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Unformatted text preview: AMS312.01 Final Exam Spring, 2010 Name __________________________________ID ______________________Signature___________________ ______ Instruction: This is a close book exam. Anyone who cheats in the exam shall receive a grade of F. Please provide complete solutions for full credit. The exam goes from 2:15 - 4:45pm. Good luck! 1. A Gallup survey portrays U.S. entrepreneurs as "... the mavericks, dreamers, and loners whose rough edges and uncompromising need to do it their own way set them in sharp contrast to senior executives in major American corporations" (Wall Street Journal, May 1985). One of the many questions put to a sample of 100 entrepreneurs about their work habits, social activities, etc., concerned the origin of the car they personally drive most frequently. The responses are given in the following table. U.S. Europe Japan 45 46 9 Do these data provide evidence of a difference in the preference of entrepreneurs for domestic cars versus foreign cars? Test at =0.05. Solution: This problem can be done in two ways using either (1) the test on one population proportion or (2) the Chi-square goodness-of-fit test with two categories. These two approaches are equivalent. (1) For the first approach, inference on one proportion, large sample, we have 100, 45 n x = = . Let p be the proportion of entrepreneurs with domestic cars, we have 45 100 p = , and we are testing: : 0.5 H p = versus : 0.5 a H p . The test statistics is: ( 29 0.5 1 0.5 1 0.5 /100 p Z- = = -- Since 0.25 1 1.96 Z Z = < = , we can not reject the null hypothesis at the significance level of 0.05. (2) Alternatively, and equivalently, you can use the Chi-square goodness-of-fit test. The above table is readily reduced to the following two-category table: Domestic cars Foreign cars 1 45 x = 2 55 x = Let 1 2 , p p be the proportion of entrepreneurs with domestic or foreign cars respectively, we are testing: 1 2 : 0.5, 0.5 H p p = = versus : a H H is not true. Hence we have 1 2 50, 50 e e = = . The test statistic is: 2 2 2 2 2 1,0.05, 0.025 1 ( ) 1 ( ) (1.96) 3.84 i i upper i i x e W Z e =- = = < = = Therefore we can not reject the null hypothesis at the significance level of 0.05. Of course you only need to show one of the two approaches above to get full credit. 2. ( 29 2 ~ , , 1,2,3 i i i X N i = and they are independent to each other, then (a) What is the distribution of 3 1 i i X = ? Prove your claim. (b) What is the distribution of 3 2 2 1 ( ) / i i i i X =- ? Prove your claim. Solution: (a) ( 29 [ ] ( 29 ( 29 ( 29 3 1 3 3 3 1 1 1 3 3 3 3 2 2 2 2 1 1 1 1 ( ) exp( ) exp( ) 1 1 exp( ) exp 2 2 i i i i i i i i i X X i i i i i i i i M t E exp X t E X t E X t M t t t t t = = = = = = = = = = = = = + = + Therefore we have shown that ( 29 ( 29 ( 29 3 3 3 2 1 1 1 ~ , i i i i i i X N = = = 1 (b) Let i i i i X Z - = , then we have...
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