exam3_A

exam3_A - EON317 Midterm Examination#3 Form A Professor...

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Unformatted text preview: EON317 Midterm Examination #3 Form A Professor Manochehr Rashidian April 22, 2008 1. (25 POINTS) Let μ 1 , μ 2 and μ 3 denote the mean tire life (in 1,000 miles) of Goodyear, Toyo, and General, respectively. The null and the alternative hypotheses are H : μ 1 = μ 2 = μ 3 , H a : The mean lives differ for at least two of the brands. Since ¯ x = 66 , SST = n 1 (¯ x 1- ¯ x ) 2 + n 2 (¯ x 2- ¯ x ) 2 + n 3 (¯ x 3- ¯ x ) = 4(62- 66) 2 + 4(65- 66) 2 + 5(70- 66) 2 = 148 , SSE = ( n 1- 1) s 2 1 + ( n 2- 1) s 2 2 + ( n 3- 1) s 2 3 = 3(11 . 5181) 2 + 3(5 . 7735) 2 + 4(6 . 2048) 2 = 652 , MST = SST p- 1 = 148 3- 1 = 74 , MSE = SSE n- p = 652 13- 3 = 65 . 2 , a test statistic F = MST MSE = 1 . 1350 . The critical value F . 10 = 2 . 92. So we FAIL to reject the null hypothesis. There is no evidence that the mean tire lives differ for at least two of the brands. 1 2. (25 POINTS) Let n 1 and n 2 denote the number of samples from ad- justable rate mortgages and fixed rate mortgages. Assuming the equal size of sampling, n 1 = n 2 = n , B = z . 01 / 2 r p 1 q 1 n + p 2 q 2 n ⇒ . 10 2 = 2 . 575 r ( . 5)( . 5) n + ( . 5)( . 5) n ⇒ n = n 1 = n 2 = (2 . 575) 2 (( . 5)( . 5) + ( . 5)( . 5)) . 05 2 = 1326 . 1 . Therefore we need n 1 = n 2 = 1327. 3. (25 POINTS) n 1 = 250 ,n 2 = 188 ,x 1 = 28 ,x 2 = 24, thus ˆ p 1 = 0 . 1120 and ˆ p 2 = 0 . 1277....
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exam3_A - EON317 Midterm Examination#3 Form A Professor...

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