advertisement

Comparing two proportions Example: Chinook Salmon Ho : Sex ratio is the same for the two capture methods during the early run Test: H0 : 1 = 2 vs. HA : 1 6= 2 Data: A 2 2 contingency table Dene 1 = Probability of catching a female with hook and line during the early run 1 = Probability of catching a female with a net during the early run Females Hook and Line Net Y1 = 172 Y2 = 165 Males n1 Y1 = 119 n2 Y2 = 202 Totals n1 = 291 n2 = 367 124 Estimated proportions: Estimated variance: Y 172 = 0:5911 p1 = 1 = n1 291 p2 = Y2 165 = = 0:4496 n2 367 Sp21 p2 = p1(1 p1) p2(1 p2) + n1 n2 = (:5911)(:4089) + (:4496)(:5504) 291 367 = :0015 Is the dierence between p1 = 0:5911 and p2 = 0:4496 greater than what can be attributed to sampling variation? V ar(p1 p2) = V ar(p1) + V ar(p2) 2Cov(p1; p2) = 1(1 1) + 2(1 2) n1 125 n2 126 Standard Error: p Sp1 p2 = :0015 = :0388 127 Test: p1 p2 Sp1 p2 = :5911 :4496 :0388 = 3:65 95% condence interval for 1 2 Z = (p1 p2) z=2Sp1 p2 % .5911 ) " .4496 (separate variance estimates) " z:025 = 1:96 .0388 0:1415 :0760 ) (:064; :218) p-value = 2(:00013) = :00026 Conclusion: Conclusion: 128 Test: (pooled variance estimates) First compute a pooled estimate of the success probability n p + n2p2 = Y1 + Y2 p= 1 1 n1 + n2 n1 + n + 2 = 172 + 165 = :51216 291 + 367 Estimate V ar(p1 p2) = V ar(p1) + V ar(p2) as = p(1 p) + p(1 p) = :00154 S2 p1 p2 Then n1 n2 129 and p1 p2 Sp1 p2 = :5911 :4496 :0392 = 3:61 Z = p-value = 2(:000156) = :00031 Note: Z 2 is equal to the Pearson chi-square statistic . p Sp1 p2 = :00154 = :03924 130 2-sample test for equality of proportions without continuity correction S-PLUS for windows: data: V1 and V2 from data set salmon2 X-square = 13.0018, df = 1, p-value = 0.0003 alternative hypothesis: two.sided 95 percent confidence interval: 0.06544146 0.21750656 sample estimates: prop'n in Group 1 prop'n in Group 2 0.5910653 0.4495913 Create a spreadsheet V1 V2 1 172 291 2-sample test for equality of proportions with continuity correction 2 165 367 Statistics ) Compare samples ) Counts and proportions ) proportion parameters data: V1 and V2 from data set salmon2 X-square = 12.4417, df = 1, p-value = 0.0004 alternative hypothesis: two.sided 95 percent confidence interval: 0.06236085 0.22058717 sample estimates: prop'n in Group 1 prop'n in Group 2 0.5910653 0.4495913 Fill in the boxes and check o Yates continuity correction 131 S-PLUS code # # # # This code compares proportions for two independent binomial samples It is stored in the file binomial2.ssc cdat<-read.table("c:/courses/st557/sas/hdata.dat", sep=c(1, 5, 7, 9, 10, 11, 12, 14, 15 ), col.names=c("year","month","day","biweek","run", "gear","age", "sex", "length" )) # Construct a 2x2 contingency table # for the sex by gear categories # for the two runs 132 # Construct vector of total counts # (all fish caught for the two gear # types in the early run) ny <- apply(ytab[,,1],2,sum) # # # # # # Apply the prop.test function to test the null hypothesis of equal proportions. The alt="two.sided" option specifies a two-sided alternative. The correct="F" option turns off the continuity correction prop.test(y,ny,alt="two.sided",correct="F") ytab<-table(cdat$sex,cdat$gear,cdat$run) # Repeat the test with continuity correction prop.test(y,ny,alt="two.sided",correct="T") # Construct vector of success counts # (female counts for the two gear types # in the early run) # Make a 99% confidence interval prop.test(y,ny,conf.level=.99)$conf.int y <- ytab[1,,1] 133 134 SAS code /* Program to analyze the 1999 Chinook salmon data. This program is stored in the file chinook2.sas */ data set1; infile 'c:\courses\st557\sas\hdata.dat'; input (year month day biweek run gear age sexa length) (4. 2. 2. 1. 1. 1. 2. $1. 4.); rage=int(age/10); oage=age-(10*rage); if(sexa = 'F') then sex=1; else sex=2; run; /* Attach labels to categories */ proc format; value run 1 = 'Early' 2 = 'Late'; value sex 1 = 'Female' 2 = 'Male'; value gear 1 = 'Hook' 2 = 'Net'; run; /* Examine partial association between sex and method of capture within each run. */ proc sort data=set1; by run; run; proc freq data=set1; by run; table gear*sex / chisq Fisher nopercent nocol expected; format sex sex. gear gear. run run.; run; 136 135 Frequency Expected Row Pct Female Male Sample Size Determination Hook 172 119 149.04 141.96 59.11 40.89 291 Net 165 202 187.96 179.04 44.96 55.04 367 Total 337 (n responses in each group) Total 321 95% condence interval for 1 2 p1 p 2 658 ] % (p1 j Statistic DF Value Prob Chi-Square 1 13.0018 0.0003 Likelihood Ratio Chi-Square 1 13.0545 0.0003 Continuity Adj. Chi-Square 1 12.4417 0.0004 p2) + z=2Sp1 p2 ] " L = half length Sample size needed for each group z=2 #2 n= [1(1 1) + 2(1 2)] L " Fisher's Exact Test Table Probability (P) Two-sided Pr <= P j % [ 9.352E-05 4.022E-04 137 138 Sample Size Determination Two sided alternative: (n responses in each group) Ha : 1 6= 2 Test of the null hypothesis Compute Ho : 1 2 p = Signicance level ( = :05) R = Power (power = 1 = :80) 1 2 v u u t 2 2p(1 p) 1(1 1) + 2(1 2) Sample size needed for each group 2 z + Rz=2 [1(1 1) + 2(1 2)] n= (1 2)2 Size of alternative j1 2j = 0:08 140 139 One sided alternative: Ha : 1 > 2 or Ha : 1 < 2 Correction given by Fleiss, J. L. (1981), Statisitcal Methods for Rates and Proportions, Wiley, New York. (pages 42-43 and Table A3) Compute p = R = 1 2 v u u t 2 2p(1 p) 1(1 1) + 2(1 2) Sample size needed for each group z + Rz 2 [1(1 1) + 2(1 2)] n= (1 2)2 141 ncorrected = 2 n4 4 v u u t 1+ 1+ 4 njp1 p2j 32 5 142 /* This program computes sample sizes needed to obtain tests with a specified power values for detecting specified differences between two proportions. This program is stored in the file size2p.sas */ * Enter a set of power values; power = {.8 .9 .95 .99}; * Enter the type I error level; alpha = .05; options nodate nonumber linesize=68; * Compute sample sizes; proc iml; start samples; za = probit(1-alpha/2); za1 = probit(1-alpha); nb = ncol(power); np = ncol(power); size = j(1,np); size1 = j(1,np); /* Enter the probability of success for the first population */ p1 = .58; /* Enter the probability of success for the second population */ p = (p1+p2)/2; rp = sqrt(2*p*(1-p)/ (p1*(1-p1)+p2*(1-p2))); p2 = .50; 144 143 * Cycle across power levels; do i2 = 1 to np; zb = probit(power[1,i2]); size[1,i2] = ((za*rp+zb)**2)* (p1*(1-p1)+p2*(1-p2))/((p1-p2)**2); size1[1,i2] = ((za1*rp+zb)**2)* (p1*(1-p1)+p2*(1-p2))/((p1-p2)**2); end; /* Compute corrected sample sizes using the Fleiss (1981) correction */ sizec=(size/4)#(j(1,np)+sqrt(j(1,np) +4/(size*abs(p1-p2))))##2; print,,,,, 'Sample sizes for testing equality', ' of two proportions'; print, 'The number of observations needed', ' for each treatment group'; print,,, p1 p2 alpha power; sizer = int(size) + j(1,np); print 'Sample sizes (2-sided test):' sizer; sizerc = int(sizec) + j(1,np); print 'Corrected sizes (2-sided test):' sizerc; size1c=(size1/4)#(j(1,np)+sqrt(j(1,np) +4/(size1*abs(p1-p2))))##2; size1rc = int(size1c) + j(1,np); print 'Corrected sizes (1-sided test):' size1rc; size1r = int(size1) + j(1,np); print 'Sample sizes (1-sided test):' size1r; 145 146 /* Determine sample sizes for constructing confidence intervals */ /* Enter confidence level */ level=0.95; percent = level*100; print,,,,'Sample sizes for' percent 'percent', ' confidence intervals'; print,,'The number of observations for each', ' treatment group'; print,,,' Half length Sample size'; print me ' ' n; /* Enter values for margin of error */ me = {0.04, 0.03, 0.02, 0.01}; alpha = 1-level; za = probit(1-alpha/2); np = ncol(me); size = j(1,np); finish; run samples; /* Compute sample sizes */ n = ((za/me)##2)#(p1*(1-p1)+p2*(1-p2)); n = int(n) + j(np,1); percent = level*100; 148 147 Sample sizes for testing equality of two proportions S-PLUS code: The number of observations needed for each treatment group P1 P2 ALPHA 0.58 0.5 0.05 Sample sizes(2-sided test): Sample sizes(1-sided test): Corrected sizes (2-sided ): Corrected sizes (1-sided ): POWER 0.8 609 479 633 504 0.9 0.95 0.99 814 1006 1422 663 838 1220 839 1031 1447 688 863 1245 Sample sizes for 95 percent confidence intervals The number of observations for each treatment group Half length ME 0.04 0.03 0.02 0.01 Sample size N 1186 2107 4741 18962 149 # # # # # # # # This program computes sample sizes needed to obtain a test of a hypothesis about a difference between two proportion with a specified power value. It also computes the number of observations needed to obtain a confidence interval with a specified accuracy. This program is stored in the file size2p.ssc # Specify the probability of success # for the first population p1 <- c(.58) # Specify the probability of success # for the second population p2 <- c(.50) # Enter power values power <- c(.8, .9, .95, .99) 150 # Increase sample size to next largest integer # and print results # Enter the type I error level alpha <- .05 size <- ceiling(size) cat("\n \n \n p1=",p1,"p2=", p2, "alpha=", alpha,"power=", power) cat("\n Sample sizes (2-sided test): " , size) za <- qnorm(1-alpha/2) za1 <- qnorm(1-alpha) nb <- length(power) p <- (p1+p2)/2 va <- p1*(1-p1)+p2*(1-p2) rp <- sqrt(2*p*(1-p)/va) cat("Sample sizes for testing equality of two proportions") # Obtain a sample size for each of # the requested power values zb <- qnorm(power) size <- ((za*rp+zb)^2)*va/((p1-p2)^2) size1 <- ((za1*rp+zb)^2)*va/((p1-p2)^2) size1 <- ceiling(size1) cat("\n \n p1=",p1,"p2=", p2, "alpha=", alpha,"power=", power) cat("\n Sample sizes (1-sided test): " , size1) # Compute corrected sample sizes according to # Fleiss, J. L. (1981) Statistical Methods for # Rates and Proportions, Wiley, New York. sizec<-(size/4)*(1+sqrt(1+4/(size*abs(p1-p2))))^2 sizecr <- ceiling(sizec) cat("\n \n \n p1=",p1,"p2=", p2, "alpha=", alpha,"power=", power) cat("\n Corrected sizes(2-sided test): ",sizecr) 151 s1c<-(size1/4)*(1+sqrt(1+4/(size1*abs(p1-p2))))^2 s1cr <- ceiling(s1c) cat("\n \n \n p1=",p1,"p2=", p2, "alpha=", alpha,"power=", power) cat("\n Corrected sizes(1-sided test): ",s1cr) # # # # Compute sample size needed to obtain a confidence interval for the difference between the two proportions with a specified margin of error # Enter the confidence level level <- c(0.95) 152 # Compute needed sample sizes alpha2 <- (1.0-level)/2 np <- length(me) one <- rep(1,np) n <-((qnorm(one-alpha2)/me)^2)*va n <- ceiling(n) percent <- level*100; cat("\n \n \n Sample sizes for ", percent, " percent confidence intervals: \n") cat("margin of error: ", me, "\n") cat("sample size: ", n, "\n") # Enter desired margin of error me <- c(0.04, .02, .01) 153 154 Sample sizes for testing equality of two proportions p1= 0.58 p2= 0.5 alpha= 0.05 power= 0.8 0.9 0.95 0.99 Sample sizes (2-sided test): 609 814 1006 1422 p1= 0.58 p2= 0.5 alpha= 0.05 power= 0.8 0.9 0.95 0.99 Sample sizes (1-sided test): 479 663 838 1220 p1= 0.58 p2= 0.5 alpha= 0.05 power= 0.8 0.9 0.95 0.99 Corrected sizes(2-sided test): 634 839 1031 1447 Sample sizes for 95 percent confidence intervals: margin of error: 0.04 0.02 0.01 sample size: 1186 4741 18962 p1= 0.58 p2= 0.5 alpha= 0.05 power= 0.8 0.9 0.95 0.99 Corrected sizes(1-sided test): 504 688 863 1245 156 155 S-PLUS for windows Statistics ) Power and Sample size ) Binomial Proportion ) change Sample Type to and specify values of two proportions Two Sample *** Power Table *** p1 p2 delta alpha 1 0.5 0.58 0.08 0.05 2 0.5 0.58 0.08 0.05 3 0.5 0.58 0.08 0.05 4 0.5 0.58 0.08 0.05 5 0.5 0.58 0.08 0.05 6 0.5 0.58 0.08 0.05 power 0.800 0.850 0.900 0.950 0.975 0.990 n1 634 721 839 1031 1214 1447 n2 634 721 839 1031 1214 1447 Comparing Several Proportions: Several independent binomial experiments (or samples) Y1 = number of successful outcomes in n1 trials for experiment 1 Bin(n1; 1) Y2 = number of successful outcomes in n2 trials for experiment 2 Bin(n2; 2) .. .. Yk = number of successful outcomes in nk trials for experiment k Bin(nk; k) 157 158 Example: Test: Sunower seeds were stored in three types of storage facilities. 100 seeds were taken from each of the three facilities and the germination of each seed was monitored. Dene H0 : 1 = 2 = = k vs. HA : i 6= j for some i 6= j Note: H0 : 1 = 2 = = k is referred to as the hypothesis of homogeneity or independence i = proportion of seeds from the i-th storage facility that will germinate % the "success rate" i does not depend on the experiment (or population) Are there dierences in germination rates? 159 160 A 3 2 contingency table Storage Number of Germination Facility Seeds Tested Rates Type 1 100 69% Type 2 100 58% Type 3 100 41% Total 300 56% Storage Number that Facility Germinated Number that failed to Germinate Totals Type 1 Y11 = 69 Y12 = 31 n1 = 100 Type 2 Y21 = 58 Y22 = 42 n2 = 100 Type 3 Y31 = 41 Y32 = 59 n3 = 100 Totals Y+1 = 168 Y+2 = 132 161 Y+j \Expected counts" m ^ ij = YiY+++ Pearson statistic: m ^ 11 = 56 m ^ 12 = 44 m ^ 21 = 56 m ^ 22 = 44 m ^ 31 = 56 m ^ 32 = 44 Here 0 1 of seeds A m ^ i1 = @ number of type i 0 (Yij m ^ ij )2 m ^ ij i=1 j =1 I X X2 = 1 A overall germination rate @ J X 2 2 = (69 56) + (31 44) 56 44 2 2 + (58 56) + (42 44) 56 44 2 2 + (41 56) + (59 44) 56 44 = 16:15 163 162 Deviance Reject H0 : 1 = 2 = 3 if G2 = 2 J X i=1 j =1 Yij log(Yij =m ^ ij ) = 2 69 log(69=56) + 31 log(31=44) +58 log(58=56) + 42 log(42=44) Note that d:f: I X 2 6 4 X 2 > 22; 22;:05 22;:005 (log-likelihood ratio test) 3 7 5 +41 log(41=56) + 59 log(59=44) = 5:99 = 10:60 = 20:12 = (number of rows 1) (of columns 1) 164 Reject H0 : 1 = 2 = 3 if G2 > 22; X 2 and G2 have the same limiting cen- tral chi-squared distribution when the null hypthesis is true. 165 The joint probability function (or likelihood function) is Maximum likelihood estimation: Y11 Bin(n1; 1) Y21 Bin(n2; 2) Y31 Bin(n3; 3) Pr (Y11 = y11; Y21 = y21; Y31 = y31) 0 1 = @ yn1 A 1y11 (1 1)n1 y11 11 are independent binomial random variables. 3 Y 0 @ 0 @ 1 n2 A y21(1 )n2 y21 2 y21 2 1 n3 A y31 n3 y31 y31 3 (1 3) 20 1 3 = [email protected] yni A iyi1(1 i)ni yi15 i1 i=1 = L(1; 2; 3; y11; y21; y31) 166 167 Since the natural logarithm is a monotone increasing function, the value that maximizes L(; data) is obtained by maximizing Consider the hypothesized model where 1 = 2 = 3 = ; 01 `(; data) = log[L(; data)] = Then the likelihood function becomes L(; data) = 3 Y i=1 2 4 3 ni! yi1(1 )ni yi15 yi1!(ni yi1)! 168 3 X [log(ni!) log(yi1!) i=1 log((ni yi1)!)] + log() 3 X i=1 + log(1 ) yi1 3 X i=1 (ni yi1) 169 Solve the likelihood equation 0 = @`( ; data) @ 3 X = i=1 yi1 Maximum likelihood estimators of mean (or expected) counts: 3 X ni yi1 i=1 1 The solution is the maximum likelihood estimate for for the \independence" or \homogeneity" model. 3 X ^ = i=1 yi1 3 X i=1 ni = y+1 y++ For this model: Yi1 Bin(ni; ) and mi1 = E (Yi1) = ni Then E (Yi2) = E (ni Yi1) = ni(1 ) m ^ i1 = ni^ = Yi+Y+1 Y++ m ^ i2 = ni(1 ^) = Yi+Y+2 Y++ Formula for \expected counts 0 10 1 total for total for @ [email protected] A i th row j th column 0 1 m ^ ij = total for the @ A entire table The formula for the \expected" counts corresponds to the maximum likelihood estimators for the means of the counts for the specied model. 171 170 Maximum likelihood estimates for the alternative model (model A): Yi1 Bin(ni; i) 0 i 1 i = 1; 2; 3 0 = @` @i = yi1 ni yi1 ; i = 1; 2; 3 i 1 i log-likelihood function `(1; 2; 3; data) = 3 X i=1 flog(ni!) log(y1i!) log((ni y1i)!)g + + 3 X i=1 3 X i=1 Likelihood equations: yi1 log(i) Solutions: y ^i = i1 = pi ; i = 1; 2; 3 ni (ni yi1) log(1 i) 172 173 Maximum likelihood estimates of expected counts for the alternative model m ^ i1 = ni^i = 1 0 yi 1 A @ ni ni m ^ i2 = ni(1 ^i) = 1 yi1 A ni nested models; Model A: Yi1 Bin(ni; i) i = 1; 2; 3 are independent random counts. = yi1 0 n ni @ i Likelihood ratio tests for comparing = yi2 Maximum likelihood estimators are Y11 n1 Y ^2 = p2 = 21 n2 Y ^3 = p3 = 31 n3 ^1 = p1 = are the observed counts. This is called a \saturated" model because it contains as many parameters (1; 2; 3) as independent counts (Y11; Y21; Y31). 175 174 Ratio of Likelihoods: ^ = Model B: Model A with the added restrictions that 1 = 2 = 3 = 3 X ^ = i=1 Yi1 3 X i=1 ni Lmodel B(^; data) Lmodel A(^1; ^2; ^3; data) ni! ^yi1(1 ^)ni yi1 y !( n y )! i 1 i i 1 i =1 = Y3 ni ! ^iyi1(1 ^i)ni yi1 i=1 yi1!(ni yi1)! 3 Y = 3 Y i=1 2 3y 2 i1 4 4 5 ^ ^i 3 1 ^ 5ni yi2 1 ^i Note that 0 ^ 1 and small values of ^ suggest that model A is substantially better than model B. 176 177 2 3 of parameter 5 d.f. = 4 dimension space for model A 2 3 dimension of parameter 4 5 space for model B When the null hypothesis (model B) is correct and each ni is \large", then = 3 1= 2 and in this case 2 log(^ ) 2d:f : 2 log(^ ) = 2 8 3 <X : ^i ! yi1 log ^ i=1 9 1 ^i != + yi2 log 1 ^ ; i=1 3 X =2 3 X 2 X i=1 j =1 yij 1 0 m A;ij A log @ ^ m ^ B;ij 179 178 \Expected" counts: Model A: m ^ A;i1 = ni^i = yi1 m ^ B;i2 = ni(1 ^i) = (ni yi1) = yi2 Model B: m ^ B;i1 = ni^ = m ^ i1 m ^ B;i1 = ni(1 ^) = m ^ i2 Then 0 1 y yij log @ ij A m ^ ij i j = G2: 2 log(^ ) = 2 X X 180 The general form for the Pearson chi-squared statistic is X X (m ^ A;ij m ^ B;ij )2 X2 = : m ^ B;ij i j When model A (the alternative hypothesis) places no restrictions on the parameters it is called the general alternative and m ^ A;ij = Yij : Simplifying the notation, m ^ ij = m ^ B;ij; the Pearson statistic becomes X X (Yij m ^ ij )2 X2 = m ^ ij i j 181