Question

Find a confidence interval for μ assuming that each sample is from a normal population. (Round the value of t to 3 decimal places and your final answers to 2 decimal places.)

(a) x ¯ = 24, s = 3, n = 7, 90% confidence. The 90% confidence interval is from?

(b) x ¯ = 42, s = 6, n = 18, 99% confidence. The 99% confidence interval is from?

(c) x ¯ = 119, s = 14, n = 28, 95% confidence. The 95% confidence interval is from?

Answer 1

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 7- 1 ) = 1.943
24 ± t(0.1/2, 7 -1) * 3/√(7)
Lower Limit = 24 - t(0.1/2, 7 -1) 3/√(7)
Lower Limit = 21.80
Upper Limit = 24 + t(0.1/2, 7 -1) 3/√(7)
Upper Limit = 26.20
90% Confidence interval is ( 21.80 , 26.20 )

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 18- 1 ) = 2.898
42 ± t(0.01/2, 18 -1) * 6/√(18)
Lower Limit = 42 - t(0.01/2, 18 -1) 6/√(18)
Lower Limit = 37.90
Upper Limit = 42 + t(0.01/2, 18 -1) 6/√(18)
Upper Limit = 46.10
99% Confidence interval is ( 37.90 , 46.10 )

Part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 28- 1 ) = 2.052
119 ± t(0.05/2, 28 -1) * 14/√(28)
Lower Limit = 119 - t(0.05/2, 28 -1) 14/√(28)
Lower Limit = 113.57
Upper Limit = 119 + t(0.05/2, 28 -1) 14/√(28)
Upper Limit = 124.43
95% Confidence interval is ( 113.57 , 124.43 )