Question

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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.

13.3	11.6	11.9	13.1	12.5	11.4	12.0
11.7	11.8	13.3

Appendix A Statistical Tables

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

90% confidence interval: enter the lower limit of the 90% confidence interval  ≤ μ ≤ enter the upper limit of the 90% confidence interval
95% confidence interval: enter the lower limit of the 95% confidence interval  ≤ μ ≤ enter the upper limit of the 95% confidence interval
99% confidence interval: enter the lower limit of the 99% confidence interval  ≤ μ ≤ enter the upper limit of the 99% confidence interval

The point estimate is enter the point estimate  .

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	13.3	1.0816
	11.6	0.4356
	11.9	0.1296
	13.1	0.7056
	12.5	0.0576
	11.4	0.7396
	12	0.0676
	11.7	0.3136
	11.8	0.2116
	13.3	1.0816
Total	122.6	4.824

Mean X̅ = Σ Xi / n
X̅ = 122.6 / 10 = 12.26
Sample Standard deviation SX = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 4.824 / 10 -1 ) = 0.7321

Point Estimate X̅ = 12.26

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
12.26 ± t(0.1/2, 10 -1) * 0.7321/√(10)
Lower Limit = 12.26 - t(0.1/2, 10 -1) 0.7321/√(10)
Lower Limit = 11.8356 ≈ 11.84
Upper Limit = 12.26 + t(0.1/2, 10 -1) 0.7321/√(10)
Upper Limit = 12.6844 ≈ 12.68
90% Confidence interval is ( 11.84 , 12.68 )

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
12.26 ± t(0.05/2, 10 -1) * 0.7321/√(10)
Lower Limit = 12.26 - t(0.05/2, 10 -1) 0.7321/√(10)
Lower Limit = 11.7363 ≈ 11.74
Upper Limit = 12.26 + t(0.05/2, 10 -1) 0.7321/√(10)
Upper Limit = 12.7837 ≈ 12.78
95% Confidence interval is ( 11.74 , 12.78 )

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
12.26 ± t(0.01/2, 10 -1) * 0.7321/√(10)
Lower Limit = 12.26 - t(0.01/2, 10 -1) 0.7321/√(10)
Lower Limit = 11.5076 ≈ 11.51
Upper Limit = 12.26 + t(0.01/2, 10 -1) 0.7321/√(10)
Upper Limit = 13.0124 ≈ 13.01
99% Confidence interval is ( 11.51 , 13.01 )