Introduction to Probability and Statistics for Engineers and Scientists

588 Chapter 14*:Life Testing

That is, 2τ/θhas a chi-square distribution with 2rdegrees of freedom. Hence,

P

{

χ 12 −α/2,2r< 2 τ/θ<χα^2 /2,2r

}

= 1 −α

and so a 100(1−α) percent confidence interval forθis

θ∈

(

2 τ

χα^2 /2,2r

,

2 τ

χ 12 −α/2,2r

)

(14.3.6)

One-sided confidence intervals can be similarly obtained.

EXAMPLE 14.3a A sample of 50 transistors is simultaneously put on a test that is to be
ended when the 15th failure occurs. If the total time on test of all transistors is equal to
525 hours, determine a 95 percent confidence interval for the mean lifetime of a transistor.
Assume that the underlying distribution is exponential.

SOLUTION From Program 5.8.1b,

χ.025,30^2 =46.98, χ.975,30^2 =16.89

and so, using Equation 14.3.6, we can assert with 95 percent confidence that

θ∈(22.35, 62.17) ■

In testing a hypothesis aboutθ, we can use Equation 14.3.6 to determine thep-value
of the test data. For instance, suppose we are interested in the one-sided test of

H 0 :θ≥θ 0

versus the alternative

H 1 :θ<θ 0

This can be tested by first computing the value of the test statistic 2τ/θ 0 — call this
valuev— and then computing the probability that a chi-square random variable with 2r
degrees of freedom would be as small asv. This probability is thep-value in the sense that
it represents the (maximal) probability that such a small value of 2τ/θ 0 would have been
observed ifH 0 were true. The hypothesis should then be rejected at all significance levels
at least as large as thisp-value.

EXAMPLE 14.3b A producer of batteries claims that the lifetimes of the items it manufac-
tures are exponentially distributed with a mean life of at least 150 hours. To test this claim,
100 batteries are simultaneously put on a test that is slated to end when the 20th failure
occurs. If, at the end of the experiment, the total test time of all the 100 batteries is equal
to 1,800, should the manufacturer’s claim be accepted?