Topic 19 Extensions on the Likelihood Ratio

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1 Topic 19 Extensions on the Likelihood Ratio Two-Sided Tests 1 / 12

2 Outline Overview Normal Observations Power Analysis 2 / 12

3 Overview The likelihood ratio test is a popular choice for composite hypothesis when Θ 0 is a subspace Θ the parameter space. The rationale for this approach is that the null hypothesis is unlikely to be true if the maximum likelihood on Θ 0 is sufficiently smaller that the likelihood maximized over Θ. Let ˆθ0 be the parameter value that maximizes the likelihood for θ Θ 0 and ˆθ be the parameter value that maximizes the likelihood for θ Θ. The likelihood ratio Λ(x) = L(ˆθ 0 x) L(ˆθ x). 3 / 12

4 Overview We have two optimization problems - maximize L(θ x) on the parameter space Θ and on the null hypothesis space Θ 0. The critical region for an α-level likelihood ratio test is {Λ(x) λ α }. As with any α level test, λ α is chosen so that P θ {Λ(X ) λ α } α for all θ Θ 0. NB. This ratio is the reciprocal from the version given by the Neyman-Pearson lemma. Thus, the critical region consists of those values that are below a critical value. 4 / 12

5 Normal Observations Consider the two-sided hypothesis H 0 : µ = µ 0 versus H 1 : µ µ 0. Here the data are n independent N(µ, σ 0 ) random variables with known variance σ0 2. The parameter space Θ is one dimensional giving the value µ for the mean. As we have seen before ˆµ = x. Θ 0 is the single point {µ 0 } and so ˆµ 0 = µ 0. L(ˆµ 0 x) = and ( Notice that ) n 1 exp 1 2πσ 2 0 2σ0 2 ( n (x i µ 0 ) 2, L(ˆµ x) = i=1 ) n 1 exp 1 2πσ 2 0 2σ0 2 ( n ) Λ(x) = exp 1 2σ0 2 ((x i µ 0 ) 2 (x i x) 2 ) = exp n 2σ 2 ( x µ 0 ) 2. i=1 0 2 ln Λ(x) = n σ0 2 ( x µ 0 ) 2 = ( ) 2 x µ0 σ 0 /. n n (x i x) 2 i=1 5 / 12

6 Then, critical region, {Λ(x) λ α } = Normal Observations { ( ) } x 2 µ0 σ 0 / 2 ln λ α. n Under the null hypothesis, ( X µ 0 )/(σ 0 / n) is a standard normal random variable, and thus 2 ln Λ(X ) is the square of a single standard normal. This is the defining property of a χ-square random variable with 1 degree of freedom. Naturally we can use both ( ) x 2 µ0 σ 0 / and n x µ 0 σ 0 / n. as a test statistic. We have seen the second choice in the example of a possible invasion of a model butterfly by a mimic. 6 / 12

7 For the two-sided two-sample α-level likelihood ratio test for population proportions p 1 and p 2, based on the hypothesis H 0 : p 1 = p 2 versus H 1 : p 1 p 2, 1 we maximize the likelihood over the subspace Θ 0 = {(p 1, p 2 ); p 1 = p 2 } (the blue line) and over the entire parameter space, Θ = [0, 1] [0, 1], shown as the square, and then take the ratio, simplify and make appropriate approximations. p p 1 The data are observations on n 1 Bernoulli trials, x 1,1, x 1,2,..., x 1,n1 from the first population and, independently, n 2 Bernoulli trials, x 2,1, x 2,2,..., x 2,n2 from the second. 7 / 12

8 The likelihood ratio test is approximately equivalent to the critical region z z α/2 where z = ˆp 1 ˆp 2 ( ) ˆp 0 (1 ˆp 0 ) n1 n2 with ˆp i, the sample proportion of successes from the observations from population i and ˆp 0, the pooled proportion ˆp 0 = 1 n 1 + n 2 ((x 1,1 + + x 1,n1 ) + (x 2,1 + + x 2,n2 )) = n 1ˆp 1 + n 2ˆp 2 n 1 + n 2. 8 / 12

9 The subsequent winter had 167 out of 250 hives surviving. To test if the two survival probabilities are significantly different: > prop.test(c(250,167),c(332,250)) 2-sample test for equality of proportions with continuity correction data: c(250, 167) out of c(332, 250) X-squared = 4.664, df = 1, p-value = alternative hypothesis: two.sided 95 percent confidence interval: sample estimates: prop 1 prop / 12

10 Exercise. 1. Give the 95% confidence interval for the difference in proportions. Does it contain the value 0? 2. Modify the R command to find a 98% confidence interval for the difference in proportions. Does it contain the value 0? 3. Explain your answers to the first two parts - why does one contain 0 and the other not? Could you have guessed this in advance by looking at the p-value. 4. Assume that the second winter was more severe. State a hypothesis for appropriate for this situation and modify the R commands and give the p-value. Could you have given this p-value based on the information in the output on the previous slide? 10 / 12

11 Power analyses can be executed in R using the power.prop.test command. If we want to be able to detect a difference between two proportions p 1 = 0.7 and p 2 = 0.6 in a one-sided test with a significance level of α = 0.05 and power 1 β = 0.8. > power.prop.test(p1=0.70,p2=0.6,sig.level=0.05,power=0.8, alternative = c("one.sided")) Two-sample comparison of proportions power calculation n = p1 = 0.7 p2 = 0.6 sig.level = 0.05 power = 0.8 alternative = one.sided NOTE: n is number in *each* group We will need a sample of n = 281 from each group. 11 / 12

12 If we vary p 2 and determine the power. > power.prop.test(n=250,p1=0.70,p2=c(0.6,0.65),sig.level=0.05, alternative = c("one.sided")) p2 = 0.60, 0.65 power = , Now, let s vary sample size. > power.prop.test(n=c(250,350,450,550),p1=0.70,p2=0.60,sig.level=0.05, alternative = c("one.sided")) n = 250, 350, 450, 550 power = , , , Exercise. Determine the reduction in power when the significance level α = 0.02 for the sample sizes above. Why is the power reduced? 12 / 12

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