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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

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 \({\rm{n = 20}}\)?

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.

\({{\rm{X}}_{\rm{1}}}{\rm{,}}.....{\rm{,}}{{\rm{X}}_{\rm{n}}}\)be a random sample from a gamma distribution with parameters \({\rm{\alpha }}\) and \({\rm{\beta }}\). a. Derive the equations whose solutions yield the maximum likelihood estimators of \({\rm{\alpha }}\) and \({\rm{\beta }}\). Do you think they can be solved explicitly? b. Show that the mle of \({\rm{\mu = \alpha \beta }}\) is \(\widehat {\rm{\mu }}{\rm{ = }}\overline {\rm{X}} \).

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.)

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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