Discussion
Suppose only the first 6 failure times of 18 items kept on test are observed. Suppose the failure times are independent and follow an exponential distribution with mean \(\theta\). If the data are :
1.5, 2, 2.5, 3, 4, 5
the maximum likelihood estimate of \(\theta\) :
Answer: D
The maximum likelihood estimate is 13.
This is type II censoring: r = 6 failures are observed out of n = 18, and the other 12 items are still working at the sixth failure time, 5. The MLE of the exponential mean is the total time on test divided by the number of failures:
\[ \begin{aligned} T &= (1.5 + 2 + 2.5 + 3 + 4 + 5) + 12 \times 5 \\ &= 18 + 60 = 78 \\ \hat{\theta} &= \frac{78}{6} = 13 \end{aligned} \]
Averaging only the six observed times gives 3, which ignores the 12 surviving items. Censored data still give an MLE, so 'cannot be obtained' is wrong.
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