Data on the numbers of hospital admissions resulting from motor vehicle crashes are given below for Fridays on the 6 th of a month and Fridays on the following 13th of the same month. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Construct a $95 \%$ confidence interval estimate of the mean of the population of differences between hospital admissions. Use the confidence interval to test the claim that when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected. Friday the 6 th Friday the 13 th 12 13 7 14 3 15 12 15 7 14 In this example, $\mu_{d}$ is the mean value of the differences $d$ for the population of all pairs of data, where each individual difference $d$ is defined as the number of hospital admissions on Friday the 6th minus the number of hospital admissions on Friday the 13th. Find the $95 \%$ confidence interval. (Round to two decimal places as needed.)

Step 1 ：Given the number of hospital admissions on Friday the 6th and Friday the 13th of the same month, we first calculate the differences between the number of admissions on these two days for each pair of data. The differences are calculated as the number of admissions on Friday the 6th minus the number of admissions on Friday the 13th. The differences are: \(-1, -7, -12, -3, -7\).

Step 2 ：We then calculate the mean and standard deviation of these differences. The mean difference is \(-6.0\) and the standard deviation of the differences is approximately \(4.24\).

Step 3 ：We have 5 pairs of data, so the degrees of freedom is \(5 - 1 = 4\). Looking up the t-distribution table for a 95% confidence interval and 4 degrees of freedom, we find the critical t-value to be approximately \(2.78\).

Step 4 ：We can now calculate the 95% confidence interval for the mean difference using the formula: mean difference ± (t-value * standard deviation / sqrt(n)). Substituting the values we have, the confidence interval is approximately \(-11.27, -0.73\).

Step 5 ：Since this confidence interval does not include 0, it does not support the claim that the numbers of hospital admissions are not affected when the 13th day of a month falls on a Friday. In fact, it suggests that there are significantly fewer hospital admissions on Friday the 13th compared to Friday the 6th.

Step 6 ：Final Answer: The 95% confidence interval for the mean difference in hospital admissions on Friday the 6th and Friday the 13th is approximately \(\boxed{(-11.27, -0.73)}\).