hw6sol - APPM 4/5520 Solutions to Problem Set Six 1. Recall...

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Unformatted text preview: APPM 4/5520 Solutions to Problem Set Six 1. Recall that we define a t-distribution with n degress of freedom by letting Z N (0 , 1) and W 2 ( n ) be independent random variables. Let T = Z radicalbig W/n . Then T t ( n ). For this problem, note that X 1 | X 2 | = X 1 radicalBig X 2 2 = X 1 / radicalbig ( X 2 / ) 2 = X 1 / radicalbig ( X 2 / ) 2 / 1 t (1) since X 1 / N (0 , 1) and X 2 / N (0 , 1) ( X 2 / ) 2 2 (1). 2. (a) From the indicator in the pdf, we note that must be positive. So, P ( c &lt; X ( n ) / &lt; 1) = P ( c &lt; X ( n ) &lt; ) = F X ( n ) ( )- F X ( n ) ( c ) . The cdf for any one of the X s is F ( x ) = integraltext x f ( u ) du = x 3 / 3 . Thus, F X ( n ) ( x ) = P ( X ( n ) x ) iid = [ P ( X 1 x )] n = x 3 n 3 n . So, F X ( n ) ( )- F X ( n ) ( c ) = 3 n 3 n- c 3 n 3 n 3 n = 1- c 3 n . (b) 1- c 12 = 0 . 90 c = 0 . 10 1 / 12 . 82540419 So, we have that . 90 = P (0 . 10 1 / 12 &lt; X (4) / &lt; 1) which implies that . 90 = P ( X (4) &lt; &lt; X (4) / . 10 1 / 12 ) . The confidence interval is ( X (4) ,X (4) / . 10 1 / 12 ) (1 . 8 , 2 . 18) . 3. Small samples, normality, S instead of use t critical values. Pooled variance S 2 p = ( n 1- 1) S 2 1 + ( n 2- 1) S 2 2 n 1 + n 2- 2 = (16- 1)(0 . 15) + (8- 1)(0 . 10) 16 + 8- 2 = 0 . 1340909 The confidence interval is X 2- X 1 t / 2 ,n 1 + n 2 1 radicalBigg S 2 p parenleftbigg 1 n 1 + 1 n 2 parenrightbigg which, plugging the numbers in gives 5 . 22- 4 . 31 2 . 074 radicalBigg . 1340909 * parenleftbigg 1 16 + 1 8 parenrightbigg which is (0 . 5811415 , 1 . 238858) 4. Let W = ( n- 1) S 2 / 2 . Since W 2 ( n- 1), we wish to find the critical values between which we can capture area 1- under the pdf of the 2 ( n- 1) pdf....
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hw6sol - APPM 4/5520 Solutions to Problem Set Six 1. Recall...

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