HW4Sol

HW4Sol - would have been v u u t n-1 2 1-,n-1 s = s 99 69 ....

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Homework 4 – 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 z 0 = 244 . 9 - 250 10 / 20 = - 2 . 28 . Since - 2 . 28 > - z 0 . 01 = 2 . 33, 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 z 0 = 244 . 9 - 250 10 / 20 = - 2 . 28 . Since - 2 . 28 < z 0 . 01 = 2 . 33, the null hypothesis cannot be rejected, and the customer will not buy the bottles. c) For the P-values we obtain, respectively, P = Φ( - 2 . 28) = 0 . 011 and P = 1 - Φ( - 2 . 28) = 0 . 989. 2. We want the 99% upper bound on the standard deviation to be no more than 0.030mm. What we have is σ v u u t n - 1 χ 2 1 - α,n - 1 s = s 49 28 . 9 · 0 . 025 = 0 . 033 , and we see that it would not be safe to use the ball bearings based on the 99% upper bound criterion. On the other hand, if the sample size had been 100, the upper bound
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Unformatted text preview: would have been v u u t n-1 2 1-,n-1 s = s 99 69 . 2 . 025 = 0 . 030 , and it would have been safe. 1 3. a) If its known that p does not exceed 0.2, we can nd the required sample size as n = z 2 / 2 E 2 p (1-p ) = 1 . 96 2 . 02 2 . 2 . 8 = 1537 . b) If nothing is known about the true proportion, we have to use the worst case ( p = 0 . 5): n = z 2 / 2 E 2 . 25 = 1 . 96 2 . 02 2 . 25 = 2401 . 4. Since the CI is symmetric with respect to x , we can nd the latter as the middle of the CI: x = 5228 + 5051 2 = 5139 . 5 . In order to nd s , note that the upper limit of the interval has the form u = x + z . 025 15 , and therefore can be found as s = u- x z . 025 15 = 5251-5139 . 5 1 . 96 15 = 220 . 2...
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HW4Sol - would have been v u u t n-1 2 1-,n-1 s = s 99 69 ....

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