Question

Help ASAP please it's my final exam.
Out of a sample of 650 high school students, 321 take the bus to school every day. Construct a 99% confidence interval for the population mean of high school students that take the bus to school every day.

CI = (45.54%, 53.23%)

CI = (46.16%, 52.61%)

CI = (43.23%, 51.18%)

CI = (44.34%, 54.43%)

Answer 1

Using the z-distribution, as we are working with a proportion, the 99% confidence interval is given by:

CI = (44.34%, 54.43%).

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.

In this problem, we have a 99% confidence level, hence $\alpha = 0.99$, z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$, so the critical value is z = 2.575.

Out of a sample of 650 high school students, 321 take the bus to school every day, hence:

$n = 650, \pi = \frac{321}{650} = 0.4938$

Then, the bounds of the interval are given by:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4938 - 2.575\sqrt{\frac{0.4938(0.5062)}{650}} = 0.4434$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4938 + 2.575\sqrt{\frac{0.4938(0.5062)}{650}} = 0.5443$

Which means that the correct option is:

CI = (44.34%, 54.43%)

More can be learned about the z-distribution at https://brainly.com/question/14398287