Chapter 11: Asking and Answering Questions About the Difference of Two Proportions

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1 Chapter 11: Askig ad Aswerig Questios About the Differece of Two Proportios These otes reflect material from our text, Statistics, Learig from Data, First Editio, by Roxy Peck, published by CENGAGE Learig, Oe proportio At oe poit, we studied a aciet coi recovered from a pirate s treasure chest, ad we wished to kow if the coi was fair or ot, that is, if it were flipped a large umber of times would the proportio of resultig heads be very close to 0.5. I this case, the relevat experimet is to flip the coi oce, ad the respose variable is categorical, the outcome is either a head or a tail. The evidece to decide the issue will be the outcomes from a certai umber of idepedet experimets. There are two ways to aalyze this evidece, by costructig a cofidece iterval or by performig a hypothesis test. The first step i both types of aalysis is to establish some defiitios ad otatio for expressig our ideas clearly. I this case, the populatio proportio, p, is the log ru probability of obtaiig a head o flippig the aciet coi. The sample proportio, ˆp, is the actual proportio of heads we observe i our sample of flips. Now, ˆp is a statistic, it varies from oe sample to the ext, so it has a samplig distributio. The stadard deviatio of that samplig distributio is called its stadard error, SE. For a proportio, the stadard error is give by SE = p(1 p)/ whe p is kow, ad SE = ˆp(1 ˆp)/ whe p is ukow. The Cetral Limit Theorem, CLT, applied to proportios shows that ˆp N(mea = p, SE = p(1 p)/) provided that certai coditios are satisfied : (1) the sample observatios are idepedet, ad (2) the sample size is sufficietly large, which is determied by the success-failure coditios (p 10 ad (1 p) 10). A cofidece iterval for a proportio ow takes the form ˆp(1 ˆp) ˆp ± z where the multiplier, z, depeds o the desired cofidece level. For istace, for sigificace level α = 0.05 ad 95% cofidece we would use z = 1.96, ad for a more geeral (1 α) level of cofidece we would calculate z with the R commad qorm(1 alpha/2). A hypothesis test for a proportio posits the ull hypothesis H 0 : p = p 0, where p 0 = 0.5 i this case because of the ull assumptio that this is a fair coi, agaist a alterative hypothesis, which may take oe of three forms: The test statistic is H 0 : p p 0, H 0 : p > p 0, H 0 : p < p 0. z = (ˆp p 0 )/SE where the formula for the stadard error takes ito accout the presumptio of the ull hypothesis, sice p 0 is the ull value. SE = p 0 (1 p 0 )/ Sprig 2016 Page 1 of 8

2 We decide betwee the ull ad alterative hypotheses by calculatig the probability, called the p value, of obtaiig the test statistic that we actually did obtai from our evidece uder the coditios specified by the ull hypothesis. We summarize the above aalyses i the followig iferece table for a sigle proportio. Cofidece Iterval Test Statistic (poit estimate - ull value) / SE proportio ˆp ± z ˆp(1 ˆp) p (ˆp p 0 )/ 0(1 p 0) Two proportios Suppose that we ow focus o two garly cois from the pirate s treasure. Is oe of them more likely tha the other to preset a head whe flipped, or do both cois share the same probability of comig up heads? This time we oce agai have a biary respose variable as a outcome, heads or tails, but the data come from two sources, the two cois. We have a biary explaatory variable with levels coi 1 ad coi 2. Let p 1 be the probability that coi 1 comes up heads, ad let p 2 be the probability that coi 2 comes up heads. The questio becomes, is p 1 = p 2? We ca ivestigate this questio with a hypothesis test whose ull hypothesis is H 0 : p 1 = p 2, or with a cofidece iterval which would put plausible bouds o the differece p 1 p 2. I both cases, the basic experimet is to flip each coi a certai umber of times ad record the outcome of each flip, head or tail. Let 1 be the umber of flips ad ˆp 1 the proportio of heads for coi 1 ad let 2 be the umber of flips ad ˆp 2 the proportio of heads for coi 2. The cofidece iterval for the differece of two proportios takes the form where the poit estimate is ˆp 1 ˆp 2, the multiplier is z as before, ad the stadard error of our ew statistic is p 1 (1 p 1 ) SE = + p 2(1 p 2 ) Notice that this formula for SE takes ito accout the variability of each coi. Therefore, the formula for the cofidece iterval for the differece of two proportios takes the form (ˆp 1 ˆp 2 ) ± z p 1 (1 p 1 ) + p 2(1 p 2 ) For a hypothesis test, the test statistic takes the form z = (poit estimate ull value)/se, where the ull value ad the SE are calculated assumig that the ull hypothesis is true. Sprig 2016 Page 2 of 8

3 The stadard error for the differece of two proportios for cofidece itervals ad for ull hypotheses of the form H 0 : p 1 p 2 0 is p 1 (1 p 1 ) SE = + p 2(1 p 2 ) is where The stadard error for the differece of two proportios for ull hypotheses of the form H 0 : p 1 p 2 = 0 ˆp pool (1 ˆp pool ) SE = + ˆp pool(1 ˆp pool ) ˆp pool = umber of successes i both groups combied All of the above cases for a categorical respose variable are summarized i the followig table. Cofidece Iterval Test Statistic (poit estimate - ull value) / SE proportio ˆp ± z ˆp(1 ˆp) (ˆp p 0 )/ differece i proportios (ˆp 1 ˆp 2 ) ± z ˆp 1(1 ˆp 1) 1 + ˆp2(1 ˆp2) ( (ˆp1 2 ˆp 2 ) 0 ) / p 0(1 p 0) ˆp(1 ˆp) 1 + ˆp(1 ˆp) 2 For hypothesis tests ivolvig a differece i proportios, the test statistic may make use of the pooled proportio, ˆp, which pools the total umber of successes ad the total umber of observatios i the two samples: p pooled = # successes # observatios. For sigificace level α, for istace α = 0.05 for a 95% cofidece level, the multiplier z ca be calculated i R with alpha < z.star <- qorm(1 - alpha/2) May proportios How ca oe compare more tha two cois? Do they all share the same probability of producig a head, or is at least oe of the cois differet? A efficiet approach to the problem of multiple comparisos goes back to the tur of the cetury with Karl Pearso s discovery i 1900 of the chi-squared test, oe of the earliest statistical tests to be clearly explaied. We will study the chi-square test for goodess of fit ad the chi-square test for idepedece i a later chapter. Sprig 2016 Page 3 of 8

4 Differece of Two Proportios Hadedess Wikipedia uses a hypothetical radom sample of left- ad right-haded people to illustrate its article o cotigecy tables. right-haded left-haded total males females totals A first questio suggested by such a table is How may people are left-haded? The totals alog the bottom row idicate that, i this sample at least, about 13% of the (hypothetical) populatio that this sample was draw from is left-haded. The table also presets data o geder, so aother questio arises, Are males ad females equally likely to be left-haded? The mosaic plot o the left illustrates the distributio of hadedess i the sample, ad the plot o the right illustrates the proportios we would expect if hadedess were idepedet of geder. There is a differece, but is the differece i proportios simply due to radom samplig from male ad female populatios that are equally likely to be left-haded, or is the differece so great that we ca take it as evidece that males ad females actually differ i their likelihood to be left-haded? Hadedess i the Sample Expected Values if Hadedess Is Idepedet of Geder m f m f left left Haded right Haded right Geder Geder A hypothesis test of the differece of two proportios addresses that questio directly. Let p 1 be the proportio of left-haded me, ad let p 2 be the proportio of left-haded wome. The hypotheses to be tested are H 0 : p 1 p 2 = 0, H a : p 1 p 2 0. The poit estimate of the differece of proportios is ˆp 1 ˆp 2 = , the pooled proportio of left-haded people is p pooled = 0.13, ad the stadard error of this statistic is ( 1 SE = p pooled (1 p pooled ) + 1 ) = , so the test statistic is z = (ˆp 1 ˆp 2 )/SE = Sprig 2016 Page 4 of 8

5 The test statistic z has a stadard ormal distributio, ad the probability of obtaiig a value this large or larger is p = Hypothesis Test (Hadedess) p = z = If males ad females are equally likely to be left-haded, we might expect to see a differece i the proportios of left- ad right-haded males ad females as large or larger tha we observed i about 18% of such samples. That is ot very much evidece agaist the ull hypothesis, so we fail to reject the ull hypothesis ad coclude that, i the light of this case at least, males ad females are equally likely to be left-haded. R Code # HT Here is supportig R code for the hypothesis test. x1 <- 9; 1 <- 52; x2 <- 4; 2 <- 48 p1.hat <- x1 / 1 p2.hat <- x2 / 2 p.pooled <- (x1 + x2) / (1 + 2) # 0.13 se <- sqrt(p.pooled * (1 - p.pooled) * (1 / / 2)) # z <- (p1.hat - p2.hat) / se # p.value <- 2 * (1 - porm(z)) # Sprig 2016 Page 5 of 8

6 Differece of Two Proportios Smokig The ad Now Here is aother hypothetical example. Suppose you come across a magazie article reportig o the icidece of smokig i Sewaee studets some years ago, say i the year Accordig to this article, a radom sample of Sewaee studets was asked Have you smoked at least oe cigarette i the last week? ad the article reports the umber of studets who respoded Yes. Hmm, you thik. I woder how that would compare with Sewaee studets today? So, of course, the atural thig to do is to take aother radom sample of today s Sewaee studets ad ask the same questio. First, let s defie some variables. Let p 1 deote the proportio of Sewaee studets who smoked i 1890, ad let p 2 deote the proportio of Sewaee studets who smoke ow. Here, whether a studet is cosidered to be a smoker or ot is determied by his aswer to that same questio. There are two differet ways to orgaize this study: calculate a cofidece iterval for the differece of populatio proportios p 2 p 1 to quatify how the proportio of smokers might have chaged, or create a hypothesis test to test whether these two populatio proportios are the same or ot. Let s create the cofidece iterval first. We record the results of the surveys symbolically: ˆp 1 is the proportio of the 1 studets reportig Yes i 1890, ad ˆp 2 is the proportio of the 2 studets reportig Yes this year. Our poit estimate for the differece i the populatio proportios is ˆp 2 ˆp 1, ad the stadard error of this statistic is ˆp 1 (1 ˆp 1 ) SE = + ˆp 2(1 ˆp 2 ), so the desired 95% cofidece iterval is (ˆp2 ˆp 1 ) ± 1.96 SE. A hypothesis test framig this questio might take the form H 0 : p 1 = p 2, H a : p 1 p 2. I effect, the ull hypothesis claims that the differece i proportios is 0, ad our poit estimate of this differece is ˆp 2 ˆp 1. But ow the stadard error of this statistic must take ito accout the ull hypothesis which claims that there is o differece i these two proportios. Therefore, we create a ew proportio by addig together all of the smokers from both samplig years, ad dividig by the total umber of sampled studets: # of smokers p pooled = # of studets. The stadard error of our poit estimate is ow ( 1 SE = p pooled (1 p pooled ) + 1 ), the z-statistic is z = (ˆp 2 ˆp 1 )/SE, ad the correspodig p-value is calculated by R as 2 * (1 - porm(z)), sice the alterative hypothesis is two-sided. Which approach seems to shed more light o the situatio, the cofidece iterval or the hypothesis test? Sprig 2016 Page 6 of 8

7 R Code Let s make up some totally fictitious data to illustrate how to tur the statistical discussio i the previous sectio o Smokig The ad Now ito umerical results. Assume that i the sample from 1890, a total of 120 Sewaee studets were iterviewed ad 23 idetified themselves as smokers. For the moder sample, assume that 142 studets were iterviewed ad 17 idetified themselves as smokers. The R fuctio prop.test is a sophisticated ad authoritative implemetatio of the above procedures. Its default is to use a cotiuity correctio for more accuracy, but if we request that the cotiuity correctio ot be used, by settig correct=false, the the results will match those that are outlied above. prop.test(x=c(23, 17), =c(120, 142), alterative="two.sided", cof.level=0.95) # 2-sample test for equality of proportios with cotiuity correctio # data: c(23, 17) out of c(120, 142) # X-squared = , df = 1, p-value = # alterative hypothesis: two.sided # 95 percet cofidece iterval: # # sample estimates: # prop 1 prop 2 # Outlie Outlie for oe-variable iferece for sample data (Peck, chapters 7 13): The goal is to geeralize from a sample to lear about a populatio categorical variable - oe proportio - cofidece iterval - oe-sample z CI for a proportio - hypothesis testig - oe-sample z HT for a proportio - differece of two proportios - cofidece iterval - two-sample z CI for a differece i proportios - hypothesis testig - two-sample z HT for a differece i proportios cotiuous variable - oe mea - cofidece iterval - oe-sample t CI for a mea - hypothesis testig - oe-sample t HT for a mea - differece of paired meas - cofidece iterval - paired t CI for a differece i meas - hypothesis testig - paired t HT for a differece i meas - differece of idepedet meas - cofidece iterval - two-sample t CI for a differece i meas - hypothesis testig - two-sample t HT for a differece i meas Sprig 2016 Page 7 of 8

8 Exercises We will attempt to solve some of the followig exercises as a commuity project i class today. Fiish these solutios as homework exercises, write them up carefully ad clearly, ad had them i at the begiig of class ext Friday. Homework 11a two proportios Exercises from Chapter 11: 11.1 (poverty), 11.5 (subur), 11.7 (distractig), (TV), (hours) Homework 11b two proportios Exercises from Chapter 11: (reservatios), (support), (support), (textig), (music) Sprig 2016 Page 8 of 8

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