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project manager wishes to estimate the worst-case scenario for the variability of material strength (assumed normally distributed) of a new type of alloy, i.e., to obtain an upper limit on . Hence, a one-sided confidence interval is suitable. (a) Find the value c (to 0.00001) so that ( ^2< ((1)^2)/) = 99%, where s and are the sample and (unknown) population standard deviations, respectively, while n = sample size. Note that (n - 1)s^2/c is therefore a 99% upper confidence limit of the true (unknown) variance ^2, since ^2 has 0.99 probability of being smaller than it. (b) A sample of seventeen pieces of the metal are tested for tensile strength, with results (in MPa): {525,528,531,549,561,570,588,591,597,603,615,606,630,639,663,714,765} Compute the sample standard deviation s, and the (one-sided) 99% upper confidence limit of the population standard deviation (to 0.01 MPa).

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