Respuesta :
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.
What is a confidence interval of proportions?
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
- [tex]\pi[/tex] is the sample proportion.
- z is the critical value.
- n is the sample size.
In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], 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:
[tex]n = 450, \pi = 0.47[/tex].
The lower and the upper bound of the interval are given, respectively, by:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.4239[/tex]
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.5161[/tex]
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 https://brainly.com/question/25890103
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