Question

Suppose that 19 measurements of the failure stress of carbon fiber specimens results in a sample mean of 562.68 MPa and sample standard deviation of 180.874 MPa and that the data is normally distributed. Calculate two-sided confidence intervals for the failure stress for the following cases:

a) 95% confidence interval for a single measurement (n = 1) of the failure stress. Assume that the population standard deviation is known to be 181 MPa.

b) 95% confidence interval for the sample mean with n = 19. Repeat for the 99% confidence interval. Assume that the population standard deviation is known to be 181 MPa

c) 95% confidence interval for the sample mean with n = 19. Repeat for the 99% confidence interval. Assume that the population standard deviation is UNKNOWN.

d) Suppose that an alternative process of producing carbon fibers results in a sample of failure stress measurements with n = 12, sample mean = 650.32 MPa and sample standard deviation of 130.470 MPa. What is the 95% confidence interval on the difference in failure stress between the two populations?

Answer 1

a) At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval for population mean is

+/- z_0.025 *

= 562.68 +/- 1.96 * 181/

= 562.68 +/- 354.76

= 207.92, 917.44

b) At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval for population mean is

+/- z_0.025 *

= 562.68 +/- 1.96 * 181/

= 562.68 +/- 81.388

= 481.292, 644.068

At 99% confidence interval the critical value is z_0.005 = 2.58

The 99% confidence interval for population mean is

+/- z_0.005 *

= 562.68 +/- 2.58 * 181/

= 562.68 +/- 107.133

= 455.547, 669.813

c) At 95% confidence interval the critical value is t^* = 2.101

The 95% confidence interval for population mean is

+/- t^* *

= 562.68 +/- 2.101 * 181/

= 562.68 +/- 87.242

= 475.438, 649.922

At 99% confidence interval the critical value is t^* = 2.878

The 99% confidence interval for population mean is

+/- t^* *

= 562.68 +/- 2.878 * 181/

= 562.68 +/- 119.507

= 443.173, 682.187

d) DF = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

          = ((180.874)^2/19 + (130.470)^2/12)^2/(((180.874)^2/19)^2/18 + ((130.470)^2/12)^2/11)

          = 28

At 95% confidence interval the critical value is t^* = 2.048

() +/- t^* * sqrt(s1^2/n1 + s2^2/n2)

= (562.68 - 650.32) +/- 2.048 * sqrt((180.874)^2/19 + (130.470)^2/12)

= -87.64 +/- 114.768

= -202.408, 27.128