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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

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

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)be a random sample from a pdf\({\rm{f(x)}}\)that is symmetric about\({\rm{\mu }}\), so that\({\rm{\backslash widetildeX}}\)is an unbiased estimator of\({\rm{\mu }}\). If\({\rm{n}}\)is large, it can be shown that\({\rm{V (\tilde X)\gg 1/}}\left( {{\rm{4n(f(\mu )}}{{\rm{)}}^{\rm{2}}}} \right)\).

a. Compare\({\rm{V(\backslash widetildeX)}}\)to\({\rm{V(\bar X)}}\)when the underlying distribution is normal.

b. When the underlying pdf is Cauchy (see Example 6.7),\({\rm{V(\bar X) = \yen}}\), so\({\rm{\bar X}}\)is a terrible estimator. What is\({\rm{V(\tilde X)}}\)in this case when\({\rm{n}}\)is large?

Let\({\rm{X}}\)have a Weibull distribution with parameters\({\rm{\alpha }}\)and\({\rm{\beta }}\), so

\(\begin{array}{l}{\rm{E(X) = \beta \times \Gamma (1 + 1/\alpha )V(X)}}\\{\rm{ = }}{{\rm{\beta }}^{\rm{2}}}\left\{ {{\rm{\Gamma (1 + 2/\alpha ) - (\Gamma (1 + 1/\alpha )}}{{\rm{)}}^{\rm{2}}}} \right\}\end{array}\)

a. Based on a random sample\({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\), write equations for the method of moments estimators of\({\rm{\beta }}\)and\({\rm{\alpha }}\). Show that, once the estimate of\({\rm{\alpha }}\)has been obtained, the estimate of\({\rm{\beta }}\)can be found from a table of the gamma function and that the estimate of\({\rm{\alpha }}\)is the solution to a complicated equation involving the gamma function.

b. If\({\rm{n = 20,\bar x = 28}}{\rm{.0}}\), and\({\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 16,500}}\), compute the estimates. (Hint:\(\left. {{{{\rm{(\Gamma (1}}{\rm{.2))}}}^{\rm{2}}}{\rm{/\Gamma (1}}{\rm{.4) = }}{\rm{.95}}{\rm{.}}} \right)\)

\({{\rm{X}}_{\rm{1}}}{\rm{,}}.....{\rm{,}}{{\rm{X}}_{\rm{n}}}\)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 \(\widehat {\rm{\mu }}{\rm{ = }}\overline {\rm{X}} \).

An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code. Having obtained a random sample of\({\rm{n}}\)students, she realizes that asking each, "Have you violated the honor code?" will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of\({\rm{100}}\)cards, of which\({\rm{50}}\)are of type I and\({\rm{50}}\)are of type II.

Type I: Have you violated the honor code (yes or no)?

Type II: Is the last digit of your telephone number a\({\rm{0 , 1 , or 2}}\)(yes or no)?

Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let\({\rm{p}}\)denote the proportion of honor-code violators (i.e., the probability of a randomly selected student being a violator), and let\({\rm{\lambda = P}}\)(yes response). Then\({\rm{\lambda }}\)and\({\rm{p}}\)are related by\({\rm{\lambda = }}{\rm{.5p + (}}{\rm{.5)(}}{\rm{.3)}}\).

a. Let\({\rm{Y}}\)denote the number of yes responses, so\({\rm{Y\sim}}\)Bin\({\rm{(n,\lambda )}}\). Thus Y / n is an unbiased estimator of\({\rm{\lambda }}\). Derive an estimator for\({\rm{p}}\)based on\({\rm{Y}}\). If\({\rm{n = 80}}\)and\({\rm{y = 20}}\), what is your estimate? (Hint: Solve\({\rm{\lambda = }}{\rm{.5p + }}{\rm{.15}}\)for\({\rm{p}}\)and then substitute\({\rm{Y/n}}\)for\({\rm{\lambda }}\).)

b. Use the fact that\({\rm{E(Y/n) = \lambda }}\)to show that your estimator\({\rm{\hat p}}\)is unbiased.

c. If there were\({\rm{70}}\)type I and\({\rm{30}}\)type II cards, what would be your estimator for\({\rm{p}}\)?
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