Nonparametric Maximum Likelihood Estimate

[skip_it2]

Thirteen electrical components were placed on a life test, yielding failure times
0.22, 0.50, 0.88, 1.00, 1.32, 1.33, 1.54, 1.76, 2.50, 3.00, 3.65, 4.20, 6.40 in hours.
(a) Give the nonparametric maximum likelihood estimate of the reliability
function.
(b) Give the point estimate and approximate 95% confidence interval for R(1.6).

 

[JohnM]

F(t) = 1 - R(t) and
R(t) = 1 - F(t)

For a nonparametric estimate, F(t) is simply the number of units that have failed at or before time t, divided by the number of units on test.

(a)
Let Y be the number of survivors by time t. Y follows a binomial distribution with parameters n and p. By time t, the unit has either survived or it hasn't...

So, the maximum likelihood function is p = Y/n where p is the probability of survival (i.e., reliability), Y is the number of surviving units, n is the total number of units on test.

(b)
since 7 units have failed by t = 1.6,
R(1.6) = 6/13 = .462

95% confidence
as n gets large, the distribution is approximately normal with mu = p and s = sqrt(pq/n)

95% conf interval = p +/- (1.96 * sqrt(pq/n))
= .462 +/- 1.96 * sqrt((.462*.538)/13))
= .462 +/- .271
= (.191, .733)