Question

A random sample of 11 items is drawn from a population whose standard deviation is unknown. The sample mean is x¯ = 920 and the sample standard deviation is s = 25. Use Appendix D to find the values of Student’s t.

(a) Construct an interval estimate of μ with 95% confidence. (Round your answers to 3 decimal places.)
  
The 95% confidence interval is from to

(b) Construct an interval estimate of μ with 95% confidence, assuming that s = 50. (Round your answers to 3 decimal places.)
  
The 95% confidence interval is from to

(c) Construct an interval estimate of μ with 95% confidence, assuming that s = 100. (Round your answers to 3 decimal places.)
  
The 95% confidence interval is from to

Answer 1

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 11- 1 ) = 2.228
920 ± t(0.05/2, 11 -1) * 25/√(11)
Lower Limit = 920 - t(0.05/2, 11 -1) 25/√(11)
Lower Limit = 903.205
Upper Limit = 920 + t(0.05/2, 11 -1) 25/√(11)
Upper Limit = 936.795
95% Confidence interval is ( 903.205 , 936.795 )

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 11- 1 ) = 2.228
920 ± t(0.05/2, 11 -1) * 50/√(11)
Lower Limit = 920 - t(0.05/2, 11 -1) 50/√(11)
Lower Limit = 886.410
Upper Limit = 920 + t(0.05/2, 11 -1) 50/√(11)
Upper Limit = 953.590
95% Confidence interval is ( 886.410 , 953.590 )

Part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 11- 1 ) = 2.228
920 ± t(0.05/2, 11 -1) * 100/√(11)
Lower Limit = 920 - t(0.05/2, 11 -1) 100/√(11)
Lower Limit = 852.819
Upper Limit = 920 + t(0.05/2, 11 -1) 100/√(11)
Upper Limit = 987.181
95% Confidence interval is ( 852.819 , 987.181 )