The Essentials of Biostatistics for Physicians, Nurses, and Clinicians

Solutions to Selected Exercises 191

(a) Since the data is complete and there are 8 event times, the median is the average

of the 4th and 5th ordered observations, which is (11 + 18.5)/2 = 14.75 months.

(b) The mean survival is just the arithmetic average of the eight event times, which

is (3 + 4.5 + 6 + 1 1 + 18.5 + 2 0 + 2 6 + 35)/8 = 15.5 months.

(c) A life table for this data using 5 - month intervals is as follows:

Time

Interval I j

No. of

deaths

in I j

No.

withdrawn

in I j

No.at

risk in I j

Avg. no.

at risk

in I j

Est.

prop. of

deaths

in I j

Est.

prop.

Surv. at

end of I j

Est.

cum.

surv. at

end of I j

[0, 5) 2 0 8 8 0.25 0.75 0.75

[5, 10) 1 0 6 6 0.167 0.833 0.585

[10 15) 1 0 5 5 0.200 .8 0.468

[15,20) 1 0 4 4 0.250 0.750 0.341

[20,25) 1 0 3 3 0.333 0.667 0.227

[25,30) 1 0 2 2 0.500 0.500 0.114

[30,35) 0 0 2 2 0.000 1.000 0.114

[35,40) 1 0 1 1 1.000 0.000 0.000

[40, ∞ ) 0 0 0 0 — — —

10. Suppose a cure model is known to have S 1 ( t ) = exp( − 0.5 t ).
    Recall S ( t ) = p + ( 1 − p ) S 1 ( t ). Suppose that we know that S (2) = 0.5259.
    Can you calculate the cure rate for this model? If so what is it?
    S ( t ) = p + ( 1 − p ) S 1 ( t ). Since we know S 1 ( t ) = exp( − 0.5 t ), we only need to deter-
    mine pS ( t ).
    We are given S (2) = 0.5259, and we can use this information to solve for p.
    S (2) = 0.5259, on the one hand, and
    Sp p p p
    p

( ) ( ) exp[ (. ) ] ( )(. )

(. ).

2 1 0 5 1 0 3679

1 0 3679 0 367

=+− −^2 =+−

=− + 9 9.

Sop(. ).1 0 3679− =−=0 5259 0 3679 0 1580...

P=−=0 1580 1 0 3679.(.). .//0 1580 0 6321 0 4295=.

So S ( t ) = 0.4295 + 0.5705exp( − 0.5 t ), and the probability of cure is 0.4295. This

means that approximately 43% of the patients receiving this treatment will be cured

of the disease based on this model.

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