Sloution to problem set 10

Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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Section 6.1 1. a. We use the sample mean, x to estimate the population mean m . 1407 . 8 27 80 . 219 ˆ = = Σ = = n x x i m b. We use the sample median, 7 . 7 ~ = x (the middle observation when arranged in ascending order). c. We use the sample standard deviation, ( 29 660 . 1 26 94 . 1860 27 8 . 219 2 2 = - = = s s d. With “success” = observation greater than 10, x = # of successes = 4, and 1481 . ˆ 27 4 = = = n x p e. We use the sample (std dev)/(mean), or 2039 . 1407 . 8 660 . 1 = = x s 2. a. With X = # of T’s in the sample, the estimator is , 10 ; ˆ = = x p n X so 50 . , 20 10 ˆ = = p . b. Here, X = # in sample without TI graphing calculator, and x = 16, so 80 . 20 16 ˆ = = p 5. N = 5,000 T = 1,761,300 6 . 374 = y 6 . 340 = x 0 . 34 = d 000 , 703 , 1 ) 6 . 340 )( 000 , 5 ( ˆ 1 = = = x N q 300 , 591 , 1 ) 0 . 34 )( 000 , 5 ( 300 , 761 , 1 ˆ 2 = - = - = d N T q 281 . 438 , 601 , 1 6 . 374 6 . 340 300 , 761 , 1 ˆ 3 = = = y x T q
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Section 6.2 20. a. We wish to take the derivative of ( 29 - - x n x p p x n 1 ln , set it equal to zero and solve for p.
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Unformatted text preview: ( 29 ( 29 ( 29 p x n p x p x n p x x n dp d---= --+ + 1 1 ln ln ln ; setting this equal to zero and solving for p yields n x p = ˆ . For n = 20 and x = 3, 15 . 20 3 ˆ = = p b. ( 29 ( 29 ( 29 p np n X E n n X E p E = = = = 1 1 ˆ ; thus p ˆ is an unbiased estimator of p. c. ( 29 4437 . 15 . 1 5 =-30. The likelihood is ( 29 ( 29 y n y p p y n p n y f-- = 1 , ; where ( 29 l l l 24 24 1 24--=-= ≥ = ∫ e dx e X P p x . We know n y p = ˆ , so by the invariance principle ( 29 [ ] 0120 . 24 ln ˆ 24 =-= ⇒ =-n y n y e l l for n = 20, y = 15....
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This homework help was uploaded on 01/31/2008 for the course ENGR 361 taught by Professor Eisenstein during the Fall '04 term at Drexel.

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Sloution to problem - 29 29 29 p x n p x p x n p x x n dp d-= 1 1 ln ln ln setting this equal to zero and solving for p yields n x p = ˆ For n =

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