Question

The following observations were made on fracture toughness of base plate of 18% nickel maraging steel in ksi √in. 69.5, 71.9, 72.6, 73.1, 73.3, 73.5, 75.5, 75.7, 75.8, 76.1, 76.2, 76.2, 77.0, 77.9, 78.1, 79.6, 79.7, 79.9, 80.1, 82.2, 83.7, 93.7, (a) Explain why the t-curve is more appropriate than the z-curve in this case. (b) Based upon this study, find the 99% lower confidence bound of the mean fracture toughness.(c) Find the 95% confidence interval.

Answer 1

a)
t-curve should be used because the population std. dev. is not given.
Also the sample size is less than 30.

b)
sample mean, xbar = 77.33
sample standard deviation, s = 5.0367
sample size, n = 22
degrees of freedom, df = n - 1 = 21

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.831

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (77.33 - 2.831 * 5.0367/sqrt(22) , 77.33 + 2.831 * 5.0367/sqrt(22))
CI = (74.29 , 80.37)

lower confidence bound = 74.29

c)
Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.08
CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (77.33 - 2.08 * 5.0367/sqrt(22) , 77.33 + 2.08 * 5.0367/sqrt(22))
CI = (75.1 , 79.56)