Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbar​, is found to be 109​, and the sample standard​ deviation, s, is found to be 10. ​

(a) Construct a 96​% confidence interval about mu if the sample​ size, n, is 13. ​

(b) Construct a 96​% confidence interval about mu if the sample​ size, n, is 29. ​

(c) Construct a 99​% confidence interval about mu if the sample​ size, n, is 13. ​

(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

Answer 1

a)
sample mean, xbar = 109
sample standard deviation, s = 10
sample size, n = 13
degrees of freedom, df = n - 1 = 12

Given CI level is 96%, hence α = 1 - 0.96 = 0.04
α/2 = 0.04/2 = 0.02, tc = t(α/2, df) = 2.3

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (109 - 2.3 * 10/sqrt(13) , 109 + 2.3 * 10/sqrt(13))
CI = (102.62 , 115.38)

b)
Given CI level is 96%, hence α = 1 - 0.96 = 0.04
α/2 = 0.04/2 = 0.02, tc = t(α/2, df) = 2.15

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (109 - 2.15 * 10/sqrt(29) , 109 + 2.15 * 10/sqrt(29))
CI = (105.01 , 112.99)

c)
Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 3.05

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (109 - 3.05 * 10/sqrt(13) , 109 + 3.05 * 10/sqrt(13))
CI = (100.54 , 117.46)

d)
No