MATH 1040 - Introduction to Statistics

19.1 Practice Problem 3

Practice

  1. A health survey of 250 randomly selected adults finds that 85 report exercising at least 3 times per week. Find the margin of error at the 99% confidence level.

First, find the sample proportion and its complement:

\[\hat{p} = \frac{x}{n} = \frac{85}{250} = 0.34 \qquad \hat{q} = 1 - \hat{p} = 0.66\]

Verify the Central Limit Theorem applies:

Find the critical value. For a 99% confidence level, the remaining 1% is in the tails, with 0.5% in each tail. As we saw in Lesson 18.1 Critical Values, the critical value for a 99% confidence level is $z_c = \pm 2.58$.

Now calculate the margin of error:

\[\begin{align*} E &= z_c\sqrt{\frac{\hat{p}\hat{q}}{n}} \\ &= 2.58\sqrt{\frac{(0.34)(0.66)}{250}} \\ &= 2.58\sqrt{\frac{0.2244}{250}} \\ &= 2.58 \times 0.02996 \\ &\approx \mathbf{0.077} \end{align*}\]

The margin of error is 0.077, or about 7.7 percentage points.

Return back to Lesson 19.1