Topic 18: Composite Hypotheses

Transcription

1 Toc 18: November, 211 Simple hypotheses limit us to a decisio betwee oe of two possible states of ature. This limitatio does ot allow us, uder the procedures of hypothesis testig to address the basic questio: Does the legth, the reactio rate, the fractio displayig a particular behavior or havig a particular oio, the temperature, the kietic eergy, the Michaelis costat, the speed of light, mutatio rate, the meltig poit, the probability that the domiat allele is expressed, the elasticity, the force, the mass, the parameter value icrease, decrease or chage at all uder uder a differet experimetal coditio? This leads us to cosider composite hypothesis. I this case, the parameter space Θ is divided ito two disjoit regios, Θ ad Θ 1. The hypothesis test is ow writte H : Θ versus H 1 : Θ 1. Agai, H is called the ull hypothesis ad H 1 the alterative hypothesis. For the three alteratives to the questio posed above, icrease would lead to the choices Θ = {; } ad Θ 1 = {; > }, decrease would lead to the choices Θ = {; } ad Θ 1 = {; < }, ad chage would lead to the choices Θ = { } ad Θ 1 = {; } for some choice of parameter value. The effect that we are meat to show, here the ature of the chage, is cotaied i Θ 1. The first two optios give above are called oe-sided tests. The third is called a two-sided test, Rejectio ad failure to reject the ull hypothesis, critical regios, C, ad type I ad type II errors have the same meaig for a composite hypotheses as it does with a simple hypothesis. Sigificace level ad power will ecessitate a extetio of the ideas for smple hypotheses. 1 Power Power is ow a fuctio of the parameter value. If our test is to reject H wheever the data fall i a critical regio C, the the power fuctio is defied as π() = P {X C}. that gives the probability of rejectig the ull hypothesis for a give value of the parameter. Cosequetly, the ideal power fuctio has π() for all Θ ad π() 1 for all Θ 1 With this property for the power fuctio, we would rarely reject the ull hypothesis whe it is true ad rarely fail to reject the ull hypothesis whe it is false. Such a power fuctio shows that we reach the correct decisio with probability early 1. c 211 Joseph C. Watkis 226

2 I reality, icorrect decisios are made. Thus, for Θ, π() is the probability of makig a type I error, i. e., rejectig the ull hypothesis whe it is ideed true, For Θ 1, 1 π() is the probability of makig a type II error, i.e., failig to reject the ull hypothesis whe it is false. The goal is to make the chace for error small. The traditioal method is aalogous to that employed i the Neyma-Pearso lemma. Fix a (sigificace) level α, ow defied to be the largest value of π() i the regio Θ defied by the ull hypothesis. I other words, by focusig o the value of the parameter i Θ that is most likely to result i a error, we isure that the probability of a type I error is o more that α irrespective of the value for Θ. The, we look for a critical regio that makes the power fuctio as large as possible for values of the parameter Θ 1 Example 1. Let X 1, X 2,..., X be idepedet N(µ, σ 2 ) radom variables with σ 2 kow ad µ ukow. For the composite hypothesis for the oe-sided test H : µ µ versus H 1 : µ > µ. We use the test statistic from the likelihood ratio test ad reject H if X is too large. Thus, the critical regio If µ is the true mea, the the power fuctio C = {x; x k(µ )}. π(µ) = P µ {X C} = P µ { X k(µ )}. As we shall see soo, the value of k(µ ) depeds o the level of the test. Note that π(µ) icreases with µ. As the actual mea µ icreases, the the probability that the sample mea X exceeds a particular value k(µ ) also icreases. To obtai level α for the hypothesis test, we use that fact that π(µ) icreases with µ to coclude that the maximum value of π o the set Θ = {µ; µ µ } takes place for the value µ, i.e., α = π(µ ) = P µ { X k(µ )}. We ow use this to fid the value k(µ ). Whe µ is the value of the mea, we stadardize to give a stadard ormal radom variable Z = X µ σ /. Choose z α so that P {Z z α } = α. I this case, Φ(z α ) = 1 α where Φ is the distributio fuctio for the stadard ormal, thus P µ {Z z α } = P µ { X µ + σ z α } ad k(µ ) = µ + (σ / )z α. If µ is the true state of ature, the Z = X µ σ / is a stadard ormal radom variable. We use this fact to determie the power fuctio for this test. π(µ) = P µ { X σ z α + µ } = P µ { X µ σ z α (µ µ )} (1) { X µ = P µ σ / z α µ µ } ( σ / = 1 Φ z α µ µ ) σ / (2) 227

3 mu mu Figure 1: Power fuctio for the oe-sided test with alterative less tha. µ = 1, σ = 3. Note, as argued i the text that π is a decreasig fuctio. (left) = 16, α =.5 (black),.2 (red), ad.1 (blue). Notice that lowerig sigificace level α reduces power π(µ) for each value of µ. (right) α =.5, = 15 (black), 4 (red), ad 1 (blue). Notice that icreasig sample size icreases power π(µ) for each value of µ µ. Exercise 2. If the alterative is less tha, show that π(µ) = Φ ( z α µ µ ) σ /. Returig to the example with a model species ad its mimic. For the plot of the power fuctio for µ = 1, σ = 3, ad = 16 observatios, > zalpha<-qorm(.95) > mu<-1 > sigma<-3 > mu<-(6:11)/1 > <-16 > z<--zalpha - (mu-mu)/(sigma/sqrt()) > <-porm(z) > plot(mu,,type="l") I Figure 1, we vary the values of the sigificace level α ad the values of, the umber of observatios i the graph of the power fuctio π Example 3 (mark ad recapture). We may use mark ad recapture to see if a populatio has reached a dagerously low level. The variables i mark ad recapture are t be the umber captured ad tagged, k be the umber i the secod capture, r the the umber i the secod capture that are tagged, ad let 228

4 N be the total populatio. If N is the level that a wildlife biologist say is dagerously low, the the atural hypothesis is oe-sided. H : N N versus H 1 : N < N. The data are the he umber i the secod capture that are tagged, r. The likelihood fuctio for N is the hypergeometric distributio. ( t N t ) r)( L(N r) = k r ( N k ) ad the maximum likelihood estimate is ˆN = [tk/r]. Thus, higher values for r lead us to lower estimates for N. Let R be the (radom) umber i the secod capture that are tagged, the, for a α level test, we look for the miimum value r α so that P N {R r α } α for all N N. (3) As N icreases, the recaptures become less likely ad the probability i (3) decreases. Thus, we should set the value of r α accordig to the parameter value N, the miimum value uder the ull hypothesis. Let s determie r α for several values of α usig the example from the toc, Maximum Likelihood Estimatio, ad cosider the case i which the critical populatio is N = 2. > N<-2 > t<-2 > k<-4 > alpha<-c(.5,.2,.1) > ralpha<-qhyper(1-alpha,t,n-t,k) > data.frame(alpha,ralpha) alpha ralpha For example, we must capture al least 49 that were tagged i order to reject H at the α =.5 level. I this case the estimate for N is ˆN = [kt/r α ] = As aticipated, r α icreases ad the critical regios shriks as the value of α decreases. Usig the level r α determied usig the value N for N, we see that the power fuctio π(n) = P N {R r α }. R is a hypergeometric radom variable with mass fuctio ( t N t ) f R (r) = P N {R = r} = r)( k r ( N ). k The plot for the case α =.5 is give usig the R commads > N<-c(13:21) > <-1-phyper(49,t,N-t,k) > plot(n,,type="l",ylim=c(,1)) We ca icrease power by icreasig the size of k, the umber the value i the secod capture. This icreases the value of r α. For α =.5, we have the table. 229

5 N N Figure 2: Power fuctio for Licol-Peterso mark ad recapture test for populatio N = 2 ad t = 2 captured ad tagged. (left) k = 4 recaptured α =.5 (black),.2 (red), ad.1 (blue). Notice that lower sigificace level α reduces power. (right) α =.5, k = 4 (black), 6 (red), ad 8 (blue). Decreased sigificace level reduces power ad icreased recapture size icreases power. > N<-2 > ralpha<-qhyper(.95,t,n-t,k) > data.frame(k,ralpha) k ralpha We show the impact o power π(n) of both sigificace level α ad the umber i the recapture k i Figure 2. Exercise 4. Determie the type II error rate for N = 16 with k = 4 ad α =.5,.2, ad.1, ad α =.5 ad k = 4, 6, ad 8. Example 5. For a two-sided test H : µ = µ versus H 1 : µ µ. I this case, the parameter values for the ull hypothesis Θ ) cosist of a sigle value, µ. We reject H if X µ is too large. Agai, with level α, we have the critical regio Z = X µ σ/ z α/2. If µ is the actual mea, the X µ σ /. 23

6 mu mu Figure 3: Power fuctio for the two-sided test. µ = 1, σ = 3. (left) = 16, α =.5 (black),.2 (red), ad.1 (blue). Notice that lower sigificace level α reduces power. (right) α =.5, = 15 (black), 4 (red), ad 1 (blue). As before, decreased sigificace level reduces power ad icreased sample size icreases power. We use this fact to determie the power fuctio for this test { X µ } π(µ) = P µ {X C} = 1 P µ {X / C} = 1 P µ σ / < z α/2 { = 1 P µ z α/2 < X } { µ σ / < z α/2 = 1 P µ z α/2 µ µ σ / < X µ σ / < z α/2 µ µ } σ / ( = 1 Φ z α/2 µ µ ) ( σ / + Φ z α/2 µ µ ) σ / If we do ot kow if the mimic is larger or smaller that the model, the we use a two-sided test. Below is the R commads for the power fuctio with α =.5 ad = 16 observatios. > zalpha = qorm(.975) > mu<-1 > sigma<-3 > mu<-(6:14)/1 > <-16 > <-1-porm(zalpha-(mu-mu)/(sigma/sqrt()))+porm(-zalpha-(mu-mu)/(sigma/sqrt())) > plot(mu,,type="l") We shall see i the the ext toc how these tests follow from extesios of the likelihood ratio test for simple hypotheses. The ext example is ulikely to occur i ay geuie scietific situatio. It is icluded because it allows us to compute the power fuctio explicitly from the distributio of the test statistic. We begi with a exercise. Exercise 6. For X 1, X 2,..., X idepedet U(, ) radom variables, Θ = (, ). The desity { 1 f X (x ) = if < x, otherwise. 231

7 Let X () deote the maximum of X 1, X 2,..., X, the X () has distributio fuctio ( x ) F X() (x) = P {X () x} =. Example 7. For X 1, X 2,..., X idepedet U(, ) radom variables, take the ull hypothesis that lads i some ormal rage of values [ L, R ]. The alterative is that lies outside the ormal rage. H : L R versus H 1 : < L or > R. If ay of our observatios X i are greater tha R, the we are certai > R ad we should reject H. O the other had, all of the observatios could be below L ad the state of ature might still lad i the ormal rage. Cosequetly, we will try to base a test based o the statistic X () = max 1 i X i ad reject H if X () > R ad too much smaller tha L, say. We shall soo see that the choice of will deped o the umber of observatios ad o α, the size of the test. The power fuctio π() = P {X () } + P {X () R } We compute the power fuctio i three cases. The secod case has the values of uder the ull hypothesis. The first ad the third cases have the values for uder the alterative hypothesis. A example of the power fuctio is show i Figure theta Figure 4: Power fuctio for the test above with L = 1, R = 3, =.9, ad = 1. The size of the test is π(1) = Case 1.. I this case all of the observatios X i must be less tha which is i tur less tha. Thus, X () is certaily less tha ad ad therefore π() = 1. Case 2. < R. P {X () } = 1 ad P {X () R } = 232

8 P {X () } = ( ) ad P {X () R } = ad therefore π() = ( /). Case 3. > R. Repeat the argumet i Case 2 to coclude that P {X () } = ( ) ad that P {X () R } = 1 P {X () < R } = 1 ( ) R ad therefore π() = ( /) + 1 ( R /). The size of the test is the maximum value of the power fuctio uder the ull hypothesis. This is case 2. Here, the power fuctio ( ) π() = decreases as a fuctio of. Thus, its maximum value takes place at L ad α = π( L ) = ( ) L To achieve this level, we solve of ad take = L α. Note that icreases with α. Cosequetly, we must reduce the critical regio i order to reduce the sigificace level. Also, icreases with ad we ca reduce the critical regio while maitaiig sigificace if we icrease the sample size. 2 The p-value The report of reject the ull hypothesis does ot describe the stregth of the evidece because it fails to give us the sese of whether or ot a small chage i the values i the data could have resulted i a differet decisio. Cosequetly, oe commo method is ot to choose, i advace, a sigificace level α of the test ad the report reject or fail to reject, but rather to report the value of the test statistic ad to give all the values for α that would lead to the rejectio of H. The p-value is the probability of obtaiig a result at least as extreme as the oe that was actually observed, assumig that the ull hypothesis is true ad measures the stregth of evidece agaist H. Cosequetly, a very low p-value idicates strog evidece agaist the ull hypothesis. If the p-value is below a give sigificace level α, the we say that the result is statistically sigificat at the level α. For example, if the test is based o havig a test statistic S(X) exceed a level k, i.e., we have decisio reject H if ad oly if S(X) k. ad if the value S(X) = k is observed, the the p-value equals to the lowest value of sigificace level that wold result i rejectio of the ull hypothesis. max{π(); Θ } = max{p {S(X) k }; Θ }. 233

9 dorm(x, 1, 3/4) Figure 5: Uder the ull hypothesis, X has a ormal distributio mea µ = 1cm, stadard deviatio 3/ 16 = 3/4cm. The p-value,.77, is the area uder the desity curve to the left of the observed value of for x, The critical value, 8.767, for a α =.5 level test is idicated by the red lie. Because the p-vlaue is greater tha the sigificace level, we caot reject H. x For the oe-sided hypothesis test to see if the mimic had ivaded, H : µ µ versus H 1 : µ < µ. with µ = 1 cm, σ = 3 cm ad = 16 observatios. Our data had sample mea x = cm. The maximum value of the power fuctio π(µ) for µ i the subset of the parameter space determied by the ull hypothesis occurs for µ = µ. Cosequetly, the p-value is P µ { X }. With the parameter value µ = 1 cm, X has mea 1 cm ad stadard deviatio 3/ 16 = 3/4. We ca compute the p-value usig R]. > porm( ,1,3/4) [1] Thus, we caot reject H at the α =.5 sigificace level. Ideed, we ca reject H at ay level below the p-value. Example 8. Returig to the example o the proportio of hives that survive the witer, the appropriate composite hypothesis test to see if more that the usual ormal of hives survive is The R output shows a p-value of 3%. H : p.7 versus H 1 : p >.7. > prop.test(88,112,.7,alterative="greater") 1-sample proportios test with cotiuity correctio 234

10 data: 88 out of 112, ull probability.7 X-squared = 3.528, df = 1, p-value =.33 alterative hypothesis: true p is greater tha.7 95 percet cofidece iterval: sample estimates: p Exercise 9. Is the hypothesis test above sigfiicat at the 5% level? the 1% level? 3 Aswers to Selected Exercises 2. I this case the critical regios is C = {x; x k(µ )} for some value k(µ ). To fid this value, ote that ad k(µ ) = (σ / )z α + µ. The power fuctio P µ {Z z α } = P µ { X σ z α + µ } π(µ) = P µ { X σ z α + µ } = P µ { X µ σ z α (µ µ )} { X µ = P µ σ / z α µ µ } ( σ / = Φ z α µ µ ) σ /. 5. The type II error rate β is 1 π(16) = P 16 {R < r α }. This is the distributio fuctio of a hypergeometric radom variable ad thus these probabilities ca be computed usig the phyper commad For varyig sigificace, we have the R commads: > t<-2;n<-16 > k<-4 > alpha<-c(.5,.2,.1) > ralpha<-c(49,51,53) > beta<-phyper(ralpha,t,n-t,k) > data.frame(alpha,beta) alpha beta Notice that the type II error probability is high for α =.5 ad icreases as α decreases. For varyig recapture size, we cotiue with the R commads: > k<-c(4,6,8) > ralpha<-c(49,7,91) > beta<-phyper(ralpha,t,n-t,k) > data.frame(k,beta) k beta

11 Notice that icreasig recapture size has a sigificat impact o type II error probabilities. 6. The i-th observatio satisfies P {X i } = 1 dx = Now, X () occurs precisely whe all of the -idepedet observatios X i satisfy X i. Because these radom variables are idepedet, F X() = P {X () x} = P {X 1 x, X 1 x,..., X x} ( x ) ( x ) ( x ) ( x ) = P {X 1 x}p {X 1 x}, P {X x} = = 9. Yes, the p-value is below.5. No, the p-value is above