According to a certain govemment agency for a large country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration (BAC) is 0.38. Suppose a random sample of 105 traffic fatalities in a certain region results in 51 that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalifies involving a positive BAC than the country at the $\alpha=0.01$ level of significance? Because n $p_{0}\left(1-P_{0}\right)=\square \square$ T 10 , the sample size is $\square$. $5 \%$ of the population size, and the sample the requirements for testing the hypothesis $\overline{\mathbf{V}}$ satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? \[ \mathrm{H}_{0}: \square \square \text { V } \square \text { versus } \mathrm{H}_{5}: \square \mathbf{\nabla} \square \] (Type integers or decimals. Do not round.) Find the test statistic, $z_{0}$. $z_{0}=\square$ (Round to two decimal places as needed.) Find the P-value. P-value $=\square$ (Round to three decimal places as needed.)

Step 1 ：Define the null and alternative hypotheses. The null hypothesis is that the proportion of fatal traffic accidents involving a positive BAC in the region is equal to the proportion in the country, which is 0.38. The alternative hypothesis is that the proportion in the region is greater than 0.38. So, \(H_{0}: p = 0.38\) versus \(H_{1}: p > 0.38\).

Step 2 ：Calculate the sample proportion. We are given a sample size of 105 and a sample proportion of 51/105, which is approximately 0.486.

Step 3 ：Calculate the test statistic. The test statistic is calculated as (sample proportion - population proportion) / sqrt[(population proportion * (1 - population proportion)) / sample size]. Using the given values, the test statistic, \(z_{0}\), is approximately 2.23.

Step 4 ：Calculate the p-value. The p-value is the probability of observing a test statistic as extreme as the one we calculated, given that the null hypothesis is true. We can find this value using a z-table or a statistical software. The p-value is approximately 0.013.

Step 5 ：Compare the p-value to the significance level. The significance level is 0.01. Since the p-value is greater than the significance level, we fail to reject the null hypothesis. This means that we do not have enough evidence to suggest that the proportion of fatal traffic accidents involving a positive BAC in the region is greater than the proportion in the country.

Step 6 ：Final Answer: The null and alternative hypotheses are \(H_{0}: p = 0.38\) versus \(H_{1}: p > 0.38\). The test statistic is \(z_{0} = 2.23\). The p-value is 0.013. Since the p-value is greater than the significance level of 0.01, we fail to reject the null hypothesis. Therefore, we do not have enough evidence to suggest that the proportion of fatal traffic accidents involving a positive BAC in the region is greater than the proportion in the country. \(\boxed{H_{0}: p = 0.38, H_{1}: p > 0.38, z_{0} = 2.23, p-value = 0.013}\).