Question

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 16, 21, 12, 23, 25, 26, 22, 17. [You may find it useful to reference the t table.]

a. Calculate the sample mean and the sample standard deviation. (Round intermediate calculations to at least 4 decimal places. Round "Sample mean" to 3 decimal places and "Sample standard deviation" to 2 decimal places.)

b. Construct the 80% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)

c. Construct the 90% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)

d. What happens to the margin of error as the confidence level increases from 80% to 90%?

Answer 1

Part a)

Mean X̅ = Σ Xi / n
X̅ = 162 / 8 = 20.25

Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 163.5 / 8 -1 ) = 4.83

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.2 /2, 8- 1 ) = 1.415
20.25 ± t(0.2/2, 8 -1) * 4.8329/√(8)
Lower Limit = 20.25 - t(0.2/2, 8 -1) 4.8329/√(8)
Lower Limit = 17.83
Upper Limit = 20.25 + t(0.2/2, 8 -1) 4.8329/√(8)
Upper Limit = 22.67
80% Confidence interval is ( 17.83 , 22.67 )

Margin of Error = t(α/2, n-1) S/√(n) = 2.4178

Part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 8- 1 ) = 1.895
20.25 ± t(0.1/2, 8 -1) * 4.8329/√(8)
Lower Limit = 20.25 - t(0.1/2, 8 -1) 4.8329/√(8)
Lower Limit = 17.01
Upper Limit = 20.25 + t(0.1/2, 8 -1) 4.8329/√(8)
Upper Limit = 23.49
90% Confidence interval is ( 17.01 , 23.49 )

Margin of Error = t(α/2, n-1) S/√(n) = 3.238

Part d)

As level of confidence increases, margin of error also increase.