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 98​% confidence interval about mu if the sample​ size, n, is 13. ​(b) Construct a 98​% confidence interval about mu if the sample​ size, n, is 28. ​(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, n - 1 = 12

For 98% Confidence level, the t-value = 2.681
CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (109 - 2.681 * 10/sqrt(13) , 109 + 2.681 * 10/sqrt(13))
CI = (101.56 , 116.44)

b)
sample size, n = 28
degrees of freedom, n - 1 = 27

For 98% Confidence level, the t-value = 2.473
CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (109 - 2.473 * 10/sqrt(28) , 109 + 2.473 * 10/sqrt(28))
CI = (104.33 , 113.67)

c)
sample size, n = 13
degrees of freedom, n - 1 = 12

For 99% Confidence level, the t-value = 3.055
CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (109 - 3.055 * 10/sqrt(13) , 109 + 3.055 * 10/sqrt(13))
CI = (100.53 , 117.47)

d)
The assumption in calculating the CI is population is normally distributed.
It is not possible to calculated CI if population is not normally distributed