1. **State the problem:** We want to construct a 99% confidence interval for the difference in proportions of ladies and men who responded in favour of compulsory motor vehicle insurance. 2. **Identify given data:** - Total ladies, $n_1 = 251$ - Ladies in favour, $x_1 = 185$ - Total men, $n_2 = 199$ - Men in favour, $x_2 = 107$ 3. **Calculate sample proportions:** $$\hat{p}_1 = \frac{185}{251} \approx 0.7371$$ $$\hat{p}_2 = \frac{107}{199} \approx 0.5377$$ 4. **Calculate the difference in sample proportions:** $$\hat{p}_1 - \hat{p}_2 = 0.7371 - 0.5377 = 0.1994$$ 5. **Find the critical value for 99% confidence:** For 99% confidence, $z_{\alpha/2} = 2.576$ 6. **Calculate the standard error (SE) of the difference:** $$SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} = \sqrt{\frac{0.7371(1-0.7371)}{251} + \frac{0.5377(1-0.5377)}{199}}$$ $$= \sqrt{\frac{0.1937}{251} + \frac{0.2487}{199}} = \sqrt{0.000771 + 0.001249} = \sqrt{0.00202} \approx 0.0449$$ 7. **Calculate the margin of error (ME):** $$ME = z_{\alpha/2} \times SE = 2.576 \times 0.0449 \approx 0.1157$$ 8. **Construct the confidence interval:** $$\left( (\hat{p}_1 - \hat{p}_2) - ME, (\hat{p}_1 - \hat{p}_2) + ME \right) = (0.1994 - 0.1157, 0.1994 + 0.1157) = (0.0837, 0.3151)$$ 9. **Interpretation:** We are 99% confident that the true difference in proportions of ladies and men who favour compulsory motor vehicle insurance is between 0.0837 and 0.3151. Since the interval is entirely positive, it suggests that a significantly higher proportion of ladies than men support compulsory insurance.