Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, is found to be 115 and the sample standard​ deviation, s, is found to be 10.

​(a) Construct a 95% confidence interval about if the sample​ size, n, is 13.

​(b) Construct a 95​% confidence interval about if the sample​ size, n, is 26.

​(c) Construct a 96​% confidence interval about if the sample​ size, n, is 13.

​

Answer 1

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 13- 1 ) = 2.179
115 ± t(0.05/2, 13 -1) * 10/√(13)
Lower Limit = 115 - t(0.05/2, 13 -1) 10/√(13)
Lower Limit = 108.9571
Upper Limit = 115 + t(0.05/2, 13 -1) 10/√(13)
Upper Limit = 121.0429
95% Confidence interval is ( 108.9571 , 121.0429 )

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 26- 1 ) = 2.06
115 ± t(0.05/2, 26 -1) * 10/√(26)
Lower Limit = 115 - t(0.05/2, 26 -1) 10/√(26)
Lower Limit = 110.9609
Upper Limit = 115 + t(0.05/2, 26 -1) 10/√(26)
Upper Limit = 119.0391
95% Confidence interval is ( 110.9609 , 119.0391 )

Part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.04 /2, 13- 1 ) = 2.303
115 ± t(0.04/2, 13 -1) * 10/√(13)
Lower Limit = 115 - t(0.04/2, 13 -1) 10/√(13)
Lower Limit = 108.6134
Upper Limit = 115 + t(0.04/2, 13 -1) 10/√(13)
Upper Limit = 121.3866
96% Confidence interval is ( 108.6134 , 121.3866 )