Question

19. In a random sample of 250 students, we found that 75 work out 4 or more times a week. Find the 95% confidence interval for the proportion of students who work out 4 or more times a week.

Answer 1

Answer:

The 95% confidence interval for the proportion of students who work out 4 or more times a week is (0.2432, 0.3568).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

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

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

In a random sample of 250 students, we found that 75 work out 4 or more times a week.

This means that $n = 250, \pi = \frac{75}{250} = 0.3$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3 - 1.96\sqrt{\frac{0.3*0.7}{250}} = 0.2432$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3 + 1.96\sqrt{\frac{0.3*0.7}{250}} = 0.3568$

The 95% confidence interval for the proportion of students who work out 4 or more times a week is (0.2432, 0.3568).