# HW4Sol - Homework 5 solution 1 a The appropriate hypothesis...

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Homework 5 – solution 1. a) The appropriate hypothesis is H 0 : μ = 250 with the alternative H 1 : μ < 250 with the rejection being a “bad” outcome for the manufacturer. To test it we compute t 0 = 244 . 9 - 250 9 . 5 / 20 = - 2 . 40 . Since - 2 . 40 > - t 0 . 01 , 19 = - 2 . 539, the null hypothesis cannot be rejected, and the customer will buy the bottles. b) If the customer is new, the appropriate hypothesis is H 0 : μ = 250 with the alternative H 1 : μ > 250 with the rejection being a “good” outcome for the manufacturer. To test it we compute t 0 = 244 . 9 - 250 9 . 5 / 20 = - 2 . 40 . Since - 2 . 40 < t 0 . 01 , 19 = 2 . 539, the null hypothesis cannot be rejected, and the customer will not buy the bottles. c) Using the site mentioned in the homework, we obtain, for the case in a), P = Pr( X < - 2 . 40) = Pr( X > 2 . 40) = 0 . 013, and for the case in b), P = Pr( X > - 2 . 40) = 0 . 987. 2. For the hypothesis in 1a, the appropriate CI is the corresponding upper bound: μ ¯ x + t 0 . 01 , 19 s 20 = 244 . 9 + 2 . 539 · 9 . 5 20 = 250 . 3 . Since the claimed value of 250 is in this CI (250 < 250 . 3), the null hypothesis cannot be rejected. For the hypothesis in 1b, the appropriate CI is the corresponding lower bound: μ ¯ x - t 0 . 01 , 19 s 20 = 244 . 9 - 2 . 539 · 9 . 5 20 = 239 . 5 . Since the claimed value of 250 is in this CI (250

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