MATH 1040 - Introduction to Statistics

Practice

1. A poll asks whether registered voters plan to vote in the upcoming election. In a random sample of 150 voters, 96 say they plan to vote. Find the margin of error at the 90% confidence level.

First, find the sample proportion and its complement:

\[\hat{p} = \frac{x}{n} = \frac{96}{150} = 0.64 \qquad \hat{q} = 1 - \hat{p} = 0.36\]

Verify the Central Limit Theorem applies:

• The sample is random — satisfied
• \(n\hat{p} = 150(0.64) = 96 \ge 10\) — satisfied
• \(n\hat{q} = 150(0.36) = 54 \ge 10\) — satisfied

Find the critical value. For a 90% confidence level, the remaining 10% is in the tails, with 5% in each tail. As we saw in Lesson 18.1 Critical Values, the critical value for a 90% confidence level is $z_c = \pm 1.645$.

Now calculate the margin of error:

\[\begin{align*} E &= z_c\sqrt{\frac{\hat{p}\hat{q}}{n}} \\ &= 1.645\sqrt{\frac{(0.64)(0.36)}{150}} \\ &= 1.645\sqrt{\frac{0.2304}{150}} \\ &= 1.645 \times 0.03919 \\ &\approx \mathbf{0.064} \end{align*}\]

The margin of error is 0.064, or about 6.4 percentage points.

Return back to Lesson 19.1