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Chapter 6: Q35SE (page 274)

Let \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) be a random sample from a pdf that is symmetric about \({\rm{\mu }}\). An estimator for \({\rm{\mu }}\) that has been found to perform well for a variety of underlying distributions is the Hodges鈥揕ehmann estimator. To define it, first compute for each \({\rm{i}} \le {\rm{j}}\) and each \({\rm{j = 1,2, \ldots ,n}}\) the pairwise average \({{\rm{\bar X}}_{{\rm{i,j}}}}{\rm{ = }}\left( {{{\rm{X}}_{\rm{i}}}{\rm{ + }}{{\rm{X}}_{\rm{j}}}} \right){\rm{/2}}\). Then the estimator is \({\rm{\hat \mu = }}\) the median of the \({{\rm{\bar X}}_{{\rm{i,j}}}}{\rm{'s}}\). Compute the value of this estimate using the data of Exercise \({\rm{44}}\) of Chapter \({\rm{1}}\). (Hint: Construct a square table with the \({{\rm{x}}_{\rm{i}}}{\rm{'s}}\) listed on the left margin and on top. Then compute averages on and above the diagonal.)

Short Answer

Expert verified

The value of \({\rm{\hat \mu = 181}}{\rm{.45}}\).

Step by step solution

01

Define median

When a set of data is sorted in ascending (more common) or descending order, the median is the middle number.

02

Explanation

The ideal technique to compute the median, as suggested in the hint, is to make a table with\({{\rm{x}}_{\rm{i}}}{\rm{'s}}\)on the left margin and on top, and compute the averages in the right triangle of the table. As stated in the exercise, no other values are required.

\({\rm{180}}{\rm{.5, 181}}{\rm{.7,180}}{\rm{.9, 181}}{\rm{.6, 182}}{\rm{.6,181}}{\rm{.6, 181}}{\rm{.3,182}}{\rm{.1, 182}}{\rm{.1, 180}}{\rm{.3, 181}}{\rm{.7,180}}{\rm{.5}}\)are the numbers given in exercise\({\rm{44}}\).

As a result, the table is,

The Sample Median is calculated with n observations ordered from least to greatest, including repeated values. As a result, sample median is,

\(\begin{array}{l}{\rm{\tilde x = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{ The only middle value if n is odd }}}\\{{\rm{ The average of the two middle values if n is even }}}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{l}}{{{\left( {\frac{{{\rm{n + 1}}}}{{\rm{2}}}} \right)}^{{\rm{th}}}}}&{{\rm{, if n = odd }}}\\{{\rm{ average of }}{{\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}^{{\rm{th}}}}{\rm{ and }}{{\left( {\frac{{\rm{n}}}{{\rm{2}}}{\rm{ + 1}}} \right)}^{{\rm{th}}}}{\rm{ ordered values }}}&{{\rm{, if n = even}}}\end{array}} \right.\end{array}\)

03

Evaluating the median

To find the median, first order the given averages. The data has been organised.

\(\begin{array}{l}{\rm{180}}{\rm{.3,180}}{\rm{.4,180}}{\rm{.4,180}}{\rm{.5,180}}{\rm{.5,180}}{\rm{.5,180}}{\rm{.6,180}}{\rm{.7,180}}{\rm{.7,180}}{\rm{.8,180}}{\rm{.9,}}\\{\rm{180}}{\rm{.9,180}}{\rm{.9,180}}{\rm{.95,180}}{\rm{.95,181,181,181}}{\rm{.05,181}}{\rm{.05,181}}{\rm{.05,181}}{\rm{.05,181}}{\rm{.1,}}\\{\rm{181}}{\rm{.1,181}}{\rm{.1,181}}{\rm{.1,181}}{\rm{.1,181}}{\rm{.2,181}}{\rm{.2,181}}{\rm{.25,181}}{\rm{.25,181}}{\rm{.3,181}}{\rm{.3,181}}{\rm{.3,}}\\{\rm{181}}{\rm{.3,181}}{\rm{.3,181}}{\rm{.3,181}}{\rm{.3,181}}{\rm{.45,181}}{\rm{.45,181}}{\rm{.45,181}}{\rm{.5,181}}{\rm{.5,181}}{\rm{.5,181}}{\rm{.5,}}\\{\rm{181}}{\rm{.55,181}}{\rm{.55,181}}{\rm{.6,181}}{\rm{.6,181}}{\rm{.6,181}}{\rm{.65,181}}{\rm{.65,181}}{\rm{.65,181}}{\rm{.65,181}}{\rm{.7,181}}{\rm{.7,}}\\{\rm{181}}{\rm{.7,181}}{\rm{.7,181}}{\rm{.7,181}}{\rm{.75,181}}{\rm{.85,181}}{\rm{.85,181}}{\rm{.85,181}}{\rm{.85,181}}{\rm{.9}}\end{array}\)

\(\begin{array}{l}{\rm{181}}{\rm{.9,181}}{\rm{.9,181}}{\rm{.9,181}}{\rm{.95,182}}{\rm{.1,182}}{\rm{.1,182}}{\rm{.1,182}}{\rm{.1,182}}{\rm{.1,182}}{\rm{.15,}}\\{\rm{182}}{\rm{.15,182}}{\rm{.35,182}}{\rm{.35,182}}{\rm{.6}}\end{array}\)

Since this sample has an even number of observations\({\rm{n = 78}}\), the values of interest are,

\(\begin{array}{l}{\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)^{{\rm{th}}}}{\rm{ = }}{\left( {\frac{{{\rm{78}}}}{{\rm{2}}}} \right)^{{\rm{th}}}}{\rm{ = 3}}{{\rm{9}}^{{\rm{th}}}}\\{\left( {\frac{{\rm{n}}}{{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{th}}}}{\rm{ = }}{\left( {\frac{{{\rm{78}}}}{{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{th}}}}{\rm{ = 4}}{{\rm{0}}^{{\rm{th}}}}\end{array}\)

In the ordered sample data, the\({\rm{3}}{{\rm{9}}^{{\rm{th}}}}\)and\({\rm{4}}{{\rm{0}}^{{\rm{th}}}}\)values are highlighted in red.

As a result, the sample median is the average of the two numbers previously indicated.

\(\begin{array}{c}{\rm{\tilde x = }}\frac{{{\rm{181}}{\rm{.45 + 181}}{\rm{.45}}}}{{\rm{2}}}\\{\rm{ = 181}}{\rm{.45}}\end{array}\)

As a result, the Hodges-Lehmann approximation is,

\({\rm{\hat \mu = 181}}{\rm{.45}}\).

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Most popular questions from this chapter

The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2.

\(\begin{array}{*{20}{r}}{{\rm{5}}{\rm{.9}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{7}}{\rm{.3}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{7}}{\rm{.0}}}\\{{\rm{7}}{\rm{.6}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{6}}{\rm{.5}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{7}}{\rm{.9}}}&{{\rm{9}}{\rm{.0}}}\\{{\rm{3}}{\rm{.2}}}&{{\rm{8}}{\rm{.7}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{9}}{\rm{.7}}}\\{{\rm{7}}{\rm{.3}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{11}}{\rm{.6}}}&{{\rm{11}}{\rm{.3}}}&{{\rm{11}}{\rm{.8}}}&{{\rm{10}}{\rm{.7}}}&{}\end{array}\)

Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used\({\rm{(Hint:\Sigma }}{{\rm{x}}_{\rm{i}}}{\rm{ = 219}}{\rm{.8}}{\rm{.)}}\)

b. Calculate a point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50 %, and state which estimator you used.

c. Calculate and interpret a point estimate of the population standard deviation\({\rm{\sigma }}\). Which estimator did you use?\({\rm{(Hint:}}\left. {{\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 1860}}{\rm{.94}}{\rm{.}}} \right)\)

d. Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds\({\rm{10MPa}}\). (Hint: Think of an observation as a "success" if it exceeds 10.)

e. Calculate a point estimate of the population coefficient of variation\({\rm{\sigma /\mu }}\), and state which estimator you used.

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)represent a random sample from a Rayleigh distribution with pdf

\({\rm{f(x,\theta ) = }}\frac{{\rm{x}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}\quad {\rm{x > 0}}\)a. It can be shown that\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = 2\theta }}\). Use this fact to construct an unbiased estimator of\({\rm{\theta }}\)based on\({\rm{\Sigma X}}_{\rm{i}}^{\rm{2}}\)(and use rules of expected value to show that it is unbiased).

b. Estimate\({\rm{\theta }}\)from the following\({\rm{n = 10}}\)observations on vibratory stress of a turbine blade under specified conditions:

\(\begin{array}{*{20}{l}}{{\rm{16}}{\rm{.88}}}&{{\rm{10}}{\rm{.23}}}&{{\rm{4}}{\rm{.59}}}&{{\rm{6}}{\rm{.66}}}&{{\rm{13}}{\rm{.68}}}\\{{\rm{14}}{\rm{.23}}}&{{\rm{19}}{\rm{.87}}}&{{\rm{9}}{\rm{.40}}}&{{\rm{6}}{\rm{.51}}}&{{\rm{10}}{\rm{.95}}}\end{array}\)

A diagnostic test for a certain disease is applied to\({\rm{n}}\)individuals known to not have the disease. Let\({\rm{X = }}\)the number among the\({\rm{n}}\)test results that are positive (indicating presence of the disease, so\({\rm{X}}\)is the number of false positives) and\({\rm{p = }}\)the probability that a disease-free individual's test result is positive (i.e.,\({\rm{p}}\)is the true proportion of test results from disease-free individuals that are positive). Assume that only\({\rm{X}}\)is available rather than the actual sequence of test results.

a. Derive the maximum likelihood estimator of\({\rm{p}}\). If\({\rm{n = 20}}\)and\({\rm{x = 3}}\), what is the estimate?

b. Is the estimator of part (a) unbiased?

c. If\({\rm{n = 20}}\)and\({\rm{x = 3}}\), what is the mle of the probability\({{\rm{(1 - p)}}^{\rm{5}}}\)that none of the next five tests done on disease-free individuals are positive?

Let\({\rm{X}}\)denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of\({\rm{X}}\)is

\({\rm{f(x;\theta ) = }}\left\{ {\begin{array}{*{20}{c}}{{\rm{(\theta + 1)}}{{\rm{x}}^{\rm{\theta }}}}&{{\rm{0拢 x拢 1}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

where\({\rm{ - 1 < \theta }}\). A random sample of ten students yields data\({{\rm{x}}_{\rm{1}}}{\rm{ = }}{\rm{.92,}}{{\rm{x}}_{\rm{2}}}{\rm{ = }}{\rm{.79,}}{{\rm{x}}_{\rm{3}}}{\rm{ = }}{\rm{.90,}}{{\rm{x}}_{\rm{4}}}{\rm{ = }}{\rm{.65,}}{{\rm{x}}_{\rm{5}}}{\rm{ = }}{\rm{.86}}\),\({{\rm{x}}_{\rm{6}}}{\rm{ = }}{\rm{.47,}}{{\rm{x}}_{\rm{7}}}{\rm{ = }}{\rm{.73,}}{{\rm{x}}_{\rm{8}}}{\rm{ = }}{\rm{.97,}}{{\rm{x}}_{\rm{9}}}{\rm{ = }}{\rm{.94,}}{{\rm{x}}_{{\rm{10}}}}{\rm{ = }}{\rm{.77}}\).

a. Use the method of moments to obtain an estimator of\({\rm{\theta }}\), and then compute the estimate for this data.

b. Obtain the maximum likelihood estimator of\({\rm{\theta }}\), and then compute the estimate for the given data.

The shear strength of each of ten test spot welds is determined, yielding the following data (psi):

\(\begin{array}{*{20}{l}}{{\rm{392}}}&{{\rm{376}}}&{{\rm{401}}}&{{\rm{367}}}&{{\rm{389}}}&{{\rm{362}}}&{{\rm{409}}}&{{\rm{415}}}&{{\rm{358}}}&{{\rm{375}}}\end{array}\)

a. Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood.

b. Again assuming a normal distribution, estimate the strength value below which\({\rm{95\% }}\)of all welds will have their strengths. (Hint: What is the\({\rm{95 th}}\)percentile in terms of\({\rm{\mu }}\)and\({\rm{\sigma }}\)? Now use the invariance principle.)

c. Suppose we decide to examine another test spot weld. Let\({\rm{X = }}\)shear strength of the weld. Use the given data to obtain the mle of\({\rm{P(X拢400)}}{\rm{.(Hint:P(X拢400) = \Phi ((400 - \mu )/\sigma )}}{\rm{.)}}\)
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