Question

Construct 90%, 95%, and 99% confidence intervals to estimate μ from the following data. State the point estimate. Assume the data come from a normally distributed population.

12.5	11.6	11.9	12.5	12.5	11.4	12.0
11.7	11.8	12.5

Appendix A Statistical Tables

(Round the intermediate values to 4 decimal places. Round your answers to 2 decimal places.)

90% confidence interval:

≤ μ ≤

95% confidence interval:

≤ μ ≤

99% confidence interval:

≤ μ ≤

The point estimate is

.

Answer 1

Mean X̅ = Σ Xi / n
X̅ = 120.4 / 10 = 12.04

Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )

SX = √ ( 1.644 / 10 -1 ) = 0.4274

Point Estimate X̅ = 12.04

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 10- 1 ) = 1.833
12.04 ± t(0.1/2, 10 -1) * 0.4274/√(10)
Lower Limit = 12.04 - t(0.1/2, 10 -1) 0.4274/√(10)
Lower Limit = 11.79
Upper Limit = 12.04 + t(0.1/2, 10 -1) 0.4274/√(10)
Upper Limit = 12.29
90% Confidence interval is ( 11.79 , 12.29 )

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 10- 1 ) = 2.262
12.04 ± t(0.05/2, 10 -1) * 0.4274/√(10)
Lower Limit = 12.04 - t(0.05/2, 10 -1) 0.4274/√(10)
Lower Limit = 11.73
Upper Limit = 12.04 + t(0.05/2, 10 -1) 0.4274/√(10)
Upper Limit = 12.35
95% Confidence interval is ( 11.73 , 12.35 )

Part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 10- 1 ) = 3.25
12.04 ± t(0.01/2, 10 -1) * 0.4274/√(10)
Lower Limit = 12.04 - t(0.01/2, 10 -1) 0.4274/√(10)
Lower Limit = 11.60
Upper Limit = 12.04 + t(0.01/2, 10 -1) 0.4274/√(10)
Upper Limit = 12.48
99% Confidence interval is ( 11.60 , 12.48 )