Using the z-distribution, it is found that the 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

If we had increased the confidence level, the margin of error also would have increased.

<h3>What is a confidence interval of proportions?</h3>

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.

In this problem, we have a 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 the critical value is z = 1.96. Increasing the confidence level, z also increases, hence the margin of error also would have increased.

The sample size and the estimate are given as follows:

$n = 450, \pi = 0.47$ .

The lower and the upper bound of the interval are given, respectively, by:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.4239$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.5161$

The 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

5 0

1 year ago