Introduction to Probability and Statistics for Engineers and Scientists

240 Chapter 7: Parameter Estimation

To estimatesi, the probability that a patient who has survived the firsti−1 months will
also survive monthi, we should take the fraction of those patients who began theirith
month of drug taking and survived the month. For instance, because 11 of the 12 patients
survived month 1,ˆs 1 =11/12. Because all 11 patients who began month 2 survived,
ˆs 2 =11/11. Because 10 of the 11 patients who began month 3 survived,ˆs 3 =10/11.
Because 8 of the 9 patients who began their fourth month of taking the drug (all but the
ones labelled 1, 3, and 3∗) survived month 4,ˆs 4 =8/9. Similar reasoning holds for the
others, giving the following survival rate estimates:

ˆs 1 =11/12

ˆs 2 =11/11

ˆs 3 =10/11

ˆs 4 =8/9

ˆs 5 =7/8

ˆs 6 =7/7

ˆs 7 =6/7

ˆs 8 =4/5

ˆs 9 =3/4

ˆs 10 =3/3

ˆs 11 =3/3

ˆs 12 =1/2

ˆs 13 =1/1

ˆs 14 =1/2

We can now use

∏j
i= 1 ˆsito estimate the probability that a drug taker survives at leastj
time periods,j=1,..., 14. For instance, our estimate ofP{X> 6 }is 35/54.

7.3Interval Estimates

Suppose thatX 1 ,...,Xnis a sample from a normal population having unknown meanμ
and known varianceσ^2. It has been shown thatX=

∑n
i= 1 Xi/nis the maximum likelihood
estimator forμ. However, we don’t expect that the sample meanXwill exactly equalμ,
but rather that it will “be close.” Hence, rather than a point estimate, it is sometimes more
valuable to be able to specify an interval for which we have a certain degree of confidence
thatμlies within. To obtain such an interval estimator, we make use of the probability
distribution of the point estimator. Let us see how it works for the preceding situation.