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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–Lehmann 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

In a random sample of 80 components of a certain type, 12 are found to be defective.

a. Give a point estimate of the proportion of all such components that are not defective.

b. A system is to be constructed by randomly selecting two of these components and connecting them in series, as shown here.

The series connection implies that the system will function if and only if neither component is defective (i.e., both components work properly). Estimate the proportion of all such systems that work properly. (Hint: If p denotes the probability that a component works properly, how can P (system works) be expressed in terms of p ?)

a. A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are 103, 156, 118, 89, 125, 147, 122, 109, 138, 99. Let m denote the average gas usage during January by all houses in this area. Compute a point estimate of m. b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let t denote the total amount of gas used by all of these houses during January. Estimate t using the data of part (a). What estimator did you use in computing your estimate? c. Use the data in part (a) to estimate p, the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a). What estimator did you use?

An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code. Having obtained a random sample of\({\rm{n}}\)students, she realizes that asking each, "Have you violated the honor code?" will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of\({\rm{100}}\)cards, of which\({\rm{50}}\)are of type I and\({\rm{50}}\)are of type II.

Type I: Have you violated the honor code (yes or no)?

Type II: Is the last digit of your telephone number a\({\rm{0 , 1 , or 2}}\)(yes or no)?

Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let\({\rm{p}}\)denote the proportion of honor-code violators (i.e., the probability of a randomly selected student being a violator), and let\({\rm{\lambda = P}}\)(yes response). Then\({\rm{\lambda }}\)and\({\rm{p}}\)are related by\({\rm{\lambda = }}{\rm{.5p + (}}{\rm{.5)(}}{\rm{.3)}}\).

a. Let\({\rm{Y}}\)denote the number of yes responses, so\({\rm{Y\sim}}\)Bin\({\rm{(n,\lambda )}}\). Thus Y / n is an unbiased estimator of\({\rm{\lambda }}\). Derive an estimator for\({\rm{p}}\)based on\({\rm{Y}}\). If\({\rm{n = 80}}\)and\({\rm{y = 20}}\), what is your estimate? (Hint: Solve\({\rm{\lambda = }}{\rm{.5p + }}{\rm{.15}}\)for\({\rm{p}}\)and then substitute\({\rm{Y/n}}\)for\({\rm{\lambda }}\).)

b. Use the fact that\({\rm{E(Y/n) = \lambda }}\)to show that your estimator\({\rm{\hat p}}\)is unbiased.

c. If there were\({\rm{70}}\)type I and\({\rm{30}}\)type II cards, what would be your estimator for\({\rm{p}}\)?

A sample of \({\rm{n}}\) captured Pandemonium jet fighters results in serial numbers\({{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{x}}_{\rm{3}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}\). The CIA knows that the aircraft were numbered consecutively at the factory starting with \({\rm{\alpha }}\)and ending with\({\rm{\beta }}\), so that the total number of planes manufactured is \({\rm{\beta - \alpha + 1}}\) (e.g., if \({\rm{\alpha = 17}}\) and\({\rm{\beta = 29}}\), then \({\rm{29 - 17 + 1 = 13}}\)planes having serial numbers \({\rm{17,18,19, \ldots ,28,29}}\)were manufactured). However, the CIA does not know the values of \({\rm{\alpha }}\) or\({\rm{\beta }}\). A CIA statistician suggests using the estimator \({\rm{max}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ - min}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ + 1}}\)to estimate the total number of planes manufactured.

a. If\({\rm{n = 5, x\_}}\left\{ {\rm{1}} \right\}{\rm{ = 237, x\_}}\left\{ {\rm{2}} \right\}{\rm{ = 375, x\_}}\left\{ {\rm{3}} \right\}{\rm{ = 202, x\_}}\left\{ {\rm{4}} \right\}{\rm{ = 525,}}\)and\({{\rm{x}}_{\rm{5}}}{\rm{ = 418}}\), what is the corresponding estimate?

b. Under what conditions on the sample will the value of the estimate be exactly equal to the true total number of planes? Will the estimate ever be larger than the true total? Do you think the estimator is unbiased for estimating\({\rm{\beta - \alpha + 1}}\)? Explain in one or two sentences.

Suppose a certain type of fertilizer has an expected yield per acre of \({{\rm{\mu }}_{\rm{2}}}\)with variance \({{\rm{\sigma }}^{\rm{2}}}\)whereas the expected yield for a second type of fertilizer is with the same variance \({{\rm{\sigma }}^{\rm{2}}}\).Let \({\rm{S}}_{\rm{1}}^{\rm{2}}\) and \({\rm{S}}_{\rm{2}}^{\rm{2}}\)denote the sample variances of yields based on sample sizes \({{\rm{n}}_{\rm{1}}}\)and \({{\rm{n}}_{\rm{2}}}\),respectively, of the two fertilizers. Show that the pooled (combined) estimator

\({{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = }}\frac{{\left( {{{\rm{n}}_{\rm{1}}}{\rm{ - 1}}} \right){\rm{S}}_{\rm{1}}^{\rm{2}}{\rm{ + }}\left( {{{\rm{n}}_{\rm{2}}}{\rm{ - 1}}} \right){\rm{S}}_{\rm{2}}^{\rm{2}}}}{{{{\rm{n}}_{\rm{1}}}{\rm{ + }}{{\rm{n}}_{\rm{2}}}{\rm{ - 2}}}}\)

is an unbiased estimator of \({{\rm{\sigma }}^{\rm{2}}}\)
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