1. **Problem Statement:** Test if there has been an increase in the proportion of drivers who text while driving from last year (17%) to this year (29%) using a 99% confidence interval. 2. **Formula and Explanation:** We use a two-proportion z-test to compare two population proportions. The test statistic is: $$z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{p(1-p)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$$ where: - $\hat{p}_1 = 0.29$ (this year), $\hat{p}_2 = 0.17$ (last year) - $n_1 = n_2 = 1000$ - $p = \frac{x_1 + x_2}{n_1 + n_2}$ is the pooled proportion 3. **Calculate pooled proportion:** $$x_1 = 0.29 \times 1000 = 290$$ $$x_2 = 0.17 \times 1000 = 170$$ $$p = \frac{290 + 170}{1000 + 1000} = \frac{460}{2000} = 0.23$$ 4. **Calculate standard error:** $$SE = \sqrt{0.23 \times (1 - 0.23) \times \left(\frac{1}{1000} + \frac{1}{1000}\right)} = \sqrt{0.23 \times 0.77 \times 0.002} = \sqrt{0.0003542} \approx 0.0188$$ 5. **Calculate z-statistic:** $$z = \frac{0.29 - 0.17}{0.0188} = \frac{0.12}{0.0188} \approx 6.38$$ 6. **Determine critical z-value for 99% CI:** For a one-tailed test at 99% confidence, critical value $z_{0.01} = 2.33$ 7. **Decision:** Since $6.38 > 2.33$, we reject the null hypothesis. 8. **Conclusion:** There is sufficient evidence at the 99% confidence level to conclude that the proportion of drivers who text while driving has increased this year compared to last year.