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Chapter 6: Q37SE (page 275)

When the sample standard deviation S is based on a random sample from a normal population distribution, it can be shown that \({\rm{E(S) = }}\sqrt {{\rm{2/(n - 1)}}} {\rm{\Gamma (n/2)\sigma /\Gamma ((n - 1)/2)}}\)

Use this to obtain an unbiased estimator for \({\rm{\sigma }}\) of the form \({\rm{cS}}\). What is \({\rm{c}}\) when \({\rm{n = 20}}\)?

Short Answer

Expert verified

The value is \({\rm{c = 1}}{\rm{.0132}}\).

Step by step solution

01

Define standard deviation

The standard deviation is a metric that indicates how much variation (such as spread, dispersion, and spread) there is from the mean. A "typical" divergence from the mean is indicated by the standard deviation. Since it returns to the data set's original units of measure, it's a common measure of variability.

02

Explanation

It is well knowledge that,

\({\rm{E(S) = }}\frac{{\sqrt {\frac{{\rm{2}}}{{{\rm{n - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}}{{{\rm{\Gamma }}\left( {\frac{{{\rm{n - 1}}}}{{\rm{2}}}} \right)}}{\rm{\sigma }}\)

As a result, the following must be true in order to generate an unbiased estimator\({\rm{\theta }}\):

\({\rm{E(cS) = \sigma }}\)

Property is the reason.

\({\rm{E(cS) = cE(S)}}\)

and because the\({\rm{E(S)}}\)is well-known, it's easy to see that c has to be,

\({\rm{c = }}\frac{{{\rm{\Gamma }}\left( {\frac{{{\rm{n - 1}}}}{{\rm{2}}}} \right)}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{n - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}}\)

\({\rm{cS}}\)must be an unbiased estimator in order for it. Take a look.

\(\begin{array}{c}{\rm{E(cS) = cE(S)}}\\{\rm{ = }}\frac{{{\rm{\Gamma }}\left( {\frac{{{\rm{n - 1}}}}{{\rm{2}}}} \right)}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{n - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}}{\rm{ \times }}\frac{{\sqrt {\frac{{\rm{2}}}{{{\rm{n - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}}{{{\rm{\Gamma }}\left( {\frac{{{\rm{n - 1}}}}{{\rm{2}}}} \right)}}{\rm{\sigma }}\\{\rm{ = \sigma }}\end{array}\)

03

Evaluating the values

The c becomes when\({\rm{n = 20}}\)is substituted in the formula for c.

\(\begin{array}{c}{\rm{c = }}\frac{{{\rm{\Gamma }}\left( {\frac{{{\rm{n - 1}}}}{{\rm{2}}}} \right)}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{n - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}}\\{\rm{ = }}\frac{{{\rm{\Gamma }}\left( {\frac{{{\rm{20 - 1}}}}{{\rm{2}}}} \right)}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{20 - 1}}}}} {\rm{ \times \Gamma }}\left( {\frac{{{\rm{20}}}}{{\rm{2}}}} \right)}}\\{\rm{ = }}\frac{{{\rm{\Gamma (9}}{\rm{.5)}}}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{19}}}}} {\rm{ \times \Gamma (10)}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{8}}{\rm{.5 \times 7}}{\rm{.5 \times \ldots \times 1}}{\rm{.5 \times 0}}{\rm{.5 \times \Gamma (0}}{\rm{.5)}}}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{19}}}}} {\rm{ \times (10 - 1)!}}}}\\{\rm{ = }}\frac{{{\rm{8}}{\rm{.5 \times 7}}{\rm{.5 \times \ldots \times 1}}{\rm{.5 \times 0}}{\rm{.5 \times \pi }}}}{{\sqrt {\frac{{\rm{2}}}{{{\rm{19}}}}} {\rm{ \times 9!}}}}\\{\rm{ = 1}}{\rm{.0132}}\end{array}\)

  1. ; the gamma function's characteristics!

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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 ?)

Of \({{\rm{n}}_{\rm{1}}}\)randomly selected male smokers, \({{\rm{X}}_{\rm{1}}}\) smoked filter cigarettes, whereas of \({{\rm{n}}_{\rm{2}}}\) randomly selected female smokers, \({{\rm{X}}_{\rm{2}}}\) smoked filter cigarettes. Let \({{\rm{p}}_{\rm{1}}}\) and \({{\rm{p}}_{\rm{2}}}\) denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes.

a. Show that \({\rm{(}}{{\rm{X}}_{\rm{1}}}{\rm{/}}{{\rm{n}}_{\rm{1}}}{\rm{) - (}}{{\rm{X}}_{\rm{2}}}{\rm{/}}{{\rm{n}}_{\rm{2}}}{\rm{)}}\) is an unbiased estimator for \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\). (Hint: \({\rm{E(}}{{\rm{X}}_{\rm{i}}}{\rm{) = }}{{\rm{n}}_{\rm{i}}}{{\rm{p}}_{\rm{i}}}\) for \({\rm{i = 1,2}}\).)

b. What is the standard error of the estimator in part (a)?

c. How would you use the observed values \({{\rm{x}}_{\rm{1}}}\) and \({{\rm{x}}_{\rm{2}}}\) to estimate the standard error of your estimator?

d. If \({{\rm{n}}_{\rm{1}}}{\rm{ = }}{{\rm{n}}_{\rm{2}}}{\rm{ = 200, }}{{\rm{x}}_{\rm{1}}}{\rm{ = 127}}\), and \({{\rm{x}}_{\rm{2}}}{\rm{ = 176}}\), use the estimator of part (a) to obtain an estimate of \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\).

e. Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator.

The mean squared error of an estimator \({\rm{\hat \theta }}\) is \({\rm{MSE(\hat \theta ) = E(\hat \theta - \hat \theta }}{{\rm{)}}^{\rm{2}}}\). If \({\rm{\hat \theta }}\) is unbiased, then \({\rm{MSE(\hat \theta ) = V(\hat \theta )}}\), but in general \({\rm{MSE(\hat \theta ) = V(\hat \theta ) + (bias}}{{\rm{)}}^{\rm{2}}}\) . Consider the estimator \({{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = K}}{{\rm{S}}^{\rm{2}}}\), where \({{\rm{S}}^{\rm{2}}}{\rm{ = }}\) sample variance. What value of K minimizes the mean squared error of this estimator when the population distribution is normal? (Hint: It can be shown that \({\rm{E}}\left( {{{\left( {{{\rm{S}}^{\rm{2}}}} \right)}^{\rm{2}}}} \right){\rm{ = (n + 1)}}{{\rm{\sigma }}^{\rm{4}}}{\rm{/(n - 1)}}\) In general, it is difficult to find \({\rm{\hat \theta }}\) to minimize \({\rm{MSE(\hat \theta )}}\), which is why we look only at unbiased estimators and minimize \({\rm{V(\hat \theta )}}\).)

At time \({\rm{t = 0}}\), there is one individual alive in a certain population. A pure birth process then unfolds as follows. The time until the first birth is exponentially distributed with parameter \({\rm{\lambda }}\). After the first birth, there are two individuals alive. The time until the first gives birth again is exponential with parameter \({\rm{\lambda }}\), and similarly for the second individual. Therefore, the time until the next birth is the minimum of two exponential (\({\rm{\lambda }}\)) variables, which is exponential with parameter \({\rm{2\lambda }}\). Similarly, once the second birth has occurred, there are three individuals alive, so the time until the next birth is an exponential \({\rm{rv}}\) with parameter \({\rm{3\lambda }}\), and so on (the memoryless property of the exponential distribution is being used here). Suppose the process is observed until the sixth birth has occurred and the successive birth times are \({\rm{25}}{\rm{.2,41}}{\rm{.7,51}}{\rm{.2,55}}{\rm{.5,59}}{\rm{.5,61}}{\rm{.8}}\) (from which you should calculate the times between successive births). Derive the mle of l. (Hint: The likelihood is a product of exponential terms.)

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.
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