We have H_0: p = .5 versus H_1: p > .5. We choose alpha=.01.
Question: Find the critical value, x*, if the sample size is n=10; is n=19.
Answer: For n = 10:
qbinom(.99,10,.5)
yields an answer of 9, but notice the following:
1-pbinom(8,10,.5) [1] 0.01074219 1-pbinom(9,10,.5) [1] 0.0009765625These results would suggest that we use x*=10, since that is the smallest value for which the true alpha is less than or equal to .01; but we will break convention here since .01074219 is so close to .01 and use x*=9.
The n=19 case is more straightforward:
qbinom(.99,19,.5) [1] 14 1-pbinom(13,19,.5) [1] 0.03178406 1-pbinom(14,19,.5) [1] 0.009605408Which says that x* for this test would be 15.
Question: Now, suppose that the true proportion of students who prefer Hersheys is .8. What would be the probability the test rejects H_0 in favor of H_1; such a result would be a correct decision, since H_1 is true if p=.8. Compute this probability for both the n=10 test and the n=19 test.
Answer: Here are the answers through R:
1-pbinom(8,10,.8) [1] 0.3758096 1-pbinom(14,19,.8) [1] 0.6732881
The language we use is this: "With n=10, the .01-level test has a power of .376 for a true p of .8. With n=19, the .01-level test has a power of .673 for a true p of .8."
Pictures and R code to go along with n=10 and n=19 Type II error discussion.
x <- 0:20 y10.0 <- dbinom(x,10,.5) y10.1 <- dbinom(x,10,.8) y19.0 <- dbinom(x,19,.5) y19.1 <- dbinom(x,19,.8) par(mfrow=c(2,1)) # for two graphs in window plot(x,y10.0,type='h') points(x+.1,y10.1,type='h',lty=3) # fudged the plot right a tad plot(x,y19.0,type='h') points(x+.1,y19.1,type='h', lty=3) # fudged the plot right a tad par(mfrow=c(1,1)) # end of code for these plots
Power curves
Here is a simple example of having R draw power curves. Here we have n Bernoulli trials, p=success probability, we are using an alpha of approximately .01. When n=10, the critical value is 9. When n=19, the critical value is 15. The true alphas for these 2 cases are .01074 and .0096, which are close enough to consider them .01-level tests. We want to compare two tests with the same alpha; hence the rare appearance of the number 19 in a classroom example.
# R code for drawing power curves with n=10 and n=19 p <- seq(.5, 1, .01) # set up the p axis. pi <- 1-pbinom(8,10,p) # Compute power values (pi) at each p. plot(p,pi,type='l',tck=1) pi2 <- 1 - pbinom(14,19,p) # The power function with n=19. lines(p,pi2,lty=2) # lines adds a figure to an existing plot. # end of code for these plots