Hw3Sol

# Hw3Sol - (1-p on the interval[0 4 the derivative f p is...

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Homework Solutions Assignment Number 3 Due February 5, 2007 Section 4.3 Problem 10. (a) Let X = number of shots made in next 100 attempts. Since p = P (attempt is successful) = 0 . 70 , P (75 X 80) = 80 X k =75 ± 100 k ² (0 . 70) k (0 . 30) 100 - k (b) With np = 100(0 . 70) = 70 and np (1 - p ) = 100(0 . 70)(0 . 30) = 21, P (74 . 5 X 80 . 5) = P ± 74 . 5 - 70 21 X - 70 21 80 . 5 - 70 21 ² = P (0 . 98 Z 2 . 29) = 0 . 1525 . Section 5.3 Problem 8 G . From Theorem 5.3.1, the conﬁdence interval is 179 220 - 1 . 64 r (179 / 220)(1 - 179 / 220) 220 , 179 220 + 1 . 64 r (179 / 220)(1 - 179 / 220) 220 ! = (0 . 771 , 0 . 857) Problem 10. Let p be the probability that a viewer would watch less than a quarter of the advertisements during Super Bowl XXIX. The conﬁdence interval for p is 281 1015 - 1 . 64 r (281 / 1015)(1 - 281 / 1015) 1015 , 281 1015 - 1 . 64 r (281 / 1015)(1 - 281 / 1015) 1015 ! = (0 . 254 , 0 . 300) Problem 24 G . As in the proof of Theorem 5.3.2, we have that n z 2 α/ 2 p (1 - p ) d 2 . Now in the proof, nothing is known about the value of p except that it is in the interval [0 , 1]. We, however know that it is in the interval [0 . 0 . 4], so we can ﬁnd a better bound for n that the theorem would give us. Namely, if we consider f ( p ) = p

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Unformatted text preview: (1-p ) on the interval [0 , . 4], the derivative f ( p ) is never zero, so the only critical points are the endpoints p = 0 and p = 0 . 4, so the largest that p (1-p ) can be is 0 . 4(1-. 4). Hence, n ≥ z 2 . 005 p (1-p ) (0 . 05) 2 = 2 . 58 2 (0 . 40)(0 . 60) (0 . 05) 2 = 639 . 01 , so we take n = 640 . 1 Section 5.4 Problem 10 G . E ( Y 2 ) = Z θ y 2 1 θ dy = θ 2 3 , so since we want E ( ˆ θ ) = θ 2 , and E ( Y 2 ) = θ 2 / 3, we can simply let ˆ θ = 3 Y 2 to get an unbiased ˆ θ . Problem 12. For each i , E [( Y i-μ ) 2 ] = σ 2 by the deﬁnition of σ 2 . Therefore, E " 1 n n X i =1 ( Y i-μ ) 2 # = 1 n n X i =1 E ± ( Y i-μ ) 2 ² = 1 n nσ 2 = σ 2 Problem 15. From Example 3.9.9, we know that Var( W ) = σ 2 n , so: E ( W 2 ) = Var( W ) + E ( W ) 2 = σ 2 n + μ 2 , so lim n →∞ E ( W 2 ) = lim n →∞ ³ σ 2 n + μ 2 ´ = μ 2 . 2...
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Hw3Sol - (1-p on the interval[0 4 the derivative f p is...

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