n=300 30% taking A relief from pain 23% taking B relief from pain Question; If there is no difference are we likely to get a 7% difference?
Hypothesis H0: p1-p2=0 H1: p1-p2!=0 (not equal to) 1>Weighed average of two sample proportion 300(0.30)+300(0.23) ------------------- = 0.265 300+300 2>Std Error estimate of the difference between two independent proportions sqrt((0.265 *0.735)*((1/300)+(1/300))) = 0.03603 3>Evaluation of the difference between sample proportion as a deviation from the hypothesized difference of zero ((0.30-0.23)-(0))/0.03603 = 1.94 z did not approach 1.96 hence H0 is not rejected. This is what I was trying to do using prop.test. prop.test(c(30,23),c(300,300)) What function should I use? -----Original Message----- From: [EMAIL PROTECTED] on behalf of Nordlund, Dan (DSHS/RDA) Sent: Thu 8/9/2007 1:26 PM To: r-help@stat.math.ethz.ch Subject: Re: [R] small sample techniques > -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Nair, > Murlidharan T > Sent: Thursday, August 09, 2007 9:19 AM > To: Moshe Olshansky; Rolf Turner; r-help@stat.math.ethz.ch > Subject: Re: [R] small sample techniques > > Thanks, that discussion was helpful. Well, I have another question > I am comparing two proportions for its deviation from the hypothesized > difference of zero. My manually calculated z ratio is 1.94. > But, when I calculate it using prop.test, it uses Pearson's > chi-squared > test and the X-squared value that it gives it 0.74. Is there > a function > in R where I can calculate the z ratio? Which is > > > ('p1-'p2)-(p1-p2) > Z= ---------------- > S > ('p1-'p2) > > Where S is the standard error estimate of the difference between two > independent proportions > > Dummy example > This is how I use it > prop.test(c(30,23),c(300,300)) > > > Cheers../Murli > > Murli, I think you need to recheck you computations. You can run a t-test on your data in a variety of ways. Here is one: > x<-c(rep(1,30),rep(0,270)) > y<-c(rep(1,23),rep(0,277)) > t.test(x,y) Welch Two Sample t-test data: x and y t = 1.0062, df = 589.583, p-value = 0.3147 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.02221086 0.06887752 sample estimates: mean of x mean of y 0.10000000 0.07666667 Hope this is helpful, Dan Daniel J. Nordlund Research and Data Analysis Washington State Department of Social and Health Services Olympia, WA 98504-5204 ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.