Find expressions for α and β that are interms of

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Find expressions for α and β that are interms of the moments of the joint distribution of x and y . In a skiing competition, a competitor’s overall time is found by adding his time in the downhill section to his time in the slalom section. Downhill times have an expected value of 1 minute and 15 seconds with a standard deviation of 5 seconds. Slalom times have an expected value of 1 minute and 45 seconds with a standard deviation of 6 seconds. Overall times have a standard deviation of 11 seconds. What is the expected overall time of a competitor who has recorded 1 minute and 10 seconds in the downhill section? Answer. The downhill time is x , the slalom time is y .We have E ( x ) = 75 , V ( x ) = 25 , E ( y ) = 105 , V ( y ) = 36 , and V ( x + y ) = V ( x ) + V ( y ) + 2 C ( x, y ) = 121 . Hence C ( x, y ) = 121 36 25 2 = 30 , and β = C ( x, y ) V ( x ) = 30 25 . If x = 70, then E ( y | x ) = E ( y ) + β © x E ( x ) = 105 + 30 25 { 70 75 } = 99 . Also E ( x + y | x ) = x + E ( y | x ) = 70 + 99 i.e. 2mins. 49secs. 5. Describe (a) the di ff erence between a one-tailed test of a statistical hy- pothesis and a two-tailed test, and (b) the di ff erence between a Type I error and a Type II error. Two guns in the same emplacement are firing in the same direction. Gun A is firing with a range of 5 km with a standard deviation of 200 m and gun B is firing with a range of 6 km with a standard derivation of 400 m. A shell falls 5.4 km from the emplacement. Test the hypothesis that it is fired by gun A, using a 5% level of significance, and evaluate the probability of committing a Type II error. 4
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EC2019 SAMPLING AND INFERENCE, January 2011 Answer. Let H A : x N ( μ A = 5 , σ A = 0 . 2) denote the hypothesis that the shell is from gun A and let H B : x N ( μ B = 6 , σ B = 0 . 4) denote the hypothesis that the shell is from gun B . Under H A , there is z A = ( x μ A ) / σ A N (0 , 1) and, with x = 5 . 4, there is z A = (5 . 4 5) / 0 . 2 = 0 . 2, which is a significant value given that P ( z A > c = 1 . 65) = 0 . 05. Therefore, we might be prepared to
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  • Spring '12
  • D.S.G.Pollock
  • Probability, Probability theory, Skiing, library card

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