Question

An insurance company collects data on​ seat-belt use among drivers in a country. Of 1400 drivers 30-39 years​ old, 18​% said that they buckle​ up, whereas 379 of 1100 drivers 55-64 years old said that they did. Find a 98​% confidence interval for the difference between the proportions of​ seat-belt users for drivers in the age groups 30-39 years and 55-64 years.
Construct a 98​% confidence interval.
The 98​% confidence interval for P1−P2 is from _ to _

Answer 1

Question #71511, Math / Statistics and Probability

An insurance company collects data on seat-belt use among drivers in a country. Of 1400 drivers 30-39 years old, 18% said that they buckle up, whereas 379 of 1100 drivers 55-64 years old said that they did. Find a 98% confidence interval for the difference between the proportions of seat-belt users for drivers in the age groups 30-39 years and 55-64 years.

Construct a 98% confidence interval.

The 98% confidence interval for P1-P2 is from _ to _

Answer.

$$\widehat{\boldsymbol{p}}_1 = \mathbf{0.18}, \quad \boldsymbol{n}_1 = \mathbf{1400}, \quad \widehat{\boldsymbol{p}}_2 = \frac{379}{1100} = \mathbf{0.34}, \quad \boldsymbol{n}_1 = \mathbf{1100}.$$ $$\begin{array}{l} 98\% CI = \\ = \left(\widehat{\boldsymbol{p}}_1 - \widehat{\boldsymbol{p}}_2 - \mathbf{z}_{0.01} \sqrt{\frac{\widehat{p}_1(1 - \widehat{p}_1)}{n_1} + \frac{\widehat{p}_2(1 - \widehat{p}_2)}{n_2}}, \widehat{\boldsymbol{p}}_1 - \widehat{\boldsymbol{p}}_2 + \mathbf{z}_{0.01} \sqrt{\frac{\widehat{p}_1(1 - \widehat{p}_1)}{n_1} + \frac{\widehat{p}_2(1 - \widehat{p}_2)}{n_2}}\right) = \\ = (0.18 - 0.34 - 2.326 \sqrt{\frac{0.18 \cdot 0.82}{1400} + \frac{0.34 \cdot 0.66}{1100}}, 0.18 - 0.34 + 2.326 \sqrt{\frac{0.18 \cdot 0.82}{1400} + \frac{0.34 \cdot 0.66}{1100}}) = \\ = (-0.16 - 0.04, -0.16 + 0.04) = (-0.20, -0.12). \end{array}$$

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