# hw13 - √ n ≤ μ ≤ ¯ X n 2 μ √ n ≈ 95 The inner...

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MATH 471 HW 13 Solutions Pengsheng Ji December 7, 2006 5.2.4 (a) Let X i = 1 if she answers question i correctly, =0 otherwise. S 48 = X 1 + ··· + X 48 is the number of questions she answers correctly. The X i have the Bernoulli distribution with p=3/4, σ = q p (1 - p ) = q 3 / 16. Using the histogram correction and (2.2) P ( S 48 > 37 . 5) = P S n - σ n > 37 . 5 - 48(3 / 4) q 48 · 3 / 16 P ( N (0 , 1) > 1 . 5 / 3) = 1 - Φ(0 . 5) = 0 . 3085 . (b) Here p = μ = 1 / 2 = q p (1 - p ) = 1 / 2. Similarly, P ( S 24 > 13 . 5) P ( N (0 , 1) > 0 . 61) = 0 . 2709 . 5.3.12 Since μ = σ = 1 , using Equation (3.1), we have P ˆ ¯ X n - 2 μ
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Unformatted text preview: √ n ≤ μ ≤ ¯ X n + 2 μ √ n ! ≈ -. 95 . The inner inequality is equivalent to ¯ X n √ n √ n +2 ≤ μ ≤ ¯ X n √ n √ n-2 , i.e. μ ∈ [2 . 727 , 3 . 333]. You may convert it to the conﬁdence interval for λ . 5.4.7 Similarly to Example 4.3, 2 s σ 2 1 n 1 + σ 2 2 n 2 = 2 s 12 2 216 + 10 2 75 = 2 . 8 while | ¯ X 1-¯ X 2 | = 2 . 5. Therefore, we do not reject H ....
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