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 110​, and the sample standard​ deviation, s, is found to be 10. ​(a) Construct an 80​% confidence interval about mu if the sample​ size, n, is 14. ​(b) Construct an 80​% confidence interval about mu if the sample​ size, n, is 23. ​(c) Construct a 99​% confidence interval about mu if the sample​ size, n, is 14. ​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

Answer 1

mean, x_bar = 110

standard deviation, s = 10

a) 80​% confidence interval about mu if the sample​ size, n, is 14

80​% confidence interval about mu = x_bar +/- z_critical*s/sqrt(n)

z_critical at 80​% = 1.28

80​% confidence interval about mu = 110 +/- 1.28*10/sqrt(14) = 110 +/- 1.28*10/3.74 = 110 +/- 3.42 = ( 106.58, 113.42 )

b) 80​% confidence interval about mu if the sample​ size, n, is 23

80​% confidence interval about mu = x_bar +/- z_critical*s/sqrt(n)

z_critical at 80​% = 1.28

80​% confidence interval about mu = 110 +/- 1.28*10/sqrt(23) = 110 +/- 1.28*10/4.79 = 110 +/- 2.67 = ( 107.33, 112.67 )

c) 99% confidence interval about mu if the sample​ size, n, is 14

99​% confidence interval about mu = x_bar +/- z_critical*s/sqrt(n)

z_critical at 99​% = 2.58

899% confidence interval about mu = 110 +/- 2.58*10/sqrt(14) = 110 +/- 2.58*10/3.74 = 110 +/- 6.89 = ( 103.11, 116.89 )

d) if the distribution of population is not known be it either normal or some other population then it is not possible to obtain the confidence interval