4.(a) The exponential distribution with mean > 0 has pdf1-x/0fe(x):= efo(x) =2This distribution has mean and variance 02. Show that the sample meanX = n¹₁ X, is a minimum-variance unbiased estimator of 0. Verifythat X achieves the CRLB for estimating 0.-1(b) The Pareto distribution with mean 0 > 1 has pdf=x > 0.x-(207)(1²1)A-"x > 1.Suppose X₁,..., Xn are iid with common pdf fo(·).1 n(i)The sample mean X = 1X₂ is an unbiased estimator of 0. Derivea formula for its variance as a function of 0.n(ii) Write down the log-likelihood lo(0; X) based on X = (X₁,..., Xn).(iii) Write down the score function(0; X). Can it be written in the form2% [B(X) – 0] ?-

Answered: Image /qna-images/answer/9663a46a-4ba6-438f-9147-348c87e9e2f1.jpg

Answered: (a) The exponential distribution with…

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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IfXhas exponential distribution with mean 1 (X~EXP(1)), then P(X>2|X>1)equals to
In the past, a chemical company produced 880 pounds of a certain type of plastic per day. Now, using a newly developed and less expensive process, the mean daily yield of plastic for the first 50 days of production is 871 pounds; the standard deviation is 21 pounds. Do the data provide sufficient evidence to indicate that the mean daily yield for the new process is less than that of the old procedure? (Use α=0.05) (a) Alternative hypothesis for the problem above is μ 871 Ομ 880 <
Suppose insurance loss X follows a Pareto distribution with mean 200 and variance 60,000. (a) Find α, and θ. (b) Find eX(100). (c) Find VaR0.95(X). (c) Find TVaR0.95(X)
N= 150 observations were collected on a time series that was identified as a AR(2) time series. The following statistics were computed from the data. Mean - 45.0 Variance 15.6 Autocorrelation function (up to lag 5) r1 = 0.80, r2 = .50, rз = .26, r4 = -.10, rs = 0.08: == Estimate the parameters of the model using the method of moments.
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A normal distribution has a mean of μμ= 54 and a standard deviation of σσ=6 What is the probability of randomly selecting a score GREATER than 51? A normal distribution has a mean of μμ= 54 and a standard deviation of σσ=6 What is the probability of selecting a sample of n = 4 scores with a mean more than M = 51? A normal distribution has a mean of μμ= 54 and a standard deviation of σσ=6 What is the probability of selecting a sample of n = 36 scores with a mean more than M = 51?
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In a survey, each participant indicated how much they agreed with the statement, "Hard work is the way to get what you want in life." Which of the following would you expect to have zero correlation, or close to zero correlation, with the response to the above statement? A) Agreement with the statement, "Happiness mostly depends on luck." B) Conscientiousness score on a personality inventory test. C) High school GPA. D) A random number between 0 and 10, chosen by the participant.
4
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1. In the past, a chemical company produced 880 pounds of a certain type of plastic per day. Now, using a newly developed and less expensive process, the mean daily yield of plastic for the first 50 days of production is 871 pounds; the standard deviation is 21 pounds. Do the data provide sufficient evidence to indicate that the mean daily yield for the new process is less than that of the old procedure? (Use α=0.05) (d) Conclusion of the test above is Reject the null hypothesis and the mean daily yield for the new process is less than that of the old procedure. Reject the null hypothesis and the mean daily yield for the new process is not less than that of the old procedure. Do not reject the null hypothesis and the mean daily yield for the new process is less than that of the old procedure. Do not reject the null hypothesis and the mean daily yield for the new process is not less than that of the old procedure.
Vehicle speeds at a certain highway location are believed to have approximately a normal distribution with mean µ = 50 mph and standard deviation σ = 5 mph. The speeds for a randomly selected sample of n = 25 vehicles will be recorded. (a) Give numerical values for the mean and standard deviation of the sampling distribution of possible sample means for randomly selected samples of n = 25 from the population of vehicle speeds. Mean = s.d.(x) = (b) Use the Empirical Rule to find values that fill in the blanks in the following sentence. For a random sample of n = 25 vehicles, there is about a 95% chance that the mean vehicle speed in the sample will be between and mph. (c) Sample speeds for a random sample of 25 vehicles are measured at this location, and the sample mean is 59 mph. Given the answer to part (b), explain whether this result is consistent with the belief that the mean speed at this location is μ = 50 mph. A sample mean of 59 mph (when n = 25) --Select--- be consistent with…

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