Dasgupta_Module5HW

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Northeastern University *

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7340

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Statistics

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Jan 9, 2024

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pdf

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Module 5 Homework Problem 1 When we have a random sample of size 6 with the observations as 1.433, 0.524, 0.384, 4.515, 1.852 and 0.429, a) The Maximum Likelihood Estimator using the Numerical method is: $minimum [1] 0.6566746 $objective [1] 8.523436 b) The Maximum Likelihood Estimator using the Analytical formula is: MLE = n/sum(observations) sum(observations) = 1.433 + 0.524 + 0.384 + 4.515 + 1.852 + 0.429 = 9.137 n = 6 MLE = 0.6566707 Problem 2 a) The point estimator m using Method of Moments is the sample mean itself. Hence, Method of Moments estimator is m = sample mean = 98.6 b) The one-sided 90% lower confidence interval of m: (97.19645, Inf) Problem 3 a) ‘ALL’ patients 95% bootstrap confidence interval for mean of Zyxin gene expression -0.580738750 -0.008846435 95% bootstrap confidence interval for variance of Zyxin gene expression

0.9812951 0.3240441 ‘AML’ patients 95% bootstrap confidence interval for mean of Zyxin gene expression 1.339698 1.833638 95% bootstrap confidence interval for variance of Zyxin gene expression 0.41621602 0.06597815 b) ‘ALL’ patients 95% parametric confidence interval for mean of Zyxin gene expression -0.580738750 -0.008846435 95% parametric confidence interval for variance of Zyxin gene expression 0.9812951 0.3240441 ‘AML’ patients 95% parametric confidence interval for mean of Zyxin gene expression 1.339698 1.833638 95% parametric confidence interval for variance of Zyxin gene expression 0.41621602 0.06597815 c) 95% bootstrap CI for the median gene expression in ‘ALL’ patients 2.5% 97.5% -0.73507 0.31432 95% bootstrap CI for the median gene expression in ‘AML’ patients 2.5% 97.5% 1.22814 1.82829 Problem 4 b) "When lambda = 0.1 : coverage for first CI is 0.877 , coverage for second CI is 0.547 ."

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