Answer
$\mu_d$ is between -6.46 and -0.14.
Work Step by Step
The corresponding critical value using the table with df=6-1=5: $t_{\alpha/2}=t_{0.025}=2.571.$ The margin of error: $E=t_{\alpha/2}\frac{s_d}{\sqrt{n}}=2.571\frac{3.0111}{\sqrt{6}}=3.16.$ Hence the confidence interval $\mu_d$ is between $\overline{d}-E$=-3.3-3.16=-6.46 and $\overline{d}+E$=-3.3+3.16=-0.14. The interval is completely under 0, hence if the 13th is a Friday, the number of admissions in a hospital are less than usually.
