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

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