Introduction to Probability and Statistics for Engineers and Scientists

586 Chapter 14*:Life Testing

and so

∂

∂θ

logL(x 1 ,...,xr,i 1 ,...,ir)=−

r

θ

+

∑r

i= 1

xi

θ^2

+

(n−r)xr

θ^2

Equating to 0 and solving yields thatθˆ, the maximum likelihood estimate, is given by

θˆ=

∑r

i= 1

xi+(n−r)xr

r

Hence, if we letX(i)denote the time at which theith failure occurs (X(i)is called theith
order statistic), then the maximum likelihood estimator ofθis

θˆ=

∑r

i= 1

X(i)+(n−r)X(r)

r

(14.3.3)

=

τ

r

whereτ, defined to equal the numerator in Equation 14.3.3, is called thetotal-time-on-test
statistic. We call it this since theith item to fail functions for a timeX(i)(and then fails),
i=1,...,r, whereas the othern−ritems function throughout the test (which lasts for
a timeX(r)). Hence the sum of the times that all the items are on test is equal toτ.
To obtain a confidence interval forθ, we will determine the distribution ofτ, the total
time on test. Recalling thatX(i)is the time of theith failure,i=1,...,r, we will start
by rewriting the expression forτ. To write an expression forτ, rather than summing the
total time on test of each of the items, let us ask how much additional time on test was
generated between each successive failure. That is, let us denote byYi,i=1,...,r, the
additional time on test generated between the (i−1)st andith failure. Now up to the first
X(1)time units (as allnitems are functioning throughout this interval), the total time on
test is

Y 1 =nX(1)

Between the first and second failures, there are a total ofn−1 functioning items, and so

Y 2 =(n−1)(X(2)−X(1))