Question

A, Eighteen subjects were randomly selected and given proficiency tests. the mean of this group is 492.3 and the standard deviation is 37.6. Construct the 95% confidence interval for the population mean.

B, use the sample data given in the above problem to construct the 98% confidence interval for the population standard deviation.

Answer 1

a)
sample mean, xbar = 492.3
sample standard deviation, s = 37.6
sample size, n = 18
degrees of freedom, df = n - 1 = 17

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.11

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (492.3 - 2.11 * 37.6/sqrt(18) , 492.3 + 2.11 * 37.6/sqrt(18))
CI = (473.6 , 511)

b)
Here s = 37.6 and n = 18
df = 18 - 1 = 17

α = 1 - 0.98 = 0.02
The critical values for α = 0.02 and df = 17 are Χ^2(1-α/2,n-1) = 6.408 and Χ^2(α/2,n-1) = 33.409

CI = (sqrt(17*37.6^2/33.409) , sqrt(17*37.6^2/6.408))
CI = (26.8213 , 61.2422)