13_AS_2_lec_a

42 dj so that p value pr 2 242 012 005 1 and hence

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Unformatted text preview: or Group 1 (in months): 2, 4*, 5, 6, 9, 9, 12, 12*, 15*, 17. Survival times for Group 2 (in months): 6*, 7, 9*, 10, 13, 15*, 17*, 18 where ∗ denotes the censored. What can you say about the hazard rates for the 2 groups? 36/42 Actuarial Statistics – Module 2: Non-parametric methods Comparing survival functions Special cases For k = 10 tj 2 5 6 7 9 10 12 13 17 18 sum 37/42 n1j 10 8 7 6 6 4 4 2 1 0 d1j 1 1 1 0 2 0 1 0 1 0 7 n2j 8 8 8 7 6 5 4 4 2 1 d2j 0 0 0 1 0 1 0 1 0 1 4 nj 18 16 15 13 12 9 8 6 3 1 dj 1 1 1 1 2 1 1 1 1 1 11 Actuarial Statistics – Module 2: Non-parametric methods Comparing survival functions Special cases We have χ2 = LR (Z1 )2 LR var (Z1 ) = k j =1 (d1j n1j k j =1 nj 1− − e1j ) n1j nj 2 nj −dj nj −1 = 2.42 dj so that p-value = Pr (χ2 > 2.42) = 0.12 > 0.05 1 and hence we can not reject the null hypothesis that the hazard rates for the 2 groups are same. 38/42 Actuarial Statistics – Module 2: Non-parametric methods Comparing survival functions Special cases Wilcoxon statistic: w (tj ) = nj (Gehan, 1965) In this particular case k W Z1 = nj (d1j − e1j ) j =1 and k (nj )2 W var (Z1 ) = j =1 n1j nj 1− n1j nj nj − dj nj − 1 Again, under the null hypothesis, W (Z1 )2 W var (Z1 ) 39/42 is Chi-square distributed (with 1 degree of freedom). dj . Actuarial Statistics – Module 2: Non-parametric methods Comparing survival functions Special cases Example A medical study was performed to investigate the difference (or otherwise) in the effectiveness of 2 alternative treatments (A and B) to a disease. The survival time (in months) are Survival times for Group 1 (in months): 2, 3∗ , 4, 6, 6, 12∗ , 12 Survival times for Group 2 (in months): 6∗ , 7, 9∗ , 10, 13, 15∗ , 17∗ where ∗ denotes a censored observation. Suppose the variance of the Wilcoxon statistic was calculated as 160.40. Calculate the Wilcoxon statistic and perform the Wilcoxon test. 40/42 Actuarial Statistics – Module 2: Non-parametric methods Comparing survival functions Special cases Solution j 1 2 3 4 5 6 7 tj 2 4 6 7 510 12 13 d1j 1 1 2 0 0 1 0 n1j 7 5 4 2 2 2 0 d2j 0 0 0 1 1 0 1 n2j 7 7 7 6 4 3 3 dj 1 1 2 1 1 1 1 nj 14 12 11 8 6 5 3 n e1j = n1jj dj 0.5 0.416666667 0.727272727 0.25 0.333333333 0.4 0 nj (d1j − e1j ) 7 7 14 -2 -2 3 0 and hence the Wilcoxon statistic is 27. The null hypothesis is that there is no difference in all points of the survival function vs the alternative of at least 1 point of difference. 272 The Wilcoxon chi-squared statistic is 160.4 = 4.545 which is significant. Hence on the basis of the Wilcoxin test result we reject the null and conclude that there is evidence of a different survival function. 41/42 Actuarial Statistics – Module 2: Non-parametric methods Comparing survival functions Special cases Discussion Some other weight functions may be appropriate. The choice of weight function depends on the investigator’s desire to give different weights to different types of error. For instance, when comparing 1 vs nj (log-rank vs Wilcoxon), the latter gives more weight to early times (because n1 > n2 > n3 > · · · ) A practical note: log-rank statistic is more powerful for detecting differences in the hazard rates when the hazard rates are proportional (h1 (t ) = rh2 (t )) i.e., S1 (t ) = [S2 (t )]r for some constant r . The above tests can be generalised to involve more than 2 groups. 42/42...
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