Step 1: Verify the Central Limit Theorem
Find the sample proportion and its complement:
\[\hat{p} = \frac{x}{n} = \frac{63}{180} = 0.35 \qquad \hat{q} = 1 - \hat{p} = 0.65\]The Central Limit Theorem applies. We can continue.
Step 2: 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 $z_c = \pm 1.645$.
Step 3: Find the Margin of Error
\[\begin{align*} E &= z_c\sqrt{\frac{\hat{p}\hat{q}}{n}} \\ &= 1.645\sqrt{\frac{(0.35)(0.65)}{180}} \\ &= 1.645\sqrt{\frac{0.2275}{180}} \\ &= 1.645 \times 0.03555 \\ &\approx 0.058 \end{align*}\]Step 4: Find the Confidence Interval
\[\hat{p} + E = 0.35 + 0.058 = 0.408\] \[\hat{p} - E = 0.35 - 0.058 = 0.292\]The confidence interval is \((0.292,\ 0.408)\).
Step 5: Interpret the Confidence Interval
We are 90% confident that the true proportion of students who experience high levels of stress during finals week is between 0.292 and 0.408 (between 29.2% and 40.8%).