Question

6. The following are IQ scores for some medical school students:

               120        106        121        112        114        134        140        121        168        127

Construct a 99% confidence interval for the population mean IQ score of a medical student.

7. Construct a 95% CI for σ in #6.

Answer 1

6)

sample mean, xbar = 126.3
sample standard deviation, s = 17.7954
sample size, n = 10
degrees of freedom, df = n - 1 = 9

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 3.25

ME = tc * s/sqrt(n)
ME = 3.25 * 17.7954/sqrt(10)
ME = 18.289

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (126.3 - 3.25 * 17.7954/sqrt(10) , 126.3 + 3.25 * 17.7954/sqrt(10))
CI = (108.01 , 144.59)

7)

Here s = 17.7954 and n = 10
df = 10 - 1 = 9

α = 1 - 0.95 = 0.05
The critical values for α = 0.05 and df = 9 are Χ^2(1-α/2,n-1) = 2.7 and Χ^2(α/2,n-1) = 19.023

CI = (sqrt(9*17.7954^2/19.023) , sqrt(9*17.7954^2/2.7))
CI = (12.24 , 32.49)