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 90​% confidence interval about mu if the sample​ size, n, is 21. ​(b) Construct a 90​% confidence interval about mu if the sample​ size, n, is 14. ​(c) Construct an 80​% confidence interval about mu if the sample​ size, n, is 21. ​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

Answer 1

Answer: 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.

Solution:

Mean x̄ = 110

S.D, s . 10

​(a) Construct a 90​% confidence interval about μ if the sample​ size, n, is 21.

At 90% confidence interval, α = 0.10

df = n-1 = 21 - 1 = 20

t critical = t(α/2,df) = t(0.05,20)

t critical = 1.7247

the 90% confidence interval for μ:

CI = x̄ ± t critical * s/√n

CI = 110 ± 1.7247 * 10 / √21

CI = 110 ± 3.8565

CI = (106.1435, 113.8565)

Therefore, the 90% confidence interval is

106.1435 < μ < 113.8565.

​(b) Construct a 90​% confidence interval about μ if the sample​ size, n, is 14.

At 90% confidence interval, α = 0.10

df = n-1 = 14 - 1 = 13

t critical = t(α/2,df) = t(0.05,13)

t critical = 1.7709

the 90% confidence interval for μ:

CI = x̄ ± t critical * s/√n

CI = 110 ± 1.7709 * 10 / √14

CI = 110 ± 4.7329

CI = (105.2671, 114.7329)

Therefore, the 90% confidence interval is

105.2671 < μ < 114..7329

​(c) Construct an 80​% confidence interval about μ if the sample​ size, n, is 21.

At 80% confidence interval, α = 0.20

df = n-1 = 21 - 1 = 20

t critical = t(α/2,df) = t(0.10,20)

t critical = 1.3253

the 80% confidence interval for μ:

CI = x̄ ± t critical * s/√n

CI = 110 ± 1.3253 * 10 / √21

CI = 110 ± 2.8920

CI = (107.108, 112.892)

Therefore, the 80% confidence interval is

107.108 < μ < 112..892

​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

No, if the population had not been normally​ distributed, we had not computed the confidence interval in parts (a)-(c) as sample size is small.