A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data for a particular clinic follows (the reported variable is the number of CAT scans each month expressed as the number of CAT scans per thousand members of the health plan): 2.31 2.09 2.36 1.95 1.98 2.25 2.16 2.07 1.88 1.94 1.97 2.02 (a) Find a 95% two-sided confidence interval on the mean number of CAT scans performed each month at this clinic. (b) Historically, the mean number of scans performed by all clinics in the system has been 1.95. Is there any evidence that this particular clinic performs more CAT scans on overage than the overall system average?

Answer: a. 1.981 < μ < 2.18               b. Yes.Step-by-step explanation:A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.First, we calculate mean of the sample:[tex]\overline{x}=\frac{\Sigma x}{n}[/tex][tex]\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}[/tex][tex]\overline{x}=[/tex] 2.08Now, we estimate standard deviation:[tex]s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }[/tex][tex]s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }[/tex]s = 0.1564For t-score, we need to determine degree of freedom and [tex]\frac{\alpha}{2}[/tex]:df = 12 - 1 df = 11[tex]\alpha[/tex] = 1 - 0.95α = 0.05[tex]\frac{\alpha}{2}=[/tex] 0.025Then, t-score is[tex]t_{11,0.025}[/tex] = 2.201The interval will be[tex]\overline{x}[/tex] ± [tex]t.\frac{s}{\sqrt{n} }[/tex]2.08 ± [tex]2.201\frac{0.1564}{\sqrt{12} }[/tex]2.08 ± 0.099The 95% two-sided CI on the mean is 1.981 < μ < 2.18.B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.