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Chapter 6: Q25E (page 273)

The shear strength of each of ten test spot welds is determined, yielding the following data (psi):

\(\begin{array}{*{20}{l}}{{\rm{392}}}&{{\rm{376}}}&{{\rm{401}}}&{{\rm{367}}}&{{\rm{389}}}&{{\rm{362}}}&{{\rm{409}}}&{{\rm{415}}}&{{\rm{358}}}&{{\rm{375}}}\end{array}\)

a. Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood.

b. Again assuming a normal distribution, estimate the strength value below which\({\rm{95\% }}\)of all welds will have their strengths. (Hint: What is the\({\rm{95 th}}\)percentile in terms of\({\rm{\mu }}\)and\({\rm{\sigma }}\)? Now use the invariance principle.)

c. Suppose we decide to examine another test spot weld. Let\({\rm{X = }}\)shear strength of the weld. Use the given data to obtain the mle of\({\rm{P(X拢400)}}{\rm{.(Hint:P(X拢400) = \Phi ((400 - \mu )/\sigma )}}{\rm{.)}}\)

Short Answer

Expert verified

a) The maximum likelihood estimates of mean value and maximum likelihood estimate of standard deviation is \({\rm{\hat \mu = 384}}{\rm{.4, \hat \sigma = 18}}{\rm{.86}}{\rm{.}}\)

b) The estimate would be a function the estimates or equality \({\rm{\hat \mu + 1}}{\rm{.645 \times \hat \sigma = 415}}{\rm{.42}}\)

c) The mle of the given data is\({\rm{P(X拢400) = 0}}{\rm{.7967}}\).

Step by step solution

01

Introduction

An estimator is a rule for computing an estimate of a given quantity based on observable data: the rule (estimator), the quantity of interest (estimate), and the output (estimate) are all distinct.

02

Explanation

a)

The maximum likelihood estimators for \({\rm{\mu }}\)and\({{\rm{\sigma }}^{\rm{2}}}\), when assuming normality, are

\({\rm{\hat \mu = \bar X,}}\)

and

\(\widehat {{{\rm{\sigma }}^{\rm{2}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - \bar X}}} \right)}^{\rm{2}}}} \)

Therefore, the maximum likelihood estimates of mean value

\({\rm{\hat \mu = }}\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{(392 + 376 + \ldots + 375) = 384}}{\rm{.4}}{\rm{.}}\)

the maximum likelihood estimate of variance is

\(\begin{array}{c}\widehat {{{\rm{\sigma }}^{\rm{2}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{10}}}}\left( {{{{\rm{(392 - 384}}{\rm{.4)}}}^{\rm{2}}}{\rm{ + (376 - 384}}{\rm{.4}}{{\rm{)}}^{\rm{2}}}{\rm{ + \ldots + (375 - 384}}{\rm{.4}}{{\rm{)}}^{\rm{2}}}} \right)\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{(57}}{\rm{.76 + 70}}{\rm{.56 + \ldots + 88}}{\rm{.36)}}\\{\rm{ = 355}}{\rm{.65}}\end{array}\)

and maximum likelihood estimate of standard deviation is

\({\rm{\hat \sigma = }}\sqrt {{\rm{355}}{\rm{.65}}} {\rm{ = 18}}{\rm{.86}}\)

Note that the mle of standard deviation differs from the sample standard deviation\({\rm{s}}\)!

03

Explanation

b)

The \({95^{{\rm{th }}}}\)percentile in terms of \(\mu \)and \(\sigma \)is

\({\rm{\mu + }}{{\rm{z}}_{{\rm{1 - 0}}{\rm{.05}}}}{\rm{\sigma }}\)

Where \({{\rm{z}}_{{\rm{1 - 0}}{\rm{.05}}}}\)is \({\rm{z - }}\)score which can be found in the appendix, the standard normal distribution and

\({{\rm{z}}_{{\rm{1 - 0}}{\rm{.05}}}}{\rm{ = 1}}{\rm{.645}}\)The Invariance Principle:

Let \({{\rm{\hat \theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,n}}\)be maximum likelihood estimates of parameters\({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,n}}\).

The mle of any function of parameters \({{\rm{\theta }}_{\rm{i}}}\)is the function of the mle's\({{\rm{\hat \theta }}_{\rm{i}}}\).

Since, \({\rm{\mu + }}{{\rm{z}}_{{\rm{1 - 0}}{\rm{.05}}}}{\rm{\sigma }}\) is a function of parameters, the estimate would be a function the estimates or equality

\(\begin{array}{l}{\rm{\hat \mu + 1}}{\rm{.645 \times \hat \sigma = 384}}{\rm{.4 + 1}}{\rm{.645 \times 18}}{\rm{.86}}\\{\rm{ = 415}}{\rm{.42}}\end{array}\)

Where \({\rm{\hat \mu }}\)and \({\rm{\hat \sigma }}\)are maximum likelihood estimates computed in\({\rm{(a)}}\).

04

Explanation

c)

The Invariance Principle:

Let \({{\rm{\hat \theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,n}}\)be maximum likelihood estimates of parameters \({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,n}}{\rm{.}}\)The mle of any function of parameters \({{\rm{\theta }}_{\rm{i}}}\)is the function of the mle's\({{\rm{\hat \theta }}_{\rm{i}}}\).

Therefore, by the invariance principle, the maximum likelihood estimates of function \({\rm{P(X拢400)}}\)is

\(\begin{array}{c}{\rm{P(X拢400) = P}}\left( {\frac{{{\rm{X - \hat \mu }}}}{{{\rm{\hat \sigma }}}}{\rm{拢}}\frac{{{\rm{400 - 384}}{\rm{.4}}}}{{{\rm{18}}{\rm{.86}}}}} \right)\\{\rm{ = P(Z拢0}}{\rm{.83)}}\\{\rm{ = \Phi (0}}{\rm{.83)}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0}}{\rm{.7967,}}\end{array}\)

(1): From the appendix's normal probability table a programmer can also be used to calculate the probability.

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Most popular questions from this chapter

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\({\rm{\delta }}{\rm{.)}}\)

At time \({\rm{t = 0, 20}}\) identical components are tested. The lifetime distribution of each is exponential with parameter \({\rm{\lambda }}\). The experimenter then leaves the test facility unmonitored. On his return \({\rm{24}}\) hours later, the experimenter immediately terminates the test after noticing that \({\rm{y = 15}}\) of the \({\rm{20}}\) components are still in operation (so \({\rm{5}}\) have failed). Derive the mle of \({\rm{\lambda }}\). (Hint: Let \({\rm{Y = }}\) the number that survive \({\rm{24}}\) hours. Then \({\rm{Y}} \sim {\rm{Bin(n,p)}}\). What is the mle of \({\rm{p}}\)? Now notice that \({\rm{p = P(}}{{\rm{X}}_{\rm{i}}} \ge {\rm{24)}}\), where \({{\rm{X}}_{\rm{i}}}\) is exponentially distributed. This relates \({\rm{\lambda }}\) to \({\rm{p}}\), so the former can be estimated once the latter has been.)

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: \({\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ = \mu }}\), \({\rm{X}}\))

A vehicle with a particular defect in its emission control system is taken to a succession of randomly selected mechanics until\({\rm{r = 3}}\)of them have correctly diagnosed the problem. Suppose that this requires diagnoses by\({\rm{20}}\)different mechanics (so there were\({\rm{17}}\)incorrect diagnoses). Let\({\rm{p = P}}\)(correct diagnosis), so\({\rm{p}}\)is the proportion of all mechanics who would correctly diagnose the problem. What is the mle of\({\rm{p}}\)? Is it the same as the mle if a random sample of\({\rm{20}}\)mechanics results in\({\rm{3}}\)correct diagnoses? Explain. How does the mle compare to the estimate resulting from the use of the unbiased estimator?

An estimator \({\rm{\hat \theta }}\) is said to be consistent if for any \( \in {\rm{ > 0}}\), \({\rm{P(|\hat \theta - \theta |}} \ge \in {\rm{)}} \to {\rm{0}}\) as \({\rm{n}} \to \infty \). That is, \({\rm{\hat \theta }}\) is consistent if, as the sample size gets larger, it is less and less likely that \({\rm{\hat \theta }}\) will be further than \( \in \) from the true value of \({\rm{\theta }}\). Show that \({\rm{\bar X}}\) is a consistent estimator of \({\rm{\mu }}\) when \({{\rm{\sigma }}^{\rm{2}}}{\rm{ < }}\infty \) , by using Chebyshev鈥檚 inequality from Exercise \({\rm{44}}\) of Chapter \({\rm{3}}\). (Hint: The inequality can be rewritten in the form \({\rm{P}}\left( {\left| {{\rm{Y - }}{{\rm{\mu }}_{\rm{Y}}}} \right| \ge \in } \right) \le {\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{/}} \in \). Now identify \({\rm{Y}}\) with \({\rm{\bar X}}\).)
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