Appendix 1. Answers to Selected Problems

Chapter 1.

  1. Their conclusions are identical.
  2. No, these are opposite conclusions.
  3. (Note: ~ is 'not' and ? is 'maybe'):  ~M,~F; ~M,?F; ?M,~F; ?M,?F; ~(MF).
  4. Answer: 35
  5. Answer:  1/3
  6. Answer: 6.
  7. Answer: 60.
  8. Answer: 3/4.
  9. 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.
  10. Answer: P = 1.0.
  11. Answers:  b. p4   c. p2q2   d. 1/4  e. p    f. p2     g. 3p2q2/(1-q2)   h. p2   i. 1-p2
  12. Answer:  
  13. Answers:  20! / (5! 6! 4! 3!); 18! / (4! 5! 4! 3!)

Chapter 2.

  1. Mean, 3.704; Variance, 4.427
  1. The required probability is P(no females) + P(all females), or P=2×0.56=0.03125.
  2. 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.
  3. 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.
  4. 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.
  5. 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.
  6. 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.
  7. The distribution will be the sum of the three Poisson distributions, with the heterozygote accounting for 1/2 the density.

Chapter 3.

  1. 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.
  2. The mean number of clones per filter is 19.8. Using the bootstrap method, we obtain 95% confidence limits of (17.4, 22.2).
  3. The mean (sd) for the three mutants are 0.458 (0.104), 1.82 (0.490), 0.149 (0.119), respectively.
  4. 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.

  1. The power is 0.98.
  2. 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.

  1. Using the Wilcoxon signed rank test, Ws= 8 for the negative values and P=0.348.
  2. For the one-sided test, W=28 and P=0.0318.
  3. Using the Kruskal-Wallis test, X2 = 17.42, P=0.0006
  4. Using the one-sided Wilcoxon rank sum test, P<6×10-3.
  5. Using the Jonckheere-Terpstra test, z=4.20 and the one-sided P-value is 1.3×10-5.

Chapter 6.

  1. Using a two-sided Fisher's exact test, P=0.193.
  2. X2=2.63, P=0.104
  3. X2=2.84, 2 degrees of freedom; P=0.242.
  4. X2=12.37, 6 degrees of freedom; P=0.055.
  5. 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.
  6. 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.

  1. X2 = 9.71, P = 0.0018.
  2. X2 = 1.67, P = 0.20

Chapter 8.

  1. K=11, n=7, P=0.068.
  2. K=11, n=6, P=0.0146.

Chapter 9.

  1. 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.
  2. Using the Mantel-Haenszel test, we get a test statistic of -2.49 and a two-sided P-value of 0.013.

Chapter 10.

  1. Using the normal approximation P(X>Y)=0.966; approximately 10 wells per group would be required for the two-tailed test.
  2. 8 observations per group.
  3. From the table, 76 per group for the one-sided test.