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}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which:
is the sample proportion.
In this problem, we have a 95% confidence level, hence
, z is the value of Z that has a p-value of
, 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:
.
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](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.47%20-%201.96%5Csqrt%7B%5Cfrac%7B0.47%280.53%29%7D%7B450%7D%7D%20%3D%200.4239)
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.5161](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.47%20%2B%201.96%5Csqrt%7B%5Cfrac%7B0.47%280.53%29%7D%7B450%7D%7D%20%3D%200.5161)
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
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