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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

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.

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:

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

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 ?)

Consider randomly selecting \({\rm{n}}\) segments of pipe and determining the corrosion loss (mm) in the wall thickness for each one. Denote these corrosion losses by \({{\rm{Y}}_{\rm{1}}}{\rm{,}}.....{\rm{,}}{{\rm{Y}}_{\rm{n}}}\). The article 鈥淎 Probabilistic Model for a Gas Explosion Due to Leakages in the Grey Cast Iron Gas Mains鈥� (Reliability Engr. and System Safety (\({\rm{(2013:270 - 279)}}\)) proposes a linear corrosion model: \({{\rm{Y}}_{\rm{i}}}{\rm{ = }}{{\rm{t}}_{\rm{i}}}{\rm{R}}\), where \({{\rm{t}}_{\rm{i}}}\) is the age of the pipe and \({\rm{R}}\), the corrosion rate, is exponentially distributed with parameter \({\rm{\lambda }}\). Obtain the maximum likelihood estimator of the exponential parameter (the resulting mle appears in the cited article). (Hint: If \({\rm{c > 0}}\) and \({\rm{X}}\) has an exponential distribution, so does \({\rm{cX}}\).)

We defined a negative binomial\({\rm{rv}}\)as the number of failures that occur before the\({\rm{rth}}\)success in a sequence of independent and identical success/failure trials. The probability mass function (\({\rm{pmf}}\)) of\({\rm{X}}\)is\({\rm{nb(x,r,p) = }}\)\(\left( {\begin{array}{*{20}{c}}{{\rm{x + r - 1}}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{r}}}{{\rm{(1 - p)}}^{\rm{x}}}\quad {\rm{x = 0,1,2, \ldots }}\)

a. Suppose that. Show that\({\rm{\hat p = (r - 1)/(X + r - 1)}}\)is an unbiased estimator for\({\rm{p}}\). (Hint: Write out\({\rm{E(\hat p)}}\)and cancel\({\rm{x + r - 1}}\)inside the sum.)

b. A reporter wishing to interview five individuals who support a certain candidate begins asking people whether\({\rm{(S)}}\)or not\({\rm{(F)}}\)they support the candidate. If the sequence of responses is SFFSFFFSSS, estimate\({\rm{p = }}\)the true proportion who support the candidate.
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