

The smallest sample size required to obtain the desired margin of error is of 2401.

You can learn more about the z-distribution at https://brainly.com/question/25844663