According to statistics reported on \(\mathrm{CNBC}\), a surprising number of motor vehicles are not covered by insurance (CNBC, February 23,2006 ). Sample results, consistent with the CNBC report, showed 46 of 200 vehicles were not covered by insurance. a. What is the point estimate of the proportion of vehicles not covered by insurance? b. Develop a \(95 \%\) confidence interval for the population proportion.

To calculate the point estimate of the proportion of vehicles not covered by insurance, we divide the number of vehicles not covered by insurance by the total number of vehicles in the sample: \[ \hat{p} = \frac{\text{number of vehicles not covered}}{\text{total number of vehicles}} = \frac{46}{200} = 0.23\] So the point estimate is \(\hat{p} = 0.23\).

To develop a 95% confidence interval for the population proportion, we will use the formula for a confidence interval of a proportion: \[ CI = \hat{p} \pm z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] We already have the point estimate \(\hat{p} = 0.23\) and the sample size n = 200. Now, we need to find the z-score corresponding to the 95% confidence level. This value can be found in a standard normal distribution table or using a calculator, and it is approximately 1.96. Now, plug these values into the formula: \[ CI = 0.23 \pm 1.96 \times \sqrt{\frac{0.23(1-0.23)}{200}}\] \[ CI = 0.23 \pm 1.96 \times \sqrt{\frac{0.1769}{200}}\] \[ CI = 0.23 \pm 1.96 \times 0.0291\] \[ CI = 0.23 \pm 0.057 \] So, the 95% confidence interval for the population proportion of vehicles not covered by insurance is (0.173, 0.287).