N0 2000 t 200 k 400 alpha c005002001 ralpha qhyper1

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> N0<-2000; t<-200; k<-400 > alpha<-c(0.05,0.02,0.01) > ralpha<-qhyper(1-alpha,t,N0-t,k) > data.frame(alpha,ralpha) alpha ralpha 1 0.05 49 2 0.02 51 3 0.01 53 For example, we must capture al least 49 that were tagged in order to reject H 0 at the = 0 . 05 level. In this case the estimate for N is ˆ N = [ kt/r ] = 1632 . As anticipated, r increases and the critical regions shrinks as the value of decreases. Using the level r determined using the value N 0 for N , we see that the power function ( N ) = P N { R r } . R is a hypergeometric random variable with mass function f R ( r ) = P N { R = r } = ( t r )( N - t k - r ) ( N k ) . The plot for the case = 0 . 05 is given using the R commands > N<-c(1300:2100) > pi<-1-phyper(49,t,N-t,k) > plot(N,pi,type="l",ylim=c(0,1)) We can increase power by increasing the size of k , the number the value in the second capture. This increases the value of r . For = 0 . 05 , we have the table. > k<-c(400,600,800) > N0<-2000 > ralpha<-qhyper(0.95,t,N0-t,k) > data.frame(k,ralpha) k ralpha 1 400 49 2 600 70 3 800 91 We show the impact on power ( N ) of both significance level and the number in the recapture k in Figure 18.3. Exercise 18.4. Determine the type II error rate for N = 1600 with k = 400 and = 0 . 05 , 0 . 02 , and 0.01, and = 0 . 05 and k = 400 , 600 , and 800. 281
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Introduction to the Science of Statistics Composite Hypotheses 1400 1600 1800 2000 0.0 0.2 0.4 0.6 0.8 1.0 N pi 1400 1600 1800 2000 0.0 0.2 0.4 0.6 0.8 1.0 N pi Figure 18.3: Power function for Lincoln-Peterson mark and recapture test for population N 0 = 2000 and t = 200 captured and tagged. (left) k = 400 recaptured = 0 . 05 (black), 0.02 (red), and 0.01 (blue). Notice that lower significance level reduces power. (right) = 0 . 05 , k = 400 (black), 600 (red), and 800 (blue). As expected, increased recapture size increases power. Example 18.5. For a two-sided test H 0 : μ = μ 0 versus H 1 : μ 6 = μ 0 . In this case, the parameter values for the null hypothesis 0 consist of a single value, μ 0 . We reject H 0 if | ¯ X - μ 0 | is too large. Under the null hypothesis, Z = ¯ X - μ 0 σ / p n is a standard normal random variable. For a significance level , choose z / 2 so that P { Z z / 2 } = P { Z  - z / 2 } = 2 . Thus, P {| Z | z / 2 } = . For data x = ( x 1 , . . . , x n ) , this leads to a critical region C = x ; ¯ x - μ 0 σ / p n z / 2 . If μ is the actual mean, then ¯ X - μ σ 0 / p n is a standard normal random variable. We use this fact to determine the power function for this test ( μ ) = P μ { X 2 C } = 1 - P μ { X / 2 C } = 1 - P μ ¯ X - μ 0 σ 0 / p n < z / 2 = 1 - P μ - z / 2 < ¯ X - μ 0 σ 0 / p n < z / 2 = 1 - P μ - z / 2 - μ - μ 0 σ 0 / p n < ¯ X - μ σ 0 / p n < z / 2 - μ - μ 0 σ 0 / p n = 1 - Φ z / 2 - μ - μ 0 σ 0 / p n + Φ - z / 2 - μ - μ 0 σ 0 / p n 282
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