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Exam1_4502_sol

# Exam1_4502_sol - STA4502 Section 2906 EXAM 1 Solutions 1...

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STA4502 Section 2906 EXAM 1. Solutions 1. Below are recorded the white blood cell counts for a sample of 6 adult males with leukemia (recorded in thousands of cells). 19.3 32.0 1.8 43.0 14.1 1.4 Making no assumptions about the distribution of white cell counts in this population, find a point estimate of the median white blood cell count for the population of adult male with leukemia. Also, find roughly 80% confidence intervals for the population median cell count and interpret it. (10 pts) ˆ θ 0 . 5 = 14 . 1 + 19 . 3 2 = 16 . 7 From R , a roughly 80% confidence interval for the population median is given by [ X (2) , X (5) ) = [1 . 8 , 32 . 0) and the achieve confidence level is 4 summationdisplay k =2 parenleftbigg 6 k parenrightbigg parenleftbigg 1 2 parenrightbigg 6 = 0 . 78125 . We are 78 . 13% confident that the median white blood cell count for the population of adult male with leukemia falls in the interval [1 . 8 , 32 . 0). 2. Suppose X 1 , . . . , X 10 iid F . Let θ 0 . 5 = median of F . Find Pr bracketleftbig X (2) θ 0 . 5 < X (9) bracketrightbig . (5 pts) Pr bracketleftbig X (2) θ 0 . 5 < X (9) bracketrightbig = 8 summationdisplay k =2 parenleftbigg 10 k parenrightbigg parenleftbigg 1 2 parenrightbigg 10 0 . 9785 3. Describe the general principle that related a 90% confidence interval to a test of hypothesis. (5 pts) The 90% confidence interval consists of all the θ 0 values such that the test of H 0 : θ = θ 0 H a : θ negationslash = θ 0 would not reject H 0 in a two-sided α = 0 . 1 test of hypothesis.

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\$435 -\$1,735 \$822 \$572 -\$308 \$116 -\$786 \$51 4. The Internal Revenue Service takes a sample of tax returns and records the refund amount requested from the government ( A negative refund means that person owed the government.) A sample of eight returns showed the refunds below: Does this data indicate that the population median tax refund amount is positive? Make no assumption about the nature of the underlying population in conducting this test. Find the p value . (10 pts) H 0 : θ 0 . 5 = 0 H a : θ 0 . 5 > 0 Under H 0 , p = Pr ( X > 0) = 0 . 5. Thus, H 0 : p = 0 . 5 H a : p > 0 . 5 Test stat B = 8 summationdisplay i =1 I ( X i > 0) Under H 0 , B Binomial( n = 8 , p = 0 . 5). Therefore, p value = Pr ( B B obs ) = Pr ( B
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