What is 2/100000 as a decimal
2 answers:
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Answer:
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