Hw7Sol - (35 . 4-2 . 2010 7 . 2 12 , 35 . 4 + 2 . 2010 7 ....

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Homework Solutions Assignment Number 7 Due March 5, 2007 Section 7.3. Problem 4. Let Y = ( n - 1) S 2 σ 2 . Then Var( Y ) = ( n - 1) 2 σ 4 Var( S 2 ) , (1) but by theorem 7.3.2, Y = χ 2 n - 1 , so Var( Y ) = Var( χ 2 n - 1 ) = 2( n - 1) , (2) with the last equality coming from exercise 7.3.2. Now, all that needs to be done is to set (1) equal to (2), to get: ( n - 1) 2 σ 4 Var( S 2 ) = 2( n - 1) Var( S 2 ) = 2 σ 4 n - 1 Problem 5. Since E ( S 2 ) = σ 2 , it follows from Chebyshev’s inequality that P ( | S 2 - σ 2 | < ε ) > 1 - Var( S 2 ) ε 2 . Now, Var( S 2 ) = 2 σ 4 n - 1 , and 2 σ 4 n - 1 0 as n → ∞ . Therefore S 2 is consistent for σ 2 . Section 7.4. Problem 9. (a) Let μ =true average age at which scientists make their greatest discoveries. Since 12 i =1 y i = 425 and 12 i =1 y 2 i = 15627 , y = 1 12 (425) = 35 . 4 and s = r 12(15627) - (425) 2 12(11) = 7 . 2 (Note: THis is the formula used in the example in the book. It is equivalent to the formula the professor gave in class.) Also, t α/ 2 ,n - 1 = t . 025 , 11 = 2 . 2010, so the 95% conFdence interval for μ is the range
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Unformatted text preview: (35 . 4-2 . 2010 7 . 2 12 , 35 . 4 + 2 . 2010 7 . 2 12 ), or (30 . 8 years , 40 . 0 years) . 1 (b) 1500 1600 1700 1800 1900 2000 20 25 30 35 40 45 50 Year Age The graph shows no correlation between time and age. Problem 19. Let = true average GMAT increase earned by students taking the review course. The hypotheses to be tested are H : = 40 versus H 1 : &lt; 40 . Here, 15 i =1 y i = 556 and 15 i =1 y 2 i = 20966, so y = 556 15 = 37 . 1, s = r 15(20966)-(556) 2 15(14) = 5 . 0, and t = 37 . 1-40 5 . / 15 =-2 . 25 . Since-t . 05 , 14 =-1 . 7613, H should be rejected at the = 0 . 05 level of signiFcance, suggesting that the MBAs R Us advertisement may be fraudulent. 2...
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Hw7Sol - (35 . 4-2 . 2010 7 . 2 12 , 35 . 4 + 2 . 2010 7 ....

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