3. (Problem 10.41 of Freund's) To show that an estimator can be consistent with-out being unbiased nor asymptotically unbiased, consider the following estimationprocedure. Let X₁,..., Xn be a random sample from a population with mean μand finite variance o2. Consider the following estimation procedure: First, we ran-domly draw one slip of n papers numbered from 1 to n. If we get 2,3,...,n, we-1 nuse X = n ¹₁ as our estimate for u; otherwise, we use n² as the estimate fori=1μ. Show that this estimation procedure is(a) consistent.(b) neither unbiased nor asymptotically unbiased.

From the above given data the following solution is provided below:

Answered: 3. (Problem 10.41 of Freund's) To show…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Consider the information above and in question 1. It is conjectured that the mean number of people attending all 2017 Veterans Day celebrations is 145, and of interest is to test this conjecture versus the alternative that the mean number of people attending all 2017 Veterans Day celebrations is different from 145 people To test the hypothesis, a simple random sample of 41 Veterans Day celebrations was selected and the number of people attending each was recorded. The mean number of people attending for this sample of 41 celebrations was 157, with a standard deviation of 23 people. The distribution of the data was bimodal. If appropriate, use this information to test the hypotheses stated in question 3 at the a= .05 level of significance. 4 Are the assumptions met? Yes, because there is a simple random sample Yes, because there is a simple random sample and the population is normally distributed Yes, because there is a simple random sample and the sample size is large enough for the…
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4
5. A quality control engineer found that the inner and outer diameters of their ball bearings, denoted respectively as X and Y, are random variables with E(X|Y = y) = 0.7 – 0.3y inches and Var(X|Y = y) = 0.0004y for all values of y. From his past experience he also knew that the mean and variance of the outer diameters were E(Y) = 0.7 inches, and Var(Y) squared inches. = 0.0001 %3D • (i) What are the conditional random variables, E(X|Y) and Var(X|Y)? • (ii) Find the mean of the inner diameters, E(X). • (ii) Find the variance of the inner diameters, Var(X).
Solve question #9 for part D only.

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