Introduction to Probability and Statistics for Engineers and Scientists

594 Chapter 14*:Life Testing

SOLUTION This is a one-sided test of

H 0 :θ≥25 versus H 1 :θ< 25

The relevant probability for determining thep-value is the probability that there would
have been as many as 30 failures if the mean life were 25. That is,

p-value=P 25 {N(600)≥ 30 }

=P{χ 602 ≤1,200/25}

=.132 from Program 5.8.1a

Thus,H 0 would be accepted when the significance level is .10. ■

14.3.3 Simultaneous Testing — Stopping by a Fixed Time

Supposeagainthatwearetestingitemswhoselifedistributionsareindependentexponential
random variables with a common unknown meanθ. As in Section 14.3.1, thenitems are
simultaneously put on test, but now we suppose that the test is to stop either at some fixed
timeTor whenever allnitems have failed — whichever occurs first. The problem is to
use the observed data to estimateθ.
The observed data will be as follows:

Data: i 1 ,i 2 ,...,ir, x 1 ,x 2 ,...,xr

with the interpretation that the preceding results when theritems numberedi 1 ,...,irare
observed to fail at respective timesx 1 ,...,xr; and the othern−ritems have not failed by
timeT.
Since an item will not have failed by timeTif and only if its lifetime is greater thanT,
we see that the likelihood of the foregoing data is

f(i 1 ,...,ir,x 1 ,...,xr)=fXi 1 ,...,Xir(x 1 ,...,xr)P{Xj>T,j=i 1 ,...,ir}

=

1

θ

e−x^1 /θ···

1

θ

e−xr/θ(e−T/θ)n−r

=

1

θr

exp









−

∑r

i= 1

xi

θ

−

(n−r)T

θ









To obtain the maximum likelihood estimates, take logs to obtain

logf(i 1 ,...,ir,x 1 ,...,xr)=−rlogθ−

∑r

1

xi

θ

−

(n−r)T

θ