1 Chapter 1

  • p. 4, line 14 from bottom: happening they way → happening the way
  • p. 36, line 6 from bottom of first paragraph: because the one assumes → because then one assumes
  • p. 44, section heading of 1.6: Data collection and storage → The design of a factorial experiment (thanks to Maria Fionda for pointing this out)

2 Chapter 2

3 Chapter 3

On p. 4:

  • what is the smallest number of heads you can get in 500 tosses that is not significantly different from the 300-out-of-500 result? It’s 277, i.e. 0.554 or 55.4% (the result of sum(dbinom(x=0:277, size=number.tosses, prob=perc.heads)) is <0.025, the result of sum(dbinom(x=0:278, size=number.tosses, prob=perc.heads)) is >0.025);
  • what is the largest number of heads you can get in 500 tosses that is not significantly different from the 300-out-of-500 result? It’s 322, i.e. 0.644 or 64.4% (the result of sum(dbinom(x=322:number.tosses, size=number.tosses, prob=perc.heads)) is <0.025, the result of sum(x=dbinom(0:321, size=number.tosses, prob=perc.heads)) is >0.025);

Thus, computing the 95%-CI for this example in this way returns [277, 322] or, in %, [0.554, 0.644], and you can see how similar they are to the result of binom.test, which was [0.556, 0.643]. Thus, the 95%-CI includes the values that do not differ significantly from the result you found and the interpretation of this approach would be to say “the 95%-CI boundaries are the numbers of heads that wouldn’t be significantly different from the 0.6 result you got, i.e. between 277 and 322”.

  • what is the smallest number of heads you can get in 500 tosses that is not significantly different from the 300-out-of-500 result? It’s 278, i.e. 0.556 or 55.6% (the result of sum(dbinom(x=0:278, size=number.tosses, prob=perc.heads)) is >0.025, the result of sum(dbinom(x=0:277, size=number.tosses, prob=perc.heads)) is >0.025);
  • what is the largest number of heads you can get in 500 tosses that is not significantly different from the 300-out-of-500 result? It’s 321, i.e. 0.642 or 64.2% (the result of sum(dbinom(x=321:number.