Appendix 1. Answers to Selected Problems
Chapter 1.
- Their conclusions are identical.
- No, these are opposite conclusions.
- (Note: ~ is 'not' and ? is 'maybe'): ~M,~F; ~M,?F; ?M,~F; ?M,?F; ~(MF).
- Answer: 35
- Answer: 1/3
- Answer: 6.
- Answer: 60.
- Answer: 3/4.
- a. P(AB)=0.15, P(A)=0.50, P(B)=0.30, P(AB)=P(A)P(B), yes.; b. P(Bc)=0.15, P(B)=0.30, P(c)=0.70, P(Bc)≠P(B)P(c), no.; c. P(A|B) = P(AB)/P(B)=0.15/0.30=1/2.; d. 1/3; e. 0.2857; f. no.
- Answer: P = 1.0.
- Answers: b. p4 c. p2q2 d. 1/4 e. p f. p2 g. 3p2q2/(1-q2) h. p2 i. 1-p2
- Answer:
- Answers: 20! / (5! 6! 4! 3!); 18! / (4! 5! 4! 3!)
Chapter 2.
- Mean, 3.704; Variance, 4.427
- The required probability is P(no females) + P(all females), or P=2×0.56=0.03125.
- For a Poisson distribution with mean 15, the probability of getting 25 or more mutants is p=1-P(x ≤ 24|m=15)= 0.0112.
- a) Given that p=0.5, we want a value of N that will insure that P(s=0)≤0.05. For N = 4, P(0)=0.0625 and N=5, P(0)=0.03125; we should use 5 animals. b) After 5 generations, the chance that it won't be lost is (1-0.03125)5=0.853.
- The chance that an individual cell will carry B is 0.092. If the two markers were independent, the probability (from the binomial distribution) of 8 or more B+ cells with N=10 and p=0.092 is about 2×10-7.
- Under the null hypothesis, the probability that, in a given pair, the X carrier would show a higher value is 0.5. For the 8 lines tested, 6 gave a higher value in the presence of X. From the binomal distribution P(s≥6|N=8,p=0.5)=0.145.
- For m=1 and x≥2,3,and 4, the exact probabilities are 0.264, 0.080, and 0.019 while the normal approximation (with continuity correction) gives 0.309, 0.067, and 0.0062. For m=8 and x≥11,14, and 16, exact probabilities are 0.184, 0.034, and 0.008 while approximate probabilities are 0.188, 0.026 and 0.004.
- The distribution will be the sum of the three Poisson distributions, with the heterozygote accounting for 1/2 the density.
Chapter 3.
- First find the MLE of Θ = (1-r)2. You should get 0.4835. From this you can find the estimate of r to be 0.3047. Then calculate its variance which is given by = 0.002174. Then from the large sample variance formula for a transformed variable, , derive the variance of to be 0.001124.
- The mean number of clones per filter is 19.8. Using the bootstrap method, we obtain 95% confidence limits of (17.4, 22.2).
- The mean (sd) for the three mutants are 0.458 (0.104), 1.82 (0.490), 0.149 (0.119), respectively.
- The mean (variance) by the exact and approximate methods are 4.22×106 (4.80×1013) and 1.80×106 (1.08×1013), respectively.
Chapter 4.
- The power is 0.98.
- Testing on D/D, P=0.091; testing on d/d, P=0.041. For the D/D, the critical values and power are as in the example. For the d/d test, the critical value is 9 and the power is 0.69.
Chapter 5.
- Using the Wilcoxon signed rank test, Ws= 8 for the negative values and P=0.348.
- For the one-sided test, W=28 and P=0.0318.
- Using the Kruskal-Wallis test, X2 = 17.42, P=0.0006
- Using the one-sided Wilcoxon rank sum test, P<6×10-3.
- Using the Jonckheere-Terpstra test, z=4.20 and the one-sided P-value is 1.3×10-5.
Chapter 6.
- Using a two-sided Fisher's exact test, P=0.193.
- X2=2.63, P=0.104
- X2=2.84, 2 degrees of freedom; P=0.242.
- X2=12.37, 6 degrees of freedom; P=0.055.
- This is an example of Simpson's paradox. The dependence of the frequency of death penalty verdicts reverse when the combined table is compared with the tables stratified by victim race.
- a. Fisher's exact test, P=0.33; b. Fisher's exact, P=0.81; c. X2=19.86, 1 df, P<10-5; d. Fisher's exact, P=0.01; e. X2=11.2, 1 df, P=0.0008; f. McNemar's test, P=0.02; g. McNemar's test, P=0.125.
Chapter 7.
- X2 = 9.71, P = 0.0018.
- X2 = 1.67, P = 0.20
Chapter 8.
- K=11, n=7, P=0.068.
- K=11, n=6, P=0.0146.
Chapter 9.
- Using the Kruskal-Wallis test, we obtain X2 values of 6.90, 9.12, and 9.05 for the three experiments (each with 3 df); P-values are 0.075, 0.028, and 0.029. Combining these experiments, we get X2=25.06 (9 df) and P=0.0029.
- Using the Mantel-Haenszel test, we get a test statistic of -2.49 and a two-sided P-value of 0.013.
Chapter 10.
- Using the normal approximation P(X>Y)=0.966; approximately 10 wells per group would be required for the two-tailed test.
- 8 observations per group.
- From the table, 76 per group for the one-sided test.