Question

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 24, 22, 14, 26, 28, 16, 20, 21. [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.) Answer is complete but not entirely correct. Sample mean Sample standard deviation 21.380 X 42.96 X b. Construct the 95% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.) Confidence interval to

Answer 1

Part a)

21.38 mean Data
4.75    std. dev.

Partb)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 8- 1 ) = 2.365

95% confidence interval is ( 17.40 , 25.35 )

Margin of Error = t(α/2, n-1) S/√(n) = 3.97

Part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 8- 1 ) = 3.499

99% confidence interval is ( 15.50 , 27.25 )

Margin of Error = t(α/2, n-1) S/√(n) = 5.88

Part d)

As confidence level increases, margin of error becomes larger.