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Chapter 6: Q9E (page 262)

Each of 150 newly manufactured items is examined and the number of scratches per item is recorded (the items are supposed to be free of scratches), yielding the following data:

Assume that X has a Poisson distribution with parameter \({\bf{\mu }}.\)and that X represents the number of scratches on a randomly picked item.

a. Calculate the estimate for the data using an unbiased \({\bf{\mu }}.\)estimator. (Hint: for X Poisson, \({\rm{E(X) = \mu }}\) ,therefore \({\rm{E(\bar X) = ?)}}\)

c. What is your estimator's standard deviation (standard error)? Calculate the standard error estimate. (Hint: \({\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ = \mu }}\), \({\rm{X}}\))

Short Answer

Expert verified

(a) The number of scratches per item and the observed frequency is \({\rm{2}}{\rm{.11}}\)

(b) The estimated standard error is \({\rm{0}}{\rm{.119}}{\rm{.}}\)

Step by step solution

01

Concept introduction

The estimation of the accuracy of any forecasts is known as the standard error of the estimate. It's abbreviated as SEE. The total of squared deviations of prediction is depreciated by the regression line. The sum of squares mistake is another name for it.

02

Step 2: Estimating The unbiased estimator of \(\mu \)is \(\bar X\).

(a)

The following is valid for a random variable X with Poisson distribution and parameter\(mu > 0.\)

\(\begin{aligned}E(X) &= V(X)\\ &= \mu \end{aligned}\)

The unbiased estimator of\(\mu \)is\(\bar X\).Proof follows

\(\begin{aligned} E(\bar X) &= E \left( {\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{X}}_{\rm{i}}}} } \right)\\ &= \frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{E}} \left( {{{\rm{X}}_{\rm{i}}}} \right) &= {{\rm{n}}}{\rm{ \times n \times E}}\left( {{{\rm{X}}_{\rm{1}}}} \right)\end{aligned}\)

(1): all\({X_i}\)have the same distribution.

For\({\rm{n = 150}}\), the estimate for the given data is

\({\rm{\bar x = }}\frac{{\rm{1}}}{{\rm{n}}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + \ldots + }}{{\rm{x}}_{\rm{n}}}} \right)\)

\({x_i}\)Is the product of the number of scratches per item and the observed frequency, so

\(\begin{aligned} \bar x &= \frac{{\rm{1}}}{{{\rm{150}}}}{\rm{(0 \times 18 + 1 \times 37 + \ldots + 7 \times 1)}}\\ &= \frac{{{\rm{317}}}}{{{\rm{150}}}}\\ &= 2 {\rm{.11}}\end{aligned}\)

03

Calculating the estimator's variance

(b)

To begin, calculate the estimator's variance as follows:

\(\begin{aligned} V(\bar X) &= V \left( {\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{X}}_{\rm{i}}}} } \right)\\ &= \frac{{\rm{1}}}{{{{\rm{n}}^{\rm{2}}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{V}} \left( {{{\rm{X}}_{\rm{i}}}} \right)\\\ &= \frac{{\rm{1}}}{{{{\rm{n}}^{\rm{2}}}}}{\rm{ \times n \times V}}\left( {{{\rm{X}}_{\rm{1}}}} \right)\\ &= \frac{{\rm{\mu }}}{{\rm{n}}}\end{aligned}\)

2): all\({X_i}\)are independent and have the same distribution.

The estimator's standard deviation is

\({{\rm{\sigma }}_{{\rm{\bar X}}}}{\rm{ = }}\sqrt {\frac{{\rm{\mu }}}{{\rm{n}}}} \)

Estimated standard error is:\(\begin{array}{c}\sqrt {\frac{{{\rm{\hat \mu }}}}{{\rm{n}}}} {\rm{ = }}\frac{{\sqrt {{\rm{2}}{\rm{.11}}} }}{{\sqrt {{\rm{150}}} }}\\{\rm{ = 0}}{\rm{.119}}{\rm{.}}\end{array}\)

Hence, the required estimated standard error value is \({\rm{0}}{\rm{.119}}{\rm{.}}\)

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Most popular questions from this chapter

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 \({\rm{V(\hat \theta )}}\).)

Consider a random sample \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) from the pdf

\({\rm{f(x;\theta ) = }}{\rm{.5(1 + \theta x)}}\quad {\rm{ - 1拢 x拢 1}}\)

where \({\rm{ - 1拢 \theta 拢 1}}\) (this distribution arises in particle physics). Show that \({\rm{\hat \theta = 3\bar X}}\) is an unbiased estimator of\({\rm{\theta }}\). (Hint: First determine\({\rm{\mu = E(X) = E(\bar X)}}\).)

Suppose the true average growth\({\rm{\mu }}\)of one type of plant during a l-year period is identical to that of a second type, but the variance of growth for the first type is\({{\rm{\sigma }}^{\rm{2}}}\), whereas for the second type the variance is\({\rm{4}}{{\rm{\sigma }}^{\rm{2}}}{\rm{. Let }}{{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{m}}}\)be\({\rm{m}}\)independent growth observations on the first type (so\({\rm{E}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \sigma\hat 2}}\)$ ), and let\({{\rm{Y}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{Y}}_{\rm{n}}}\)be\({\rm{n}}\)independent growth observations on the second type\(\left( {{\rm{E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{Y}}_{\rm{j}}}} \right){\rm{ = 4}}{{\rm{\sigma }}^{\rm{2}}}} \right)\)

a. Show that the estimator\({\rm{\hat \mu = \delta \bar X + (1 - \delta )\bar Y}}\)is unbiased for\({\rm{\mu }}\)(for\({\rm{0 < \delta < 1}}\), the estimator is a weighted average of the two individual sample means).

b. For fixed\({\rm{m}}\)and\({\rm{n}}\), compute\({\rm{V(\hat \mu ),}}\)and then find the value of\({\rm{\delta }}\)that minimizes\({\rm{V(\hat \mu )}}\). (Hint: Differentiate\({\rm{V(\hat \mu )}}\)with respect to\({\rm{\delta }}{\rm{.)}}\)

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?

Consider a random sample \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}.....{\rm{,}}{{\rm{X}}_{\rm{n}}}\) from the shifted exponential pdf

\({\rm{f(x;\lambda ,\theta ) = }}\left\{ {\begin{array}{*{20}{c}}{{\rm{\lambda }}{{\rm{e}}^{{\rm{ - \lambda (x - \theta )}}}}}&{{\rm{x}} \ge {\rm{\theta }}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\). Taking \({\rm{\theta = 0}}\) gives the pdf of the exponential distribution considered previously (with positive density to the right of zero). An example of the shifted exponential distribution appeared in Example \({\rm{4}}{\rm{.5}}\), in which the variable of interest was time headway in traffic flow and \({\rm{\theta = }}{\rm{.5}}\) was the minimum possible time headway. a. Obtain the maximum likelihood estimators of \({\rm{\theta }}\) and \({\rm{\lambda }}\). b. If \({\rm{n = 10}}\) time headway observations are made, resulting in the values \({\rm{3}}{\rm{.11,}}{\rm{.64,2}}{\rm{.55,2}}{\rm{.20,5}}{\rm{.44,3}}{\rm{.42,10}}{\rm{.39,8}}{\rm{.93,17}}{\rm{.82}}\), and \({\rm{1}}{\rm{.30}}\), calculate the estimates of \({\rm{\theta }}\) and \({\rm{\lambda }}\).
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