Introduction to Probability and Statistics for Engineers and Scientists

600 Chapter 14*:Life Testing

SOLUTION The value of the test statisticX/Yis 42/34=1.2353. To compute the prob-
ability that anF-random variable with parameters 10, 15 is less than this value, we run
Program 5.8.3a to obtain that

P{F10,15<1.2353}=.6554

Because thep-value is equal to 2(1−.6554)=.6892, we cannot rejectH 0. ■

14.5 The Weibull Distribution in Life Testing

Whereas the exponential distribution arises as the life distribution when the hazard rate
functionλ(t) is assumed to be constant over time, there are many situations in which it is
more realistic to suppose thatλ(t) either increases or decreases over time. One example of
such a hazard rate function is given by

λ(t)=αβtβ−^1 , t> 0 (14.5.1)

whereαandβare positive constants. The distribution whose hazard rate function is given
by Equation 14.5.1 is called theWeibulldistribution with parameters (α,β). Note that
λ(t) increases whenβ>1; decreases whenβ<1; and is constant (reducing to the
exponential) whenβ=1.
The Weibull distribution function is obtained from Equation 14.5.1 as follows:

F(t)= 1 −exp

{

−

∫t

0

λ(s)ds

}

, t> 0

= 1 −exp{−αtβ}

Differentiating yields its density function:

f(t)=αβtβ−^1 exp{−αtβ}, t> 0 (14.5.2)

This density is plotted for a variety of values ofαandβin Figure 14.2.
Suppose now thatX 1 ,...,Xnare independent Weibull random variables each having
parameters (α,β), which are assumed unknown. To estimateαandβ, we can employ the
maximum likelihood approach. Equation 14.5.2 yields the likelihood, given by

f(x 1 ,...,xn)=αnβnx 1 β−^1 ···xnβ−^1 exp

{

−α

∑n

i= 1

xiβ

}

Hence,

logf(x 1 ,...,xn)=nlogα+nlogβ+(β−1)

∑n

i= 1

logxi−α

∑n

i= 1

xβi