Question

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\({\rm{P(X£400)}}{\rm{.(Hint:P(X£400) = \Phi ((400 - \mu )/\sigma )}}{\rm{.)}}\)

Short answer

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\({\rm{P(X£400) = 0}}{\rm{.7967}}\).

Step-by-step solution

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.

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\({\rm{s}}\)!

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\({\rm{(a)}}\).

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 \({\rm{P(X£400)}}\)is

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

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