MATH 1040 - Introduction to Statistics

19.2 Practice Problem 1

Practice

  1. A local news station surveys a random sample of 300 residents and finds that 174 support a proposed new park. Construct a 95% confidence interval for the true proportion of residents who support the new park.

Step 1: Verify the Central Limit Theorem

Find the sample proportion and its complement:

\[\hat{p} = \frac{x}{n} = \frac{174}{300} = 0.58 \qquad \hat{q} = 1 - \hat{p} = 0.42\]

The Central Limit Theorem applies. We can continue.

Step 2: Find the Critical Value

For a 95% confidence level, the remaining 5% is in the tails, with 2.5% in each tail. As we saw in Lesson 18.1 Critical Values, the critical value for a 95% confidence level is $z_c = \pm 1.96$.

Step 3: Find the Margin of Error

\[\begin{align*} E &= z_c\sqrt{\frac{\hat{p}\hat{q}}{n}} \\ &= 1.96\sqrt{\frac{(0.58)(0.42)}{300}} \\ &= 1.96\sqrt{\frac{0.2436}{300}} \\ &= 1.96 \times 0.02850 \\ &\approx 0.056 \end{align*}\]

Step 4: Find the Confidence Interval

\[\hat{p} + E = 0.58 + 0.056 = 0.636\] \[\hat{p} - E = 0.58 - 0.056 = 0.524\]

The confidence interval is \((0.524,\ 0.636)\).

Step 5: Interpret the Confidence Interval

We are 95% confident that the true proportion of residents who support the proposed new park is between 0.524 and 0.636 (between 52.4% and 63.6%).

Return back to Lesson 19.2