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28 06 12

Esame completo di Statistica per il corso di Mechanical Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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Esame completo di Statistica per il corso di Mechanical Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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Second Written Partial Examination of Statistics. MEC Students - BOVISA Prof.ssa A. Guglielmi 28.06.12 /EP All rights reserved. Legal action will be taken against infr ingement. Reproduction is prohibited without prior consent. Name: Student Id. Number: Properly justify all your answers. Exercises Exercise 1 A new type of bow for archery competitions must be tested to de termine its shot length. The bow is fixed on a pedestal and is operated by a mecha nical arm in order to achieve the optimal traction for the longest possible shot. During a windless day, distances (in meters) covered by n = 16 consecutive arrows are measured by a very precise GPS dev ice, which returns two quantities: the sample mean ¯ x = 173 .13 of the distances covered by the arrows, and the lower confidence limit l = 172.08 of a one-sided 95% confidence interval for the mean distanc e. Assume that the shot lengths are independent and distribute d as a Gaussian random variable. 1. Compute the sample variance of the distances covered by th e arrows. 2. Is there empirical evidence that the expected shot length of the new bow is greater than 170 at the 1% significance level? 3. The chief project engineer provides later a reliable valu e for the variance of the distance, σ2 = 5. Knowing this value, compute the minimum number n1 of shots needed to get a type II error probability lower than 20% in a test with the sam e hypotheses and significance level as at point 2., when the true value of the mean is µ = 172. Explicitly solve an inequality. Solution The data are a i.i.d sample X1, . . . , X16 from N (µ, σ2), with unknown σ2. 1. The lower confidence limit of a one-sided 95% CI is l = ¯x − tα,n−1s16/√ n, where tα,n−1 = t0.05,15 = 1.753; therefore s2 16 = ((¯x − l)√ n/tα,n−1)2 = 5.7403. 2. We want to test H0 :…

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