Question

A simple random sample of 100 8th graders at a large suburban middle school indicated that 83% of them are involved with some type of after school activity. Find the 99% confidence interval that estimates the proportion of them that are involved in an after school activity. a) (0.733, 0.927) b) (0.653, 0.927) c) (0.633, 0.877) d) (0.783, 0.788) e) (0.733, 0.727) f) None of the above

Answer 1

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

a) (0.733, 0.927).

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.

The other parameters are p = 0.83, n = 100, hence the bounds of the interval are given as follows.

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.83 - 2.575\sqrt{\frac{0.83(0.17)}{100}} = 0.733$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.83 + 2.575\sqrt{\frac{0.83(0.17)}{100}} = 0.927$

Hence option a is correct.

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