Researchers collected data on the numbers of hospital admissions resulting from motor vehicle crashes, and results are given below for Fridays on the 6 th of a month and Fridays on the following 13 th of the same month. Use a 0.05 significance level to test the claim that when the 13 th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected. \begin{tabular}{lrrrrrr} Friday the 6th: & 10 & 5 & 12 & 12 & 3 & 40 \\ Friday the 13th: & 14 & 11 & 14 & 11 & 5 & 12 \end{tabular} What are the hypotheses for this test? Let $\mu_{d}$ be the mean of the differences in the numbers of hospital admissions resulting from motor vehicle crashes for the population of all pairs of data. \[ \begin{array}{l} H_{0}: \mu_{d}=0 \\ H_{1}: \mu_{d} \neq 0 \end{array} \] Find the value of the test statistic. $t=-2.87 i$ (Round to three decimal places as needed.) Identify the critical value(s). Select the correct cholce below and fill the answer box within your choice. (Round to three decimal places as needed.) A. The critical value is $t=\square$. B. The critical values are $t= \pm \square$.

Step 1 ：Let \(\mu_{d}\) be the mean of the differences in the numbers of hospital admissions resulting from motor vehicle crashes for the population of all pairs of data. The hypotheses for this test are: \(H_{0}: \mu_{d}=0\) and \(H_{1}: \mu_{d} \neq 0\).

Step 2 ：The value of the test statistic is calculated using the formula for the t-statistic in a paired t-test. The test statistic is \(t=-2.87 i\) (rounded to three decimal places).

Step 3 ：The critical value is the value of t such that the probability of observing a value as extreme as, or more extreme than, the observed value under the null hypothesis is 0.05. This value can be found using a t-distribution table or a statistical software. The critical values are \(t= \pm 2.571\) (rounded to three decimal places).

Step 4 ：The test statistic is approximately \(0.482\), and the critical values are approximately \(\pm 2.571\). This means that if the test statistic falls outside of the range -2.571 to 2.571, we would reject the null hypothesis. However, our test statistic of 0.482 falls within this range, so we do not have enough evidence to reject the null hypothesis. Therefore, we cannot conclude that the mean difference in hospital admissions on Fridays the 6th and Fridays the 13th is different from zero.

Step 5 ：Final Answer: The test statistic is approximately \(\boxed{0.482}\) and the critical values are approximately \(\boxed{\pm 2.571}\).