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 a 96​% confidence interval about mu if the sample​ size, n, is 12. ​(b) Construct a 96​% confidence interval about mu if the sample​ size, n, is 26. ​(c) Construct a 99​% confidence interval about mu if the sample​ size, n, is 12. ​(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 = 110
sample standard deviation, s = 10
sample size, n = 12
degrees of freedom, n - 1 = 11

For 96% Confidence level, the t-value = 2.328
CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (110 - 2.328 * 10/sqrt(12) , 110 + 2.328 * 10/sqrt(12))
CI = (103.28 , 116.72)

b)
sample mean, xbar = 110
sample standard deviation, s = 10
sample size, n = 26
degrees of freedom, n - 1 = 25

For 96% Confidence level, the t-value = 2.167
CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (110 - 2.167 * 10/sqrt(26) , 110 + 2.167 * 10/sqrt(26))
CI = (105.75 , 114.25)

c)
sample mean, xbar = 110
sample standard deviation, s = 10
sample size, n = 12
degrees of freedom, n - 1 = 11

For 99% Confidence level, the t-value = 3.106
CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (110 - 3.106 * 10/sqrt(12) , 110 + 3.106 * 10/sqrt(12))
CI = (101.03 , 118.97)

d)
No, it is not possible to calculate the CI if population is not normally distributed.