Introduction to Probability and Statistics for Engineers and Scientists

14.3The Exponential Distribution in Life Testing 589

SOLUTION Since 2τ/θ 0 =3,600/150=24, thep-value is

p-value=P{χ 402 ≤ 24 }

=.021 from Program 5.8.1a

Hence, the manufacturer’s claim should be rejected at the 5 percent level of significance
(indeed at any significance level at least as large as .021). ■

It follows from Equation 14.3.5 that the accuracy of the estimatorτ/rdepends only
onrand not onn, the number of items put on test. The importance ofnresides in the
fact that by choosing it large enough we can ensure that the test is, with high probability,
of short duration. In fact, the moments ofX(r), the time at which the test ends are easily
obtained. Since, withX(0)≡0,

X(j)−X(j−1)=

Yj

n−j+ 1

, j=1,...,r

it follows upon summing that

X(r)=

∑r

j= 1

Yj

n−j+ 1

Hence, from Equation 14.3.4,X(r)is the sum ofrindependent exponentials having
respective meansθ/n,θ/(n−1),...,θ/(n−r+1). Using this, we see that

E[X(r)]=

∑r

j= 1

θ

n−j+ 1

=θ

∑n

j=n−r+ 1

1

j

(14.3.7)

Var(X(r))=

∑r

j= 1

(

θ

n−j+ 1

) 2

=θ^2

∑n

j=n−r+ 1

1

j^2

where the second equality uses the fact that the variance of an exponential is equal to the
square of its mean. For largen, we can approximate the preceding sums as follows:

∑n

j=n−r+ 1

1

j

≈

∫n

n−r+ 1

dx

x

=log

(

n

n−r+ 1

)

∑n

j=n−r+ 1

1

j^2

≈

∫n

n−r+ 1

dx

x^2

=

1

n−r+ 1

−

1

n

=

r− 1

n(n−r+1)