If  = 0, no difference in temperatureNote: n = number of pairsTwo-Sample TestsStudying underlying parameters from two different populations3 approaches1. Longitudinal—baseline and follow-up of same population. For example, differences in weight before and after intervention program. Followed overtime. Often leads to a paired-sample design. Each subject is his/her own control. Data point of 1st sample is related to a unique data point in the second sample.2. Cross-sectional—matched. The two samples are made alike on as many confounding variables as possible except for the variable of interest.3. Cross-sectional—two unrelated populations. For example, weight in males vs. females. One point in time. Data points in the two samples are unrelated.Each is a useful design. The first may be more definitive as this design controls for lots of confounders. There may be unknown confounders when using the third design. Examples of paired-sample study1. Longitudinal—20 subjects are weighed before starting an exercise program and then again after 6 weeks. Each subject is his/her own control. 2. Cross sectional—Subjects are paired by age, gender and blood pressure. Within each pair, one subject is randomly assigned to receive a new drug while the other receives the usual treatment. Even though different subject, they are “matched” on important variables.3. Cross-sectional—A camera has been developed to detect presence of cataracts more accurately. Six subjects each with cataracts in one eye. Measure “gray level” in each eye. Same subjects.Example: In a pediatric clinic, a study is carried out to see how whether aspirin changes body temperature. Twelve 5-year-old children suffering from influenza had their temperature taken immediately before and 1 hour after administration of aspirin. The results are given in the table. We will assume normality, and will conduct a two-sided test to see if aspirin changes temperature.Body Temperature oFPatient Before After1 102.4 99.62 103.2 100.13 101.9 100.24 103.0 101.15 101.2 99.86 100.7 100.27 102.5 101.08 103.1 100.19 102.8 100.710 102.3 101.111 101.9 101.312 101.4 100.2Assume that temperature of the ith child is normally distributed at baseline with mean  and variance 2 and at follow-up with mean  +  and variance 2. Therefore,  represents the mean difference in temperatureIf  = 0, no difference in temperatureIf  > 0, aspirin associated with an increase in temperatureIf  < 0, aspirin associated with a decrease in temperatureHo:  = 0 or d = 0H1:   0 or d  0Note: i is unknown and different for each childWe will consider di = Xi2 – Xi1, wheredi ~ N(, d2). Even though temperature is different for each child, differences have same underlying mean and variance over entire population of children.This leads to a one-sample t-test based on differences. Variance is unknown but estimated from the data. nd...dddn21Test statistic:1nnddswheren/sds0dtn1i2n1ii2idddNote: n = number of pairsReject Ho if t > tn-1, 1-/2 or if t < tn-1, /2p-value:1. If t < 0, p-value is 2 * area to left of n/sdtd02. If t > 0, p-value is 2 * area to right of n/sdtd0Back to the example:Body Temperature oFPatient Before After Difference1 102.4 99.6 2.82 103.2 100.1 3.13 101.9 100.2 1.74 103.0 101.1 1.95 101.2 99.8 1.46 100.7 100.2 0.57 102.5 101.0 1.58 103.1 100.1 3.09 102.8 100.7 2.110 102.3 101.1 1.211 101.9 101.3 0.612 101.4 100.2 1.2Reject Ho if t > t11, .975 = 2.201 or t < t11, .025 = -2.20175.1120.21122.1...1.38.2nddi 869207555113181175360645111202106452.......sd 11975625097511286927510~t.../..n/sdtdReject Hop-value: p < 0.0005*2 = 0.001Often useful to construct Confidence Limits for the true mean difference ()di ~ N(, d2)Therefore,  /nΔ,σ~Ndd2100% x (1 - ) CI for  =n/*stddα,n211 For the example, 3022219781552207512509020127511286920201275195.,....*../.*..%CI95% of the confidence intervals obtained from samples of 12 children contain the true population difference. We do not know if the confidence interval from this sample does.One-tailed versions:Ho:  = 0 Ho:  = 0H1:  < 0 H1:  > 0Reject Ho ift < tn-1, t > tn-1, 1-p-valuep = area to left of t p = area to right of tKashima et al. conducted research on self-help teaching skills with parents of mentally retarded children. As part of the study, 12 families were given the Behavioral Vignettes Test before and after the training program. The test assesses knowledge of behavior modification principles with a higher score indicating greater knowledge. The table below contains the pre- and post-scores. Parent Pre-score Post-score1 7 112 6 143 10 164 16 175 8 96 13 157 8 98 14 179 14 1710 16 2011 11 1212 14 12May we conclude, on the basis of these data, that the training program increase knowledge of behavior modification programs? Let  = 0.01. Let d = post – preH0:  = 0H1:   0Reject H0 if t > t11, 0.95 = 2.718Parent Pre-score Post-score di1 7 11 42 6 14 83 10 16 64 16 17 15 8 9 16 13 15 27 8 9 18 14 17 39 14 17 310 16 20 411 11 12 112 14 12 -2162d32d2ii6667.21232nddi 64.29697.61112321621nndds2n1i2n1ii2id503.37621.067.212/64.26667.2n/sds0dtddReject H0. There is evidence that the training program increases knowledge of behavior modification programs. 0.0005 < p <