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Unformatted text preview: AMS 312.01 Practice Final Exam #1 Spring, 2010 Name_____________________________ID_______________Signature_____________ Instructions: 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. Good luck! 1. Arctic and Alpine Research investigated the relationship between the mean daily air temperature and the cocoon temperature of woolybear caterpillar’s of the High Arctic. (a) According to the data, can you conclude, at the significance level of 0.05, that the caterpillar’s body temperature is higher than the outside air temperature? (b) What assumptions are necessary for the above test? Temperature (ºC) Day Air Cocoon 1 10 15 2 9 14 3 2 7 4 3 6 5 5 10 Solution: By taking the paired differences (Diff) between the cocoon and the air temperatures for each day sampled, this problem reduce to a one-sample t-test on Diff. Temperature (ºC) Day Air Cocoon Diff 1 10 15 5 2 9 14 5 3 2 7 5 4 3 6 3 5 5 10 5 (a). Sample statistics: n = 5, , 6 . 4 = x s = 0.9. Hypotheses: H : μ = 0 versus Ha: μ > 0. Test statistic: 5 . 11 5 / 9 . 6 . 4 / ≈- =- = n s x t Since , 13 . 2 5 . 11 05 . , 4 = ≈ t t we reject H 0 in favor of Ha at the 0.05 significance level. That is, we conclude, at the significance level of 0.05, that the caterpillar’s body temperature is higher than the outside air temperature. (b). The assumption is that the population distribution of “Diff” is normal. 2. Let X 1 , X 2 , …, X n be a random sample from a normal population N(μ, σ 2 ). Furthermore, the population variance σ 2 is known. For a 2-sided test of H : μ = μ 0 versus Ha: μ ≠ μ , at the significance level α, (a). Derive the one-sample z-test using the pivotal quantity method. (* Please include the derivation of the pivotal quantity, the proof of its distribution, and the derivation of the rejection region for full credit.) (b). Prove that the likelihood ratio test is equivalent to the usual one sample z-test. Solution: (a). (1) Since this is inference on one population mean, we start with its point estimator – the sample mean, . X The distribution of X is ( 29 n N / , 2 σ μ . The distribution of X is not entirely known because μ is unknown. By taking the linear transformation n X Z / σ μ- = , however, we can see that Z is a function of the sample statistics and the unknown parameter of interest (μ) only. Furthermore the distribution of Z is completely known: ( 29 1 , ~ N Z . Thus Z is a pivotal quantity for the inference on μ. (2) The following is the proof of the distribution of Z using the moment generating function method. ( 29 ( 29 ( 29 ( 29 ( 29 = + - = + - = + ∏ - = ∏ - =...
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This note was uploaded on 11/14/2010 for the course AMS 312 taught by Professor Zhu,w during the Spring '08 term at SUNY Stony Brook.
- Spring '08