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 two-sided 95% confidence interval for the standarddeviation.(b) What should you do to address any reservations about theconfidence interval you found in part (a)?

a) σ_95% = [ ± 0.09934 ]

b) Increase the number of samples taken from each clinic and find their average over course of months for the entire year.

Step-by-step explanation:

Solution:-

- The data from the health provider for the number of CAT scans performed in a month in its clinics.

- The reported variable "X" : number of CAT scans each month expressed as the number of CAT scans per thousand members of the health plan.

- The statistical results were obtained as follows:

           2.31, 2.09, 2.36, 1.95, 1.98, 2.25, 2.16, 2.07, 1.88, 1.94, 1.97,2.02

- We see that a samples were taken from n = 12 clinics were taken. The sample mean ( x_bar ) can be calculated from the following descriptive stats formula:

- Similarly, we will compute the sample standard deviation ( s ) ( Normally distributed population ) for unknown population standard deviation ( σ ), the following formula is used:

- We have two parameters ( x_bar and s ) for the approximated normal distribution of random variable X with unknown population:

                       x_bar = 2.0817

                       s = 0.15636

- The sample size n = 12 ≤ 30 and unknown population standard deviation standard normal distribution is not applicable. In such case we use t-distributions for test statistics and rejection criteria.

                       degree of freedom = n - 1 = 12 - 1 = 11

                       CI (two sided ) = 95%

                       Significance Level (α) = 1 - CI = 1-0.95 = 0.05

- To determine the t-critical value defined by the significance level of two sided test we have:

                      t-critical = t_α/2 = t _ 0.025

                      t-critical =  t _ 0.025 = ±2.201

- We will now construct a 95% confidence interval for the population standard deviation ( σ ).

                      σ_95% = [ -t-critical * ( s / √n ) , t-critical * ( s / √n ) ]

                      σ_95% = [ -2.201 * ( 0.15636 / √12 ) , 2.201 * ( 0.15636 / √12 ) ]  

                      σ_95% = [ ± 0.09934 ]                      

- To address any reservation to the standard error in standard deviation can be curtailed by increasing the sample size ( n ) at-least enough for the distribution "X" to assume standard normality. n ≥ 30

i would say c is the best answer, hope it's right!

good luck! =)

cutie~