Question

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)?

Answer 1

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.

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:

`x_b_a_r = (Sum ( x_i ))/(n)\\\\x_b_a_r=(2.31+2.09+2.36+1.95+1.98+2.25+2.16+2.07+1.88+1.94+1.97+2.02)/(12)\\\\x_b_a_r=(24.98)/(12)\\\\x_b_a_r=2.0817\\`

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

`s = \sqrt{(Sum (x_i - x_b_a_r)^2)/(n-1) } \\\\Sum (x_i - x_b_a_r)^2 = (2.31 - 2.0817)^2 + (2.09 - 2.0817)^2 + (2.36 - 2.0817)^2 + (1.95 -\\\\2.0817)^2 + (1.98 - 2.0817)^2+ (2.25 - 2.0817)^2+ (2.16 - 2.0817)^2 + (2.07 - 2.0817)^2\\\\(1.88 - 2.0817)^2 + (1.94 - 2.0817)^2 + (1.97 - 2.0817)^2 + (2.02 - 2.0817)^2\\\\= 0.26896668\\\\\\s = \sqrt{(0.26896668)/(12-1) } = 0.15636`

- 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