Question

A study suggests that about 80\%80%80, percent of undergraduate students are working at a job. Researchers plan to do a follow-up study with a new sample of students. They'll use the new data to make a one-sample zzz interval to estimate the proportion of students who are working at a job. The researchers plan on using a confidence level of 90\%90%90, percent, and they want the margin of error to be no more than \pm 1\%±1%plus minus, 1, percent. If we assume \hat p=0. 80 p ^ ​ =0. 80p, with, hat, on top, equals, 0, point, 80, what is the smallest sample size required to obtain the desired margin of error?.

Answer 1

Using the z-distribution, it is found that the smallest sample size required to obtain the desired margin of error is of 4330.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

The margin of error is given by:

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

In this problem:

• The estimate of the proportion is of $\pi = 0.8$.

• The desired margin of error is of M = 0.01.

• 90% confidence level, hence$\alpha = 0.9$, z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$, so $z = 1.645$.

Then, we solve for n to find the minimum sample size.

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

$0.01 = 1.645\sqrt{\frac{0.8(0.2)}{n}}$

$0.01\sqrt{n} = 1.645\sqrt{0.8(0.2)}$

$\sqrt{n} = \frac{1.645\sqrt{0.8(0.2)}}{0.01}$

$(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.8(0.2)}}{0.01}\right)^2$

$n = 4329.6$

Rounding up, the smallest sample size required to obtain the desired margin of error is of 4330.

To learn more about the z-distribution, you can take a look at brainly.com/question/12517818