1Stat 13, UCLA, Ivo DinovSlide 1UCLA STAT 13Introduction toStatistical Methods for the Life and Health SciencesInstructor: Ivo Dinov, Asst. Prof. of Statistics and NeurologyTeaching Assistants:Brandi Shanata & Tiffany HeadUniversity of California, Los Angeles, Fall 2007http://www.stat.ucla.edu/~dinov/courses_students.htmlStat 13, UCLA, Ivo DinovSlide 2Chapter 9 Paired DataStat 13, UCLA, Ivo DinovSlide 3z In chapter 7 we discussed how to compare two independentsamplesz In chapter 9 we discuss how to compare two samples that are paired In other words the two samples are not independent, Y1and Y2are linked in some way, usually by a direct relationship For example, measure the weight of subjects before and after a six month dietComparison of Paired SamplesStat 13, UCLA, Ivo DinovSlide 4Paired dataz To study paired data we would like to examine the differences between each pair d = Y1-Y2 each Y1, Y2pair will have a difference calculatedz With the paired t test we would like to concentrate our efforts on this difference data we will be calculating the mean of the differences and the standard error of the differencesStat 13, UCLA, Ivo DinovSlide 5Paired dataz The mean of the differences is calculated just like the one sample mean we calculated in chapter 2 it also happens to be equal to the difference in the sample means – this is similar to the t testz This sample mean differences is an estimate of the population mean difference μd= μ1 –μ221yynddd−==∑Stat 13, UCLA, Ivo DinovSlide 6Paired dataz Because we are focusing on the differences, we can use the same reasoning as we did for a single sample in chapter 6 to calculate the standard error aka. the standard deviation of the sampling distribution of z Recall: z Using similar logic: where sdis the standard deviation of the differences and ndis the sample size of the differencesdnsSE =dddnsSE =2Stat 13, UCLA, Ivo DinovSlide 7Paired dataExample: Suppose we measure the thickness of plaque (mm) in the carotid artery of 10 randomly selected patients with mild atherosclerotic disease. Two measurements are taken, thickness before treatment with Vitamin E (baseline) and after two years of taking Vitamin E daily.What makes this paired data rather than independent data?Why would we want to use pairing in this example?Subject Before After Difference 1 0.66 0.60 0.06 2 0.72 0.65 0.07 3 0.85 0.79 0.06 4 0.62 0.63 -0.01 5 0.59 0.54 0.05 6 0.63 0.55 0.08 7 0.64 0.62 0.02 8 0.70 0.67 0.03 9 0.73 0.68 0.05 10 0.68 0.64 0.04 mean 0.682 0.637 0.045 sd 0.0742 0.0709 0.0264 Stat 13, UCLA, Ivo DinovSlide 8Paired data – EDAhttp://socr.ucla.edu/htmls/SOCR_Charts.html0.6,0.65,0.79,0.63,0.54,0.55,0.62,0.67,0.68,0.64After0.66,0.72,0.85,0.62,0.59,0.63,0.64,0.7,0.73,0.68BeforeStat 13, UCLA, Ivo DinovSlide 9Paired dataCalculate the mean of the differences and the standard error for that estimate045.0=d0264.0=ds00833.0100264.0===dddnsSEStat 13, UCLA, Ivo DinovSlide 10Paired CI for μdz A 100(1 -α)% confidence interval for μdwhere df = nd-1 Very similar to the one sample confidence interval we learned in section 6.3, but this time we are concentrating on a difference column rather than a single sample)()(2dSEdftdα±Stat 13, UCLA, Ivo DinovSlide 11Paired CI for μdExample: Vitamin E (cont’)Calculate a 90% confidence interval for the true mean difference in plaque thickness before and after treatment with Vitamin E)0603.0,0297.0()00833.0)(833.1(045.0)00833.0()9(045.0)()(05.02=±=±=±tSEdftddαStat 13, UCLA, Ivo DinovSlide 12Paired CI for μdCONCLUSION: We are highly confident, at the 0.10 level, that the true mean difference in plaque thicknessbefore and after treatment with Vitamin E is between 0.03 mm and 0.06 mm.z Great, what does this really mean?z Does the zero rule work on this one?3Stat 13, UCLA, Ivo DinovSlide 13Paired t testz Of course there is also a hypothesis test for paired dataz #1 Hypotheses:Ho: μd= 0Ha: μd!= 0 or Ha: μd< 0 or Ha: μd> 0z #2 test statisticWhere df = nd–1z #3 p-value and #4 conclusion similar idea to that of the independent t testdsSEdt0−=Stat 13, UCLA, Ivo DinovSlide 14Paired t testExample: Vitamin E (cont’)Do the data provide enough evidence to indicate that there is a difference in plaque before and after treatment with vitamin E for two years? Test using α= 0.10Ho: μd= 0 (thickness in plaque is the same before and after treatment with Vitamin E )Ha: μd!= 0 (thickness in plaque after treatment is different than before treatment with Vitamin E )df = 10 – 1 = 9p < 2(0.0005) = 0.001, so we reject Ho. 402.500833.00045.0=−=stStat 13, UCLA, Ivo DinovSlide 15Paired t testCONCLUSION: These data show that the true mean thickness of plaque after two years of treatment with Vitamin E is statistically significantly different than before the treatment(p < 0.001). In other words, vitamin E appears to be a effective in changing carotid artery plaque after treatment May have been better to conduct this as an upper-tailed test because we would hope that vitamin E will reduce clogging however, researchers need to make this decision before analyzing dataStat 13, UCLA, Ivo DinovSlide 16Paired t testPaired T-Test and CI: Before, After Paired T for Before - AfterN Mean StDev SE MeanBefore 10 0.682000 0.074207 0.023466After 10 0.637000 0.070875 0.022413Difference 10 0.045000 0.026352 0.00833390% CI for mean difference: (0.029724, 0.060276)T-Test of mean difference = 0 (vs not = 0): T-Value = 5.40 P-Value = 0.000Stat 13, UCLA, Ivo DinovSlide 17Paired t test – validity?!?SOCR EDA: http://socr.ucla.edu/htmls/SOCR_Charts.html0.640.680.680.730.670.700.620.640.550.630.540.590.630.620.790.850.650.720.600.66AfterBeforeStat 13, UCLA, Ivo DinovSlide 18Results of Ignoring Pairingz Suppose we accidentally analyzed the groups independently (like an independent t-test) rather than a paired test? keep in mind this would be an incorrect way of analyzing the dataz How would this change our results?4Stat 13, UCLA, Ivo DinovSlide 19Results of Ignoring PairingExample Vitamin E (con’t)Calculate the test statistic and p-value as if this were an independent t test df = 172(0.05) < p < 2(0.1)0.10 < p < 0.2 Fail To Reject Ho!0325.0100709.0100742.02222212121=+=+=−nsnsSEyy38.10325.0637.0682.02121=−=−=− yysSEyytStat 13, UCLA, Ivo DinovSlide 20Results of Ignoring Pairingz What happens to a CI?Calculate a 90% confidence interval for
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