Question

A sample of 20 joint specimens of a particular type gave a sample mean proportional limit stress of 8.56 MPa and a sample standard deviation of 0.78 MPa.
a. Calculate and interpret a 95% lower confidence bound for the true average proportional limit stress of all such joints.
b. Calculate and interpret a 95% lower prediction bound for proportional limit stress of a single joint of this type.
c.

Answer 1

Answer:

The answer is below

Step-by-step explanation:

n = 20 joints, mean (μ) = 8.56 MPa, standard deviation (σ) = 0.78 MPa

Given that the confidence is 95% (0.95).

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

The z score of α/2 is the same as the z score of 0.475 (0.5 - 0.025) which is equal to 1.96

a) for all joints, n = 20

The margin of error (E) is:

$E=z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } \\\\E=1.96*\frac{0.78}{\sqrt{20} } =0.34$

The confidence interval is:

Confidence interval = (μ ± E) = (8.56 ± 0.34) = (8.22, 8.90)

The 95% confidence interval is between 8.22 MPa and 8.90 MPa

b) for a single joint, n = 1

The margin of error (E) is:

$E=z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } \\\\E=1.96*\frac{0.78}{\sqrt{1} } =1.53$

The confidence interval is:

Confidence interval = (μ ± E) = (8.56 ± 1.53) = (7.03, 10.09)

The 95% confidence interval is between 7.03 MPa and 10.09 MPa