Estimation, Evaluation, and Selection of Actuarial Models

124 APPENDIX A. SOLUTIONS TO EXERCISES
Weibull Fit
F(x)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.2 0.4 0.6 0.8 1
F n (x)
p − p plot for Data Set B censored at 1,000
Exercise 78 For Data Set B truncated at 50, the test statistic is 0.0887 while the critical value is
unchanged from the example (0.3120). The null hypothesis is not rejected and it is plausible that
the data came from a Weibull population. For Data Set B censored at 1,000, the test statistic is
0.0991 while the critical value is 0.3041. The null hypothesis is not rejected.
Exercise 79 The first step is to obtain the distribution function. This can be recognized as an
inverse exponential distribution, or the calculation done as
F (x) =
=
Z x
0
Z ∞
2/x
2y −2 e −2/y dy =
e −z dz = e −2/x .
Z ∞
2/x
2(2/z) −2 e −z (2z −2 )dz
In the first line the substitution z =2/y was used. The calculation then proceeds as follows:
x F (x) compare to max difference
1 0.135 0, 0.2 0.135
2 0.368 0.2, 0.4 0.168
3 0.513 0.4, 0.6 0.113
5 0.670 0.6, 0.8 0.130
13 0.857 0.8, 1 0.143
The test statistic is the maximum from the final column, or 0.168.
Exercise 80 The distribution function is
F (x) =
Z x
The calculation proceeds as follows:
0
2(1 + y) −3 dy = −(1 + y) −2¯¯x
0 =1− (1 + x)−2 .