Chapter 8 Paired observations



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Chapter 8 Paired observations Timothy Hanson Department of Statistics, University of South Carolina Stat 205: Elementary Statistics for the Biological and Life Sciences 1 / 19

Book review of two-sample t-test ingredients 2 / 19

Paired designs Paired data arise when two of the same measurements are taken from the same subject, but under different experimental conditions. Subjects often receive both a treatment Y 1 and a control Y 2. Pairing observations reduces the subject-to-subject variability in the response. The analysis focuses on the difference in response from treatment to control. Let µ D be the mean difference for the entire population. We want a confidence interval for µ D and will want to test H 0 : µ D = 0 vs. one of (a) H A : µ D 0, (b) H A : µ D < 0, or (c) H A : µ D > 0. 3 / 19

Example 8.1.1 Coffee and blood flow Doctors studying healthy subjects measured myocardial blood flow (MBF) (ml/min/g) during bicycle exercise before and after giving the subjects the equivalent of two cups of coffee (200 mg of caffeine). Some people have high blood flow both before and after caffeine. Others have low blood flow before and after. By focusing on the differences from the same individual before and after, we adjust for individuals tendancy to be high or low regardless. How does this analyis differ from those in Chapters 6 and 7? Observations are collected on the same person. 4 / 19

Example 8.1.1 blood flow data 5 / 19

Example 8.1.1 blood flow data Each subject has a connected line (control and treatment). What does caffeine do to bloodflow? 6 / 19

Paired analysis in R Null is H 0 : µ D = 0. t.test(sample1,sample2,paired=true) gives P-value for H A : µ D 0. t.test(sample1,sample2,paired=true,alternative= less ) gives P-value for H A : µ D < 0. t.test(sample1,sample2,paired=true,alternative= greater ) gives P-value for H A : µ D > 0. 7 / 19

R code for bloodflow data > baseline=c(6.37,5.69,5.58,5.27,5.11,4.89,4.70,3.53) > caffeine=c(4.52,5.44,4.70,3.81,4.06,3.22,2.96,3.20) > t.test(baseline,caffeine,paired=true) Paired t-test data: baseline and caffeine t = 5.1878, df = 7, p-value = 0.00127 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.6278643 1.6796357 sample estimates: mean of the differences 1.15375 We estimate µ D as 1.15 ml/min/g. We are 95% confident that the true mean bloodflow is between 0.63 and 1.68 ml/min/g greater in the control group. We reject H 0 : µ D = 0 at the 5% level because P-value = 0.0013 < 0.05. Caffeine significantly reduces bloodflow. 8 / 19

Validity of paired t-test (p. 306) Let n be the number of paired observations. The paired sample t-test and confidence interval are valid if (a) The sample size is large enough, n > 30, say, or (b) the differences are approximately normal. Normality can be checked with a normal probability plot. If the two samples are sample1 and sample2, type qqnorm(sample1-sample2) in R. 9 / 19

Example 8.2.4 Hunger rating During a weight loss study each of n = 9 subjects was given either the active drug m-chlorophenylpiperazine (mcpp) for two weeks and then a placebo for another two weeks, or else was given the placebo for the first two weeks and then mcpp for the second two weeks. As part of the study the subjects were asked to rate how hungry they were at the end of each two-week period. 10 / 19

Hunger rating data 11 / 19

Hunger rating dotplot & normal probability plot 12 / 19

R code for hunger rating > drug=c(79,48,52,15,61,107,77,54,5) > placebo=c(78,54,142,25,101,99,94,107,64) > qqnorm(drug-placebo) > t.test(drug,placebo,paired=true) Paired t-test data: drug and placebo t = -2.7014, df = 8, p-value = 0.02701 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -54.784709-4.326402 sample estimates: mean of the differences -29.55556 We estimate µ D as 30. We are 95% confident that the drug reduces hunger between 4 and 55 points. We reject H 0 : µ D = 0 at the 5% level because P-value = 0.027 < 0.05. The drug significantly reduces hunger. 13 / 19

8.3 Paired designs Paired analyses reduce variability and make it easier to reject H 0 : µ D = 0. Need to have the paired observations come from very similar experimental units. Examples: Ex. 8.3.1 Two plants grown in the same container. Ex. 8.3.2 Case-control data from people matched on gender, age. Ex. 8.3.3 Tryglycerides measured before and after exercise. 14 / 19

Example 8.3.4 Triglycerides and exercise Triglycerides play a role in coronary artery disease. Researchers measured blood triglycerides in seven men before and after a 10-week exercise program. Subject Before After 1 0.87 0.57 2 1.13 1.03 3 3.14 1.47 4 2.14 1.43 5 2.98 1.20 6 1.18 1.09 7 1.60 1.51 15 / 19

8.4 The sign test The paired t-test assumes that differences follow a normal distribution. If the data aren t normal and the sample size is small, e.g. n < 30, then you can use the sign test. The sign test focuses on the median difference η D rather than the mean µ D. This test looks at the number of differences D = Y 1 Y 2 that are positive N + and the number that are negative N. These numbers should be similar if H 0 : η D = 0 is true. A P-value is based on the binomial distribution. Under H 0 : η D = 0, N + bin(n, 0.5). 16 / 19

Sign test in R In R, binom.test(n +,n) tests H 0 : η D = 0 vs. H A : η D 0. Need to count the number of + s and put that as first number, second number is sample size. For H A : η D < 0 use binom.test(n +,n,alternative= less ). For H A : η D > 0 use binom.test(n +,n,alternative= greater ). Ignore all output except the P-value. 17 / 19

Example 8.3.4 Triglycerides and exercise Subject Before After Sign 1 0.87 0.57 + 2 1.13 1.03 + 3 3.14 1.47 + 4 2.14 1.43 + 5 2.98 1.20 + 6 1.18 1.09 + 7 1.60 1.51 + N + = 7 and N = 0; P-value should be small. > binom.test(7,7) number of successes = 7, number of trials = 7, p-value = 0.01563 18 / 19

Two more examples Hunger rating > binom.test(2,9) number of successes = 2, number of trials = 9, p-value = 0.1797 P-value from t-test is 0.02701; not close at all. The t-test has greater power to reject H 0 when data are really normal. Caffeine and blood flow > binom.test(8,8) number of successes = 8, number of trials = 8, p-value = 0.007812 P-value from t-test is 0.00127; fairly similar but t-test has smaller P-value (more power if differences really are normal). 19 / 19