Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Stat 491: Biostatistics Chapter 8: Hypothesis Testing–Two-Sample Inference Solomon W. Harrar The University of Montana Fall 2012 Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Two-Sample Inference In Chapter 6 and 7, we had only one-sample. Underlying µ (or p) of the population from which the sample was drawn was compared with known mean (prevalence rate) of the general population. Example: Asian immigrants mean cholesterol was compared with the general US mean cholesterol known to be 190 mg/dL. In this chapter, we do have two samples each from a different population. Interest lies in comparing the underlying unknown means of the two populations. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Randomized Clinical Trials (RCT) Patients are assigned to treatments by some random mechanism. If sample sizes are large, we expect type of patients assigned to different treatment modalities to be similar. If sample sizes is small, patient characteristics of treatment groups may not be comparable. A table of characteristics of the treatment groups are customarily presented to check that the randomization is working well. Design features of RCT Randomization: Complete, Block , Cluster (Group), Stratified (by age, sex, or overall clinical condition). Blinding: Single, Double, Triple and unblinded Example: Greek Health Project Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Two Types of Samples Paired Samples: Each data point in one samples is matched and related to a unique data point in the other sample. Independent Samples: The data points in one sample are unrelated to the data points in the other sample. Example: Suppose we are interested in studying the association between Oral Contraceptive (OC) use and blood pressure. One can start with non OC user women in the child bearing age group (16-49 years of age) and follow them for one year. For those who started using OC within the one year period, compare the blood pressure at baseline and follow-up. Alternatively, one can identify a group of OC user women and another group of non users and compare their blood pressures. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Paired Samples Arise When Having the same set of experimental units receive both treatments (Cross-Over Design) Having measurement taken before and after treatment (Repeated-Measures Design) No randomization. Matching Subjects (Matched-Pair Design) Using naturally occurring pairs such as twins or husbands and wives. Matching with respect to extraneous factors that may mask differences in the treatments. Block Randomization Matched Case-Control Study (Observational study) Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Paired or Independent Sample In repeated measures, each subject is serving as their own control. This design may benefit from having a control group as it allows to rule out other factors that may cause changes between the two time points. In matching, extraneous factors are expected to influence both members of the pair equally. Hence, paired design is definitive in that if difference is present, it is highly likely that it occurred because of the the difference in treatment. Difference in the independent samples are only suggestive. The differences in the subjects may mask true treatment or group differences. Paired design may NOT sometimes be practical and is usually expensive. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Paired t Test Let µd = µ1 − µ2 . Let n denote the number of pairs of measurements in the sample. Let di denote the difference between the first and second measurement in the ith pair. Assumption: d1 , d2 , . . . , dn constitute a random sample from a normally distributed population with mean µd and unknown variance σd2 . We can look at Q-Q plot and Box plots of the d’s to check violation of the normality assumption. Compute s Pn n X ¯ 2 1 i=1 (di − d) ¯ di and sd = . d= n n−1 i=1 Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples The Paired t-test Hypotheses: Case 1. H0 : µd = 0 Case 2. H0 : µd = 0 Case 3. H0 : µd = 0 T.S.: vs Ha : µd > 0 vs Ha : µd < 0 vs Ha : µd 6= 0 d¯ √ sd / n R.R.: For a specified value of α, Case 1. Reject H0 if t ≥ tn−1,1−α . Case 2. Reject H0 if t ≤ −tn−1,1−α . Case 3. Reject H0 if |t| ≥ tn−1,α/2 . p-Value: Case 1. P(t > tcomputed ) Case 2. P(t < tcomputed ) Case 3. 2 × P(t > |tcomputed |) for two-sided test. t= Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Confidence Interval for µd A 100(1 − α)% two-sided confidence interval estimate of the size of the difference (µd ) is sd d¯ ± tn−1,1−α/2 √ . n A 100(1 − α)% lower-sided confidence limit for the size of the difference (µd ) is sd d¯ + tn−1,1−α √ . n A 100(1 − α)% upper-sided confidence limit for the size of the difference (µd ) is sd d¯ − tn−1,1−α √ . n If n is large then the z-test is used and normality is not needed. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Example: Nutrition An important hypothesis in hypertension research is that sodium restriction may lower blood pressure. However, it is difficult to achieve sodium restriction over the long term, and dietary counseling in a group setting is sometimes used to achieve this goal. The data on overnight urinary sodium excretion (mEq/8hr) were obtained on eight individuals enrolled in a sodium-restricted group. Data was collected at baseline and after one week of dietary counseling. (d¯ = 1.14 and sd = 12.22) Person Baseline Week 1 di 1 7.85 9.59 -1.74 2 12.03 34.50 -22.47 3 21.84 4.55 17.29 4 13.94 20.78 -6.84 5 16.68 11.69 4.99 6 41.78 32.51 9.27 7 14.97 5.46 9.51 8 12.072 12.95 -0.88 Test the appropriate hypothesis and report p-value. Construct 95% CI for the true mean change in overnight sodium excretion over a one-week period. Verify the validity of the normality assumption. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Power Analysis and Sample-Size Estimation Note that di = x1i − x2i where x1i and x2i are the measurements on the ith subject at the baseline and follow-up, respectively. Assumed d1 , . . . , dn constitute a random sample from N(µd , σd2 ). If we can get a good working estimate of σd from a previous or pilot or reproducibility study, we can use the power and sample-size formulae from the one sample problem here. More specifically, for the two-sided alternative PWR(µd ) ≈ P(Z ≤ −z1−α/2 + n = σd2 |µd | √ ) σd / n and (z1−α/2 + z1−β )2 µ2d For one-sided test, replace α/2 with α, and the power is exact. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Power Analysis and Sample-Size Estimation Cont’d... However, caution has to be used when using estimate of σd from a previous study, in particular, in longitudinal studies. Know that σd2 = σ12 + σ22 − 2ρσ1 σ2 where ρ is the correlation between X1 and X2 . σd2 depends on the correlation ρ. The correlation typically decreases at the time separation increases. To use σd from a previous study, we have to make sure that the time separation between baseline and follow up in the previous study and the planned study are about the same. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Background Notations: Let us denote the population means and standard deviations from the two populations as Population 1: µ1 and σ1 Population 2: µ2 and σ2 Notations: Let us denote the means, standard deviation and sample sizes of the two independent samples from the two populations as Sample 1: x¯1 , s1 and n1 Sample 2: x¯2 , s2 and n2 We are interested in making inference about µ1 − µ2 .. ¯1 − X ¯2 . A natural estimator of µ1 − µ2 is X Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples ¯1 − X ¯2 The Sampling Distribution of X If the two populations are normally distributed then the ¯1 − X ¯2 is normal with mean sampling distribution of X µX¯1 −X¯2 = µ1 − µ2 and standard deviation σX2¯1 −X¯2 = σ12 σ22 + . n1 n2 If either of the two populations are non-normal but n1 and n2 are both large, then the above sampling distribution of ¯1 − X ¯2 hold approximately. This is a consequence of the X CLT. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples The three cases Case 1: Both populations are normally distributed with (a) σ1 = σ2 = σ (Pooled-variance t-procedures). (b) σ1 = 6 σ2 (Welch-Satterthwaite t-procedures). Case 2: Both Sample Sizes n1 and n2 are large (z procedures) Case 3: Either n1 or n2 is small and the population is non-normal. (Bootstrap or Nonparametric procedures) Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples The Equal-Variance Case The two populations are normally distributed, t= ¯1 − X ¯2 ) − (µ1 − µ2 ) (X q s n11 + n12 where S2 = ∼ tn1 +n2 −2 (n1 − 1)S12 + (n2 − 1)S22 . n1 + n2 − 2 Notice the degrees of freedom n1 + n2 − 2 comes from S 2 . We will use this quantity to construct tests and confidence intervals when the two populations are normal and the standard deviations are equal. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Large-Samples Case When the sample sizes n1 and n2 are large, we use the quantity Z= ¯1 − X ¯ ) − (µ1 − µ2 ) (X q2 2 S22 S1 n1 + n2 · ∼ N(0, 1) This is true whether or not normality or equality of variance hold. This quantity is used for tests and confidence intervals when n1 and n2 are large. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples The Independent-Samples t-test for µ1 − µ2 Hypotheses: Case 1. H0 : µ1 − µ2 ≤ 0 vs Ha : µ1 − µ2 > 0 Case 2. H0 : µ1 − µ2 ≥ 0 vs Ha : µ1 − µ2 < 0 Case 3. H0 : µ1 − µ2 = 0 vs Ha : µ1 − µ2 6= 0 T.S.: p t = (¯ x1 − x¯2 )/(s 1/n1 + 1/n2 ) R.R.: For a specified value of α, Case 1. Reject H0 if t ≥ tn1 +n2 −2,1−α . Case 2. Reject H0 if t ≤ −tn1 +n2 −2,1−α . Case 3. Reject H0 if |t| ≥ tn1 +n2 −2,1−α/2 . p-Value: Case 1. Reject H0 if P(t > tcomputed ). Case 2. Reject H0 if P(t < tcomputed ). Case 3. Reject H0 if P(t > |tcomputed |). Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples 100(1 − α)% CI for µ1 − µ2 when σ1 = σ2 A 100(1 − α)% confidence interval for µ1 − µ2 is given by r 1 1 (¯ x1 − x¯2 ) ± tn1 +n2 −2,1−α/2 s + n1 n2 Lower-sided confidence interval for µ1 − µ2 r 1 1 (¯ x1 − x¯2 ) + tn1 +n2 −2,1−α s + . n1 n2 Upper-sided confidence interval for µ1 − µ2 r 1 1 (¯ x1 − x¯2 ) − tn1 +n2 −2,1−α s + . n1 n2 In R inference for difference in means can be tested in one of the following two ways depending on how your data is organized. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples t test in R Inference for difference in means can be computed in R in one of the following two ways depending on how your data is organized. If the two samples are entered as vectors x and y then t.test(x,y,mu=0,paired=F,var.equal=T, alternative="two.sided") If the all the data form the two samples is in one vector y and the vector x contains indicators of sample, then we use t.test(y~x,mu=0,paired=F,var.equal=T, alternative="two.sided") Examples: x=c(2.3,3.4,1.2,4.4) y=c(3.2,1.5,2.6,3.3,4.5) t.test(x,y,var.eual=T) x=c(1,1,1,1,2,2,2,2,2) y=c(2.3,3.4,1.2,4.4,3.2,1.5,2.6,3.3,4.5) t.test(y~x,var.eual=T) Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Example: Veterinary Science An experiment was conducted to evaluate the effectiveness of a treatment for tapeworm in the stomachs of sheep. A random sample of 24 worm-infected lamb of approximately the same age and health was randomly divided into two groups. Twelve of the lambs were injected with the drug and the remaining twelve were left untreated. After a 6-month period, the lambs were slaughtered and the following worm counts were recorded: Drug Treated: 18, 43, 28, 50, 16, 32, 13, 35, 38, 33, 6, 7 Untreated: 40, 54, 26, 63, 21, 37, 39, 23, 48, 58, 28,39 (a) Does any of the assumptions of the pooled t-test appear to an issue? (b) Test whether the mean number of tapeworms in the stomachs of the treated lambs is less than the mean for untreated lambs. Use α = 0.05. (c) What is the level of significance for this test? (d) Place a 95% CI on µ1 − µ2 to assess the size of the difference in the two means. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Pooled-Variance t-test for µ1 − µ2 :An Example Cont’d... x¯1 = 26.58, s1 = 14.36, x¯2 = 39.67 and s2 = 13.86 Normal Q−Q Plot for Untreated 50 Normal Q−Q Plot for Drug Treated ● 60 ● ● 30 ● ● ● ● ● ● 30 ● ● 40 Sample Quantiles ● ● ● ● 50 ● 20 Sample Quantiles 40 ● ● ● 10 ● ● −1.5 10 ● 20 ● ● −1.0 −0.5 20 0.0 0.5 1.0 1.5 −1.5 −1.0 −0.5 0.0 0.5 Theoretical Quantiles Theoretical Quantiles Box Plot for Drug Treated Box Plot for Untreated 30 40 50 Chapter 8: Hypothesis Testing–Two-Sample Inference 20 30 40 Stat 491: Biostatistics 1.0 50 1.5 60 Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Test for Equality of Variances The choice between the pooled-variance and Welch-Satterthwaite procedures depends on whether the variances of the two populations are equal or not. In reality, it may not always be clear if equality holds or not. However, we can conduct a statistical test to assess the departure from equality using sample data. Assume the two populations are normally distributed. We are interested in testing H0 : σ12 = σ22 Chapter 8: Hypothesis Testing–Two-Sample Inference vs Ha : σ12 6= σ22 Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Test for Equality of Variances Cont’d... The quantity, F = S12 /σ12 S22 /σ22 ∼ Fn1 −1,n2 −1 where n1 P S12 = n2 P ¯1 )2 (X1i − X i=1 n1 − 1 and S22 = ¯2 )2 (X2i − X i=1 n2 − 1 . The Fd1 ,d2 distribution depends on two degrees of freedom known as the numerator and denominator degrees of freedom. The Fd1 ,d2 distribution is a right-skewed distribution over the interval (0, ∞). We want to reject H0 , when the test-statistic F = S12 /S22 is small or large compared to 1. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Test for Equality of Variances Cont’d... For a size-α test, we reject H0 if F ≤ Fn1 −1,n2 −1,α/2 or F ≥ Fn1 −1,n2 −1,1−α/2 In R, the quantiles of Fn1 −1,n2 −1 can be obtained as qf(alpha/2, n1-1, n2-1) qf(1-alpha/2, n1-1, n2-1) p-value, ( 2 × P(F > Fcomputed ) if Fcomputed ≥ 1 p−value = 2 × P(F < Fcomputed ) if Fcomputed < 1 Area under the curve of Fn1 −1,n2 −1 to the left of Fcomputed can be found in R by pf(F_computed,n1-1,n2-1) For the tape worm data, test the hypothesis of equality of variance. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Test for Equality of Variances Cont’d... In the test-statistic F , F = S12 S22 H0 ∼ Fn1 −1,n2 −1 , we are using using the variance of the sample from population 1 in numerator and that of population 2 in the denominator. The labeling of the population is arbitrary. We could define the test statistic as F = S22 S12 H0 ∼ Fn2 −1,n1 −1 . Do we get the same conclusion? YES. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Test for Equality of Variances Cont’d... We observe, under the null hypothesis H0 : σ12 = σ22 , that P( S22 < Fn2 −1,n1 −1,α/2 ) = α/2 S12 = P( S12 S22 1 > F ) = P( < ) n −1,n −1,1−α/2 1 2 Fn1 −1,n2 −1,1−α/2 S22 S12 Therefore, Fn2 −1,n1 −1,α/2 = 1 Fn1 −1,n2 −1,1−α/2 . In R equality of variance can be tested in one of the following two ways depending on how your data is organized. var.test(x,y,ratio=1,alternative="two.sided") var.test(y~x,ratio=1,alternative="two.sided") Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples The Behrens-Fisher Problem Assume two independent samples from normal populations. We know, by conducting a test or otherwise, σ1 6= σ2 . Inference about µ1 − µ2 in this situation is known as the Behrens-Fisher problem. The test and confidence interval procedure was developed by Welch(1938) using Satterthwaite approximation for the degrees of freedom and, hence, is referred to as Welch-Satterthwaite Method. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples The Behrens-Fisher Problem Cont’d... The quantity t0 = (¯ x1 − x¯2 ) − (µ1 − µ2 ) · q 2 ∼ td s22 s1 n1 + n2 where d= (s12 /n1 + s22 /n2 )2 . (s12 /n1 )2 /(n1 − 1) + (s22 /n2 )2 /(n2 − 1) This quantity is used for tests and confidence intervals concerning µ1 − µ2 . For example, a 100(1 − α)% CI for µ1 − µ2 is given by s s12 s2 (¯ x1 − x¯2 ) ± td,1−α/2 + 2. n1 n2 Effect of unequal variance is large for unequal sample sizes. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Strategy for Testing Equality of Means When it is not clear whether σ12 = σ22 but normality appears to hold, use the following strategy. Fail to Reject Test Reject H0 : σ21 = σ22 Use Pooled Use Welch's t Test for t Test for H0 : µ1 = µ2 H0 : µ1 = µ2 Test for equality of variance is sensitive to departure from normality. Non-parametric methods must be used in these cases. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Behrens-Fisher Problem: Example A possible important environmental determinant of lung function in children is amount of cigarette smoking in the home. Suppose this question is studied by selecting two groups: group 1 consists of 23 nonsmoking children 5-9 years of age, both of whose parents smoke, who have a mean forced expiratory volume (FEV) of 2.1 L and standard deviation of 0.7 L; group 2 consists of 20 nonsmoking children of comparable age, neither of whose parents smoke, who have mean FEV of 2.3 L and a standard deviation of 0.4 L. (a) What are the appropriate null and alternative hypothesis in this situation? (b) What is the appropriate test procedure for the hypotheses above? (c) Carry out the test and report p-value. (d) Provide 95% CI for the true mean difference in FEV between 5- to 9-year-old children whose parents smoke and comparable children whose parents do not smoke. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Power Analysis For given sample sizes n1 and n2 and significance level α, the power the study will have in detecting a difference of ∆ = |µ1 − µ2 | is ∆ PWR(∆) = P(Z < −z1−α/2 + q ) σ12 /n1 + σ22 /n2 ∆ = pnorm(−z1−α/2 + q , 0, 1) σ12 /n1 + σ22 /n2 For one-sided alternative, we replace α/2 with α. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Power Analysis : Example Suppose 100 OC users and 100 non-OC users are available for study and a true mean difference of µ1 − µ2 = 5 mm Hg is anticipated, with OC users having the higher mean SBP. How much power would such a study have if estimates of the standard deviations for OC users and non-users were obtained from a pilot study as 15.34 mm Hg and 18.23 mm Hg, respectively? Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Sample-Size Estimation The appropriate sample size to have a probability of 1 − β of finding a significant difference based on a two-sided test with significance level α when the absolute difference in mean between the two groups is ∆ = |µ1 − µ2 | is: a. Equal sample sizes anticipated n1 = n2 = (σ12 + σ22 ) (z1−α/2 + z1−β )2 . ∆2 b. A known proportion n2 = kn1 anticipated n1 = (σ12 + σ22 /k) (z1−α/2 + z1−β )2 . ∆2 For one-sided test, we replace α/2 with α. When σ1 = σ2 , the smallest total sample size for a given α and β is achieved by the equal sample size allocation. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Sample-Size Estimation: Example Suppose we anticipate twice as many non-OC users as OC users entering the study. From a pilot study, estimates of the standard deviations for OC users and non-users were obtained as 15.34 and 18.23, respectively. Project the required sample size to find a significant difference in a two-sided test with 5% significance level and 80% power when there a 5 mm Hg difference in the true SBP means of OC users and non-OC users. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Paired-Samples versus Independent-Samples t Test For one sided test H0 : µ1 ≤ µ2 versus H0 : µ1 > µ2 , the power of the Z test is given by, PWR(∆) = P(Z < −z1−α + ∆ σX 1 −X 2 ) where ∆ = µ1 − µ2 . For paired sample, σX2 1 −X 2 = σ12 σ22 σ1 σ2 + − 2ρ . n n n For independent samples, σ12 σ22 + . 1 −X 2 n n When ρ > 0, which is typically the case, paired sample will have higher power than independent samples. σX2 Chapter 8: Hypothesis Testing–Two-Sample Inference = Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Efficiency of Pairing: An Example A study was designed to measure the effect of home environment on academic achievement of 12-year-old students. Because genetic differences may also contribute to academic achievement, the researcher wanted to control for this factor. Thirty sets of identical twins were identified who had been adopted prior to their first birthday, with one twin placed in a home in which academics were emphasized (Academic) and the other twin placed in a home in which academics were not emphasize (Nonacademic). The p values for comparing the mean scores for the academic and nonacademic environments were 0.000 and 0.24 for the paired and independent sample t tests, respectively. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Efficiency of Pairing: An Example Cont’d... (a) Is there a difference in in the mean final grade between the students in an academically oriented home environment and those in a nonacademic home environments? (b) Does it appear that using twins in this study to control for variation in the final scores were effective as compared to taking a random sample of 30 students in both types of environments? Justify your answer? See scatter plot on the next page. Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics Introduction Inference about µ1 − µ2 : Paired Samples Inference about µ1 − µ2 : Independent Samples Efficiency of Pairing: An Example Cont’d... Scatter Plot of Scores of Academic and Nonacademic Twins ● 90 ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 70 Nonacademic Environment 80 ● ● ● ● ● ● 60 ● ● ● ● ● 50 ● ● 50 60 70 80 Academic Environment Chapter 8: Hypothesis Testing–Two-Sample Inference Stat 491: Biostatistics 90

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