a. Explanation Given info: A random sample of n observations, X 1 , X 2 , ... , X n has Rayleigh distribution. The pdf is: f ( x ; θ ) = x θ e − x 2 2 θ , x > 0 . Calculation: The likelihood function for the n observations is: L ( θ ) = f ( x 1 , x 2 , ... , x n | θ ) = ∏ i = 1 p x i θ e − x i 2 2 θ . The log likelihood function, l ( θ ) , is obtained by taking logarithm of the likelihood function: l ( θ ) = ln ( L ( θ ) ) = ln ( f ( x 1 , x 2 , ... , x n | θ ) ) = ln ( ∏ i = 1 p ( x i θ e − x i 2 2 θ ) ) = ln [ 1 θ n ∏ i = 1 p x i e − ∑ i = 1 n x i 2 2 θ ] = − n ln ( θ ) + ln ( x 1 x 2 ... x n ) + ln ( e − ∑ i = 1 n x i 2 2 θ ) = ln ( x 1 x 2 ... x n ) − n ln ( θ ) − ∑ i = 1 n x i 2 2 θ . The mle of θ can be obtained by differentiating l ( θ ) and equating it with zero to solve for θ . Now, d d x l ( θ ) = l ' ( θ ) = 0. Thus, 0 = d d x [ ln ( x 1 x 2 .. b. To determine Find the mle of the median of the vibratory stress distribution.

Probability and Statistics for Engineering and the Sciences

9th Edition

ISBN: 9781305251809

Author: Jay L. Devore

Publisher: Cengage Learning

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Question

Definition Definition Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.

Chapter 6.2, Problem 28E

To determine

Find the maximum likelihood estimator of θ.

Calculate the estimate for the vibratory stress data given.

Decide whether this estimator is the same as the unbiased estimator in Exercise 15.

To determine

Find the mle of the median of the vibratory stress distribution.

Expert Solution & Answer

Chapter 6 Solutions

Probability and Statistics for Engineering and the Sciences

Ch. 6.1 - The accompanying data on flexural strength (MPa)...Ch. 6.1 - The National Health and Nutrition Examination...Ch. 6.1 - Consider the following sample of observations on...Ch. 6.1 - The article from which the data in Exercise 1 was...Ch. 6.1 - As an example of a situation in which several...Ch. 6.1 - Urinary angiotensinogen (AGT) level is one...Ch. 6.1 - a. A random sample of 10 houses in a particular...Ch. 6.1 - In a random sample of 80 components of a certain...Ch. 6.1 - Each of 150 newly manufactured items is examined...Ch. 6.1 - Using a long rod that has length , you are going...

Ch. 6.1 - Of n1 randomly selected male smokers, X1 smoked...Ch. 6.1 - Suppose a certain type of fertilizer has an...Ch. 6.1 - Consider a random sample X1,..., Xn from the pdf...Ch. 6.1 - A sample of n captured Pandemonium jet fighters...Ch. 6.1 - Let X1, X2,..., Xn represent a random sample from...Ch. 6.1 - Suppose the true average growth of one type of...Ch. 6.1 - In Chapter 3, we defined a negative binomial rv as...Ch. 6.1 - Let X1, X2,..., Xn be a random sample from a pdf...Ch. 6.1 - An investigator wishes to estimate the proportion...Ch. 6.2 - A diagnostic test for a certain disease is applied...Ch. 6.2 - Let X have a Weibull distribution with parameters ...Ch. 6.2 - Let X denote the proportion of allotted time that...Ch. 6.2 - Let X represent the error in making a measurement...Ch. 6.2 - A vehicle with a particular defect in its emission...Ch. 6.2 - The shear strength of each of ten test spot welds...Ch. 6.2 - Consider randomly selecting n segments of pipe and...Ch. 6.2 - Let X1,..., Xn be a random sample from a gamma...Ch. 6.2 - Prob. 28E Ch. 6.2 - Consider a random sample X1, X2,, Xn from the...Ch. 6.2 - At time t = 0, 20 identical components are tested....Ch. 6 - An estimator is said to be consistent if for any ...Ch. 6 - a. Let X1,.., Xn be a random sample from a uniform...Ch. 6 - At time t = 0, there is one individual alive in a...Ch. 6 - The mean squared error of an estimator is MSE ()...Ch. 6 - Prob. 35SE Ch. 6 - When the population distribution is normal, the...Ch. 6 - When the sample standard deviation S is based on a...Ch. 6 - Each of n specimens is to be weighed twice on the...

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Find the maximum likelihood estimator of θ . Calculate the estimate for the vibratory stress data given. Decide whether this estimator is the same as the unbiased estimator in Exercise 15.

Solution Summary: The author calculates the maximum likelihood estimator of theta . The estimate for the vibratory stress data given is 74.5053.

Find the maximum likelihood estimator of θ.

Calculate the estimate for the vibratory stress data given.

Decide whether this estimator is the same as the unbiased estimator in Exercise 15.

Find the mle of the median of the vibratory stress distribution.

Chapter 6 Solutions

Find the maximum likelihood estimator of θ . Calculate the estimate for the vibratory stress data given. Decide whether this estimator is the same as the unbiased estimator in Exercise 15.

Solution Summary: The author calculates the maximum likelihood estimator of theta . The estimate for the vibratory stress data given is 74.5053.

Find the maximum likelihood estimator of θ. Calculate the estimate for the vibratory stress data given. Decide whether this estimator is the same as the unbiased estimator in Exercise 15.

Find the mle of the median of the vibratory stress distribution.

Chapter 6 Solutions

Find the maximum likelihood estimator of θ.

Calculate the estimate for the vibratory stress data given.

Decide whether this estimator is the same as the unbiased estimator in Exercise 15.