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 0.32 . Suppose a random sample of 110 traffic fatalities in a certain region results in 49 that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the $\alpha=0.1$ level of significance?
Because $\left.\mathrm{np}_{0}\left(1-\mathrm{p}_{0}\right)=23.9\right]> 10$, the sample size is less than $5 \%$ of the population size, and the sample is given to be random, (Round to one decimal place as needed.) the requirements for testing the hypothesis are satisfied.
What are the null and alternative hypotheses?
$\mathrm{H}_{0}: \mathrm{p}=32$ versus $\mathrm{H}_{1}: \mathrm{p}> 32$
(Type integers or decimals. Do not round.)
Find the test statistic, $z_{0}$.
$z_{0}=\square$ (Round to two decimal places as needed.)
Answer
The final answer is: The null and alternative hypotheses are \(H_{0}: p=0.32\) versus \(H_{1}: p>0.32\). The test statistic, \(z_{0}\), is approximately \(\boxed{2.82}\).
Step 1 ：The problem is asking whether the proportion of fatal traffic accidents involving a positive blood alcohol concentration (BAC) in a certain region is higher than the country's proportion. The country's proportion is given as 0.32. A random sample of 110 traffic fatalities in the region resulted in 49 that involved a positive BAC.
Step 2 ：The null hypothesis is that the proportion of fatal traffic accidents involving a positive BAC in the region is the same as the country's proportion, which is 0.32. This can be written as \(H_{0}: p=0.32\).
Step 3 ：The alternative hypothesis is that the proportion in the region is greater than the country's proportion. This can be written as \(H_{1}: p>0.32\).
Step 4 ：To find the test statistic, we can use the formula for the z-score in hypothesis testing for proportions, which is \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\), where \(\hat{p}\) is the sample proportion, \(p_0\) is the population proportion, and \(n\) is the sample size.
Step 5 ：In this case, \(\hat{p} = \frac{49}{110}\), \(p_0 = 0.32\), and \(n = 110\).
Step 6 ：Substituting these values into the formula, we get \(z = \frac{0.44545454545454544 - 0.32}{\sqrt{\frac{0.32(1-0.32)}{110}}}\).
Step 7 ：Solving this gives a test statistic, \(z_0\), of approximately 2.82. This value represents how many standard deviations the sample proportion is away from the population proportion under the null hypothesis.
Step 8 ：The final answer is: The null and alternative hypotheses are \(H_{0}: p=0.32\) versus \(H_{1}: p>0.32\). The test statistic, \(z_{0}\), is approximately \(\boxed{2.82}\).
