Question

A simple random sample of sirenie drewn from a population that is normally distributed. The sample mean, is found to be 105, and the sample standard deviations, is found to be 10 (a) Construct a 95% confidence interval about if the sample size, n, is 15. (b) Construct a 95% confidence interval about the sample sens 29 (c) Construct a confidence interval about the samples , n, is 15. (d) Could we have computed the confidence intervals in parts (a)(0) the population had not been normally distruled? Click the icon to view the table of areas under the distribution

Answer 1

a)
sample mean, xbar = 105
sample standard deviation, s = 10
sample size, n = 15
degrees of freedom, df = n - 1 = 14

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.145

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (105 - 2.145 * 10/sqrt(15) , 105 + 2.145 * 10/sqrt(15))
CI = (99.46 , 110.54)

b)
sample size, n = 29
degrees of freedom, df = n - 1 = 28

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.048

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (105 - 2.048 * 10/sqrt(29) , 105 + 2.048 * 10/sqrt(29))
CI = (101.2 , 108.8)

c)
Given CI level is 96%, hence α = 1 - 0.96 = 0.04
α/2 = 0.04/2 = 0.02, tc = t(α/2, df) = 2.264

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (105 - 2.264 * 10/sqrt(15) , 105 + 2.264 * 10/sqrt(15))
CI = (99.15 , 110.85)

d)
No, as the baseline assumption is that data should be selected from normally distributed population