Introduction to Statistics inPsychologyPSY 201Professor Greg FrancisLecture 31PowerWill your research project work?HYPOTHESIS TESTINGsuppose I want to test whether apopulation mean is different from avalueH0: µ = 455Ha: µ != 455now further suppose that thepopulation mean really is different, sayit is µ = 465.2ERRORSState of natureDecision made H0true H0falseReject H0Type I error Correct decisionDo not reject H0Correct decision Type II errorWe have mostly focused on avoidingType I errors (small α)the probability of making a Type IIerror is βprobability of rejecting H0when it isfalse (correct decision) is called thepower of the test1 − β3FACTORS AFFECTINGPOWER1. the directional nature of Ha(one-tailed versus two-tailed test).2. The level of significance (α).3. The sample size (n).4. The effect size (ES).4DIRECTIONone-tailed tests are more powerful:Power = 0.33, β =0.67400 420 440 460 480 500 520Sample Mean00.010.020.030.04Power = 0.2244, β =0.7756400 420 440 460 480 500 520Sample Mean00.010.020.030.045LEVEL OF SIGNIFICANCEpower increases with αfor α =0.1, β =0.5319, Power = 0.4681400 420 440 460 480 500 520Sample Mean00.010.020.030.04for α =0.01, β =0.8708, Power = 0.1292400 420 440 460 480 500 520Sample Mean00.010.020.030.046SAMPLE SIZElarger samples result in smaller standard error,which increases powerfor n = 144, σX=8.33, β =0.67, Power =0.33400 420 440 460 480 500 520Sample Mean00.010.020.030.04for n = 576, σX=4.17, β =0.2266,Power = 0.7734400 420 440 460 480 500 520Sample Mean00.020.040.060.087EFFECT SIZEpower increases with effect sizefor Ha: µ = 465, (ES = 10), β =0.67,Power = 0.33400 420 440 460 480 500 520Sample Mean00.010.020.030.04for Ha: µ = 475, (ES = 20), β =0.2236,Power = 0.7764400 420 440 460 480 500 520Sample Mean00.010.020.030.048GENERAL APPROACHthe same basic approach works forother situationse.g.two-sample case for the difference ofmeansH0: µ1− µ2=0Ha: µ1− µ2!=0find tcvcan calculate power for an ES of 6Ha: µ1− µ2= −6find area under sampling distributionfor Hagreater than tcv(note, need agood table of t-scores)9EXAMPLECalculating power is useful because itlets you know whether your experimentis appropriate for what you wantSuppose you are a graduate studentinvestigating 5th graders with ADDyour PhD thesis is based on the ideathat a special remedial program willimprove the reading scores of ADDstudentsyou estimate that the remedialprogram might improve the scores byas much as 10 points10EXAMPLEYour field of study expects you will useα =0.05You have a pretty good estimate of thevariability of reading scores amongADD students (s = 30)You have a pretty good estimate of themean reading scores of ADD studentswithout the remedial program(X = 80).You go to the local schools to getstudents to participate in your researchproject . You get 7 students tovolunteer to participate in yourremedial programWill you be able to graduate?11HYPOTHESIS TESTyou will want to run a test like this:H0: µ = 80Ha: µ>80α =0.05, n =7What is the probability of being ableto reject H0if there really is an effect?powerthis is, in effect, the probability thatyour research project works wellenough for you to graduate12COMPUTE THE POWERFirst, find the critical value for thecorresponding hypothesis testwe assume that s = 30 is a pretty goodestimate of σ, the population standarddeviation of reading scores for ADD studentswe find the critical value for the hypothesistest is:zcv=1.645which, in terms of reading scores is:X = µ + zcvσXσX≈s√n=30√7= 11.33So,X = 80 + (1.645)(11.33) = 98.63So, you will reject H0if X>98.63.13COMPUTE THE POWERSecond, suppose the remedial testreally did work, and the mean readingscores of ADD students in the remedialprogram is actually 10 points higherHa: µ = 90then our sampling distribution will becentered on 90What is the probability that we’ll getX>98.63? (and thus reject H0)Convert to a z-scorez =98.63 − 9011.33=0.761Note, we use the same estimate of thestandard error as we used before,because we don’t have anything better14COMPUTE THE POWERPower is the probability that we’llreject the null hyp othesis if theremedial program really does work asyou exp ect it to.That’s the area under the curve of thesampling distribution (a normaldistribution), to the right of z =0.761We look that up in standard normaldistribution table to findpower = 0.2236which means, if the remedial programactually works, there’s a 22% chancethat you will be able to prove that itworks(in the sense that you reject the H0)15BITS AND PIECESWe treated the sampling distributionas a normal distributionbut with a sample size of n = 7, itwould really be a t-distributionoften this doesn’t make a big differencebecause things are pretty similar forthe normal distribution and thet-distributionyou could repeat the calculations anduse the t-distribution (but not withthe tables in the textbook)tables like this exist on the internet,and are easy to use16CONCLUSIONSfactors that effect powerwhy you should care17NEXT TIMEdesigning efficient experimentsstandardized effect sizeHow to spend your money