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 105​, 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 24. ​(b) Construct a 90​% confidence interval about mu if the sample​ size, n, is 20. ​(c) Construct a 70​% confidence interval about mu if the sample​ size, n, is 24. ​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

Answer 1

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 24- 1 ) = 1.714
105 ± t(0.1/2, 24 -1) * 10/√(24)
Lower Limit = 105 - t(0.1/2, 24 -1) 10/√(24)
Lower Limit = 101.5013
Upper Limit = 105 + t(0.1/2, 24 -1) 10/√(24)
Upper Limit = 108.4987
90% Confidence interval is ( 101.5013 , 108.4987 )

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 20- 1 ) = 1.729
105 ± t(0.1/2, 20 -1) * 10/√(20)
Lower Limit = 105 - t(0.1/2, 20 -1) 10/√(20)
Lower Limit = 101.1338
Upper Limit = 105 + t(0.1/2, 20 -1) 10/√(20)
Upper Limit = 108.8662
90% Confidence interval is ( 101.1338 , 108.8662 )

Part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.3 /2, 24- 1 ) = 1.06
105 ± t(0.3/2, 24 -1) * 10/√(24)
Lower Limit = 105 - t(0.3/2, 24 -1) 10/√(24)
Lower Limit = 102.8363
Upper Limit = 105 + t(0.3/2, 24 -1) 10/√(24)
Upper Limit = 107.1637
70% Confidence interval is ( 102.8363 , 107.1637 )

Part d)

Yes, confidence interval can be calculated even if population had not been normally​ distributed, because we use central limit theorem to calculate confience interval.