# Two-sample tests. Binary or categorical outcomes (proportions) Outcome Variable Are the observations correlated?Alternative to the chi- square test if.

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Two-sample tests

Binary or categorical outcomes (proportions) Outcome Variable Are the observations correlated?Alternative to the chi- square test if sparse cells: independentcorrelated Binary or categorical (e.g. fracture, yes/no) Chi-square test: compares proportions between two or more groups Relative risks: odds ratios or risk ratios Logistic regression: multivariate technique used when outcome is binary; gives multivariate-adjusted odds ratios McNemar’s chi-square test: compares binary outcome between correlated groups (e.g., before and after) Conditional logistic regression: multivariate regression technique for a binary outcome when groups are correlated (e.g., matched data) GEE modeling: multivariate regression technique for a binary outcome when groups are correlated (e.g., repeated measures) Fisher’s exact test: compares proportions between independent groups when there are sparse data (some cells <5). McNemar’s exact test: compares proportions between correlated groups when there are sparse data (some cells <5).

Recall: The odds ratio (two samples=cases and controls) Smoker (E)Non-smoker (~E) Stroke (D)1535 No Stroke (~D)842 50 Interpretation: there is a 2.25-fold higher odds of stroke in smokers vs. non-smokers.

Inferences about the odds ratio… Does the sampling distribution follow a normal distribution? What is the standard error?

Simulation… 1. In SAS, assume infinite population of cases and controls with equal proportion of smokers (exposure), p=.23 (UNDER THE NULL!) 2. Use the random binomial function to randomly select n=50 cases and n=50 controls each with p=.23 chance of being a smoker. 3. Calculate the observed odds ratio for the resulting 2x2 table. 4. Repeat this 1000 times (or some large number of times). 5. Observe the distribution of odds ratios under the null hypothesis.

Properties of the OR (simulation) (50 cases/50 controls/23% exposed) Under the null, this is the expected variability of the sample OR  note the right skew

Properties of the lnOR Normal!

Properties of the lnOR From the simulation, can get the empirical standard error (~0.5) and p-value (~.10)

Properties of the lnOR Or, in general, standard error =

Inferences about the ln(OR) Smoker (E)Non-smoker (~E) Stroke (D)1535 No Stroke (~D)842 50 p=.10

Confidence interval… Smoker (E)Non-smoker (~E) Stroke (D)1535 No Stroke (~D)842 50 Final answer: 2.25 (0.85,5.92)

Practice problem: Suppose the following data were collected in a case-control study of brain tumor and cell phone usage: Brain tumorNo brain tumor Own a cell phone 2060 Don’t own a cell phone 1040 Is there sufficient evidence for an association between cell phones and brain tumor?

Answer 1. What is your null hypothesis? Null hypothesis: OR=1.0; lnOR = 0 Alternative hypothesis: OR  1.0; lnOR>0 2. What is your null distribution? lnOR~ N(0, ) ; =SD (lnOR) =.44 3. Empirical evidence: = 20*40/60*10 =800/600 = 1.33  lnOR =.288 4. Z = (.288-0)/.44 =.65 p-value = P(Z>.65 or Z<-.65) =.26*2 5. Not enough evidence to reject the null hypothesis of no association TWO-SIDED TESTTWO-SIDED TEST: it would be just as extreme if the sample lnOR were.65 standard deviations or more below the null mean

Key measures of relative risk: 95% CIs OR and RR: For an odds ratio, 95% confidence limits: For a risk ratio, 95% confidence limits:

Continuous outcome (means) Outcome Variable Are the observations independent or correlated? Alternatives if the normality assumption is violated (and small sample size): independentcorrelated Continuous (e.g. pain scale, cognitive function) Ttest: compares means between two independent groups ANOVA: compares means between more than two independent groups Pearson’s correlation coefficient (linear correlation): shows linear correlation between two continuous variables Linear regression: multivariate regression technique used when the outcome is continuous; gives slopes Paired ttest: compares means between two related groups (e.g., the same subjects before and after) Repeated-measures ANOVA: compares changes over time in the means of two or more groups (repeated measurements) Mixed models/GEE modeling: multivariate regression techniques to compare changes over time between two or more groups; gives rate of change over time Non-parametric statistics Wilcoxon sign-rank test: non-parametric alternative to the paired ttest Wilcoxon sum-rank test (=Mann-Whitney U test): non- parametric alternative to the ttest Kruskal-Wallis test: non- parametric alternative to ANOVA Spearman rank correlation coefficient: non-parametric alternative to Pearson’s correlation coefficient

The two-sample t-test

The two-sample T-test Is the difference in means that we observe between two groups more than we’d expect to see based on chance alone?

The standard error of the difference of two means ** First add the variances and then take the square root of the sum to get the standard error. Recall, Var (A-B) = Var (A) + Var (B) if A and B are independent!

Shown by simulation: One sample of 30 (with SD=5). Difference of the two samples.

Distribution of differences If X and Y are the averages of n and m subjects, respectively:

But… As before, you usually have to use the sample SD, since you won’t know the true SD ahead of time… So, again becomes a T-distribution...

Estimated standard error of the difference…. Just plug in the sample standard deviations for each group.

Case 1: un-pooled variance Question: What are your degrees of freedom here? Answer: Not obvious!

Case 1: ttest, unpooled variances It is complicated to figure out the degrees of freedom here! A good approximation is given as df ≈ harmonic mean (or SAS will tell you!):

Case 2: pooled variance If you assume that the standard deviation of the characteristic (e.g., IQ) is the same in both groups, you can pool all the data to estimate a common standard deviation. This maximizes your degrees of freedom (and thus your power). Degrees of Freedom!

Estimated standard error (using pooled variance estimate) The degrees of freedom are n+m-2

Case 2: ttest, pooled variances

Alternate calculation formula: ttest, pooled variance

Pooled vs. unpooled variance Rule of Thumb: Use pooled unless you have a reason not to. Pooled gives you more degrees of freedom. Pooled has extra assumption: variances are equal between the two groups. SAS automatically tests this assumption for you (“Equality of Variances” test). If p<.05, this suggests unequal variances, and better to use unpooled ttest.

Example: two-sample t-test In 1980, some researchers reported that “men have more mathematical ability than women” as evidenced by the 1979 SAT’s, where a sample of 30 random male adolescents had a mean score ± 1 standard deviation of 436±77 and 30 random female adolescents scored lower: 416±81 (genders were similar in educational backgrounds, socio-economic status, and age). Do you agree with the authors’ conclusions?

Data Summary nSample Mean Sample Standard Deviation Group 1: women 3041681 Group 2: men 3043677

Two-sample t-test 1. Define your hypotheses (null, alternative) H 0 : ♂-♀ math SAT = 0 Ha: ♂-♀ math SAT ≠ 0 [two-sided]

Two-sample t-test 2. Specify your null distribution: F and M have similar standard deviations/variances, so make a “pooled” estimate of variance.

Two-sample t-test 3. Observed difference in our experiment = 20 points

Two-sample t-test 4. Calculate the p-value of what you observed data _null_; pval=(1-probt(.98, 58))*2; put pval; run; 0.3311563454 5. Do not reject null! No evidence that men are better in math ;)

Example 2: Difference in means Example: Rosental, R. and Jacobson, L. (1966) Teachers’ expectancies: Determinates of pupils’ I.Q. gains. Psychological Reports, 19, 115-118.

The Experiment (note: exact numbers have been altered) Grade 3 at Oak School were given an IQ test at the beginning of the academic year (n=90). Classroom teachers were given a list of names of students in their classes who had supposedly scored in the top 20 percent; these students were identified as “ academic bloomers ” (n=18). BUT: the children on the teachers lists had actually been randomly assigned to the list. At the end of the year, the same I.Q. test was re- administered.

Example 2 Statistical question: Do students in the treatment group have more improvement in IQ than students in the control group? What will we actually compare? One-year change in IQ score in the treatment group vs. one-year change in IQ score in the control group.

“Academic bloomers” (n=18) Controls (n=72 ) Change in IQ score: 12.2 (2.0) 8.2 (2.0) Results: 12.2 points 8.2 points Difference=4 points The standard deviation of change scores was 2.0 in both groups. This affects statistical significance…

What does a 4-point difference mean? Before we perform any formal statistical analysis on these data, we already have a lot of information. Look at the basic numbers first; THEN consider statistical significance as a secondary guide.

Is the association statistically significant? This 4-point difference could reflect a true effect or it could be a fluke. The question: is a 4-point difference bigger or smaller than the expected sampling variability?

Hypothesis testing Null hypothesis: There is no difference between “academic bloomers” and normal students (= the difference is 0%) Step 1: Assume the null hypothesis.

Hypothesis Testing These predictions can be made by mathematical theory or by computer simulation. Step 2: Predict the sampling variability assuming the null hypothesis is true

Hypothesis Testing Step 2: Predict the sampling variability assuming the null hypothesis is true—math theory:

Hypothesis Testing In computer simulation, you simulate taking repeated samples of the same size from the same population and observe the sampling variability. I used computer simulation to take 1000 samples of 18 treated and 72 controls Step 2: Predict the sampling variability assuming the null hypothesis is true—computer simulation:

Computer Simulation Results Standard error is about 0.52

3. Empirical data Observed difference in our experiment = 12.2-8.2 = 4.0

4. P-value t-curve with 88 df’s has slightly wider cut-off’s for 95% area (t=1.99) than a normal curve (Z=1.96) p-value <.0001

If we ran this study 1000 times, we wouldn’t expect to get 1 result as big as a difference of 4 (under the null hypothesis). Visually…

5. Reject null! Conclusion: I.Q. scores can bias expectancies in the teachers’ minds and cause them to unintentionally treat “bright” students differently from those seen as less bright.

Confidence interval (more information!!) 95% CI for the difference: 4.0±1.99(.52) = (3.0 – 5.0) t-curve with 88 df’s has slightly wider cut- off’s for 95% area (t=1.99) than a normal curve (Z=1.96)

What if our standard deviation had been higher? The standard deviation for change scores in treatment and control were each 2.0. What if change scores had been much more variable—say a standard deviation of 10.0 (for both)?

Standard error is 0.52 Std. dev in change scores = 2.0 Std. dev in change scores = 10.0 Standard error is 2.58

With a std. dev. of 10.0… LESS STATISICAL POWER! Standard error is 2.58 If we ran this study 1000 times, we would expect to get  +4.0 or  –4.0 12% of the time. P-value=.12

Don’t forget: The paired T-test Did the control group in the previous experiment improve at all during the year? Do not apply a two-sample ttest to answer this question! After-Before yields a single sample of differences… “within-group” rather than “between-group” comparison…

Continuous outcome (means); Outcome Variable Are the observations independent or correlated? Alternatives if the normality assumption is violated (and small sample size): independentcorrelated Continuous (e.g. pain scale, cognitive function) Ttest: compares means between two independent groups ANOVA: compares means between more than two independent groups Pearson’s correlation coefficient (linear correlation): shows linear correlation between two continuous variables Linear regression: multivariate regression technique used when the outcome is continuous; gives slopes Paired ttest: compares means between two related groups (e.g., the same subjects before and after) Repeated-measures ANOVA: compares changes over time in the means of two or more groups (repeated measurements) Mixed models/GEE modeling: multivariate regression techniques to compare changes over time between two or more groups; gives rate of change over time Non-parametric statistics Wilcoxon sign-rank test: non-parametric alternative to the paired ttest Wilcoxon sum-rank test (=Mann-Whitney U test): non- parametric alternative to the ttest Kruskal-Wallis test: non- parametric alternative to ANOVA Spearman rank correlation coefficient: non-parametric alternative to Pearson’s correlation coefficient

Data Summary nSample Mean Sample Standard Deviation Group 1: Change 72+8.22.0

Did the control group in the previous experiment improve at all during the year? p-value <.0001

Normality assumption of ttest If the distribution of the trait is normal, fine to use a t-test. But if the underlying distribution is not normal and the sample size is small (rule of thumb: n>30 per group if not too skewed; n>100 if distribution is really skewed), the Central Limit Theorem takes some time to kick in. Cannot use ttest. Note: ttest is very robust against the normality assumption!

Alternative tests when normality is violated: Non-parametric tests

Continuous outcome (means); Outcome Variable Are the observations independent or correlated? Alternatives if the normality assumption is violated (and small sample size): independentcorrelated Continuous (e.g. pain scale, cognitive function) Ttest: compares means between two independent groups ANOVA: compares means between more than two independent groups Pearson’s correlation coefficient (linear correlation): shows linear correlation between two continuous variables Linear regression: multivariate regression technique used when the outcome is continuous; gives slopes Paired ttest: compares means between two related groups (e.g., the same subjects before and after) Repeated-measures ANOVA: compares changes over time in the means of two or more groups (repeated measurements) Mixed models/GEE modeling: multivariate regression techniques to compare changes over time between two or more groups; gives rate of change over time Non-parametric statistics Wilcoxon sign-rank test: non-parametric alternative to the paired ttest Wilcoxon sum-rank test (=Mann-Whitney U test): non- parametric alternative to the ttest Kruskal-Wallis test: non- parametric alternative to ANOVA Spearman rank correlation coefficient: non-parametric alternative to Pearson’s correlation coefficient

Non-parametric tests t-tests require your outcome variable to be normally distributed (or close enough), for small samples. Non-parametric tests are based on RANKS instead of means and standard deviations (=“population parameters”).

Example: non-parametric tests 10 dieters following Atkin’s diet vs. 10 dieters following Jenny Craig Hypothetical RESULTS: Atkin’s group loses an average of 34.5 lbs. J. Craig group loses an average of 18.5 lbs. Conclusion: Atkin’s is better?

Example: non-parametric tests BUT, take a closer look at the individual data… Atkin’s, change in weight (lbs): +4, +3, 0, -3, -4, -5, -11, -14, -15, -300 J. Craig, change in weight (lbs) -8, -10, -12, -16, -18, -20, -21, -24, -26, -30

Jenny Craig -30-25-20-15-10-505101520 0 5 10 15 20 25 30 P e r c e n t Weight Change

Atkin’s -300-280-260-240-220-200-180-160-140-120-100-80-60-40-20020 0 5 10 15 20 25 30 P e r c e n t Weight Change

t-test inappropriate… Comparing the mean weight loss of the two groups is not appropriate here. The distributions do not appear to be normally distributed. Moreover, there is an extreme outlier (this outlier influences the mean a great deal).

Wilcoxon rank-sum test RANK the values, 1 being the least weight loss and 20 being the most weight loss. Atkin’s +4, +3, 0, -3, -4, -5, -11, -14, -15, -300 1, 2, 3, 4, 5, 6, 9, 11, 12, 20 J. Craig -8, -10, -12, -16, -18, -20, -21, -24, -26, -30 7, 8, 10, 13, 14, 15, 16, 17, 18, 19

Wilcoxon rank-sum test Sum of Atkin’s ranks: 1+ 2 + 3 + 4 + 5 + 6 + 9 + 11+ 12 + 20=73 Sum of Jenny Craig’s ranks: 7 + 8 +10+ 13+ 14+ 15+16+ 17+ 18+19=137 Jenny Craig clearly ranked higher! P-value *(from computer) =.018 *For details of the statistical test, see appendix of these slides…

Binary or categorical outcomes (proportions) Outcome Variable Are the observations correlated?Alternative to the chi- square test if sparse cells: independentcorrelated Binary or categorical (e.g. fracture, yes/no) Chi-square test: compares proportions between two or more groups Relative risks: odds ratios or risk ratios Logistic regression: multivariate technique used when outcome is binary; gives multivariate-adjusted odds ratios McNemar’s chi-square test: compares binary outcome between two correlated groups (e.g., before and after) Conditional logistic regression: multivariate regression technique for a binary outcome when groups are correlated (e.g., matched data) GEE modeling: multivariate regression technique for a binary outcome when groups are correlated (e.g., repeated measures) Fisher’s exact test: compares proportions between independent groups when there are sparse data (some cells <5). McNemar’s exact test: compares proportions between correlated groups when there are sparse data (some cells <5).

Difference in proportions (special case of chi-square test)

Standard error of the difference of two proportions=Standard error of a proportion= Null distribution of a difference in proportions Standard error can be estimated by= (still normally distributed) Analagous to pooled variance in the ttest The variance of a difference is the sum of variances (as with difference in means).

Null distribution of a difference in proportions Difference of proportions

Difference in proportions test Null hypothesis: The difference in proportions is 0. Recall, variance of a proportion is p(1-p)/n Use average (or pooled) proportion in standard error formula, because under the null hypothesis, groups have equal proportions. Follows a normal because binomial can be approximated with normal

Recall case-control example: Smoker (E)Non-smoker (~E) Stroke (D)1535 No Stroke (~D)842 50

Absolute risk: Difference in proportions exposed Smoker (E)Non-smoker (~E) Stroke (D)1535 No Stroke (~D)842 50

Difference in proportions exposed

Example 2: Difference in proportions Research Question: Are antidepressants a risk factor for suicide attempts in children and adolescents? Example modified from: “ Antidepressant Drug Therapy and Suicide in Severely Depressed Children and Adults ”; Olfson et al. Arch Gen Psychiatry.2006;63:865- 872.

Example 2: Difference in Proportions Design: Case-control study Methods: Researchers used Medicaid records to compare prescription histories between 263 children and teenagers (6-18 years) who had attempted suicide and 1241 controls who had never attempted suicide (all subjects suffered from depression). Statistical question: Is a history of use of antidepressants more common among cases than controls?

Example 2 Statistical question: Is a history of use of antidepressants more common among heart disease cases than controls? What will we actually compare? Proportion of cases who used antidepressants in the past vs. proportion of controls who did

No (%) of cases (n=263) No (%) of controls (n=1241 ) Any antidepressant drug ever 120 (46%) 448 (36%) 46% 36% Difference=10% Results

Is the association statistically significant? This 10% difference could reflect a true association or it could be a fluke in this particular sample. The question: is 10% bigger or smaller than the expected sampling variability?

Hypothesis testing Null hypothesis: There is no association between antidepressant use and suicide attempts in the target population (= the difference is 0%) Step 1: Assume the null hypothesis.

Hypothesis Testing Step 2: Predict the sampling variability assuming the null hypothesis is true

Also: Computer Simulation Results Standard error is about 3.3%

Hypothesis Testing Step 3: Do an experiment We observed a difference of 10% between cases and controls.

Hypothesis Testing Step 4: Calculate a p-value

When we ran this study 1000 times, we got 1 result as big or bigger than 10%. P-value from our simulation… We also got 3 results as small or smaller than –10%.

P-value From our simulation, we estimate the p-value to be: 4/1000 or.004

Here we reject the null. Alternative hypothesis: There is an association between antidepressant use and suicide in the target population. Hypothesis Testing Step 5: Reject or do not reject the null hypothesis.

What would a lack of statistical significance mean? If this study had sampled only 50 cases and 50 controls, the sampling variability would have been much higher—as shown in this computer simulation…

Standard error is about 10% 50 cases and 50 controls. Standard error is about 3.3% 263 cases and 1241 controls.

With only 50 cases and 50 controls… Standard error is about 10% If we ran this study 1000 times, we would expect to get values of 10% or higher 170 times (or 17% of the time).

Two-tailed p-value Two-tailed p-value = 17%x2=34%

Practice problem… An August 2003 research article in Developmental and Behavioral Pediatrics reported the following about a sample of UK kids: when given a choice of a non-branded chocolate cereal vs. CoCo Pops, 97% (36) of 37 girls and 71% (27) of 38 boys preferred the CoCo Pops. Is this evidence that girls are more likely to choose brand-named products?

Answer 1. Hypotheses: H 0 : p ♂ -p ♀ = 0 Ha: p ♂ -p ♀ ≠ 0 [two-sided] 2. Null distribution of difference of two proportions: 3. Observed difference in our experiment =.97-.71=.26 4. Calculate the p-value of what you observed: data _null_; pval=(1-probnorm(3.06))*2; put pval; run; 0.0022133699 5. p-value is sufficiently low for us to reject the null; there does appear to be a difference in gender preferences here. Null says p’s are equal so estimate standard error using overall observed p

Key two-sample Hypothesis Tests… Test for H o : μ x - μ y = 0 (σ 2 unknown, but roughly equal): Test for H o : p 1- p 2 = 0:

Corresponding confidence intervals… For a difference in means, 2 independent samples (σ 2 ’s unknown but roughly equal): For a difference in proportions, 2 independent samples:

Appendix: details of rank-sum test…

Wilcoxon Rank-sum test

Example For example, if team 1 and team 2 (two gymnastic teams) are competing, and the judges rank all the individuals in the competition, how can you tell if team 1 has done significantly better than team 2 or vice versa?

Answer Intuition: under the null hypothesis of no difference between the two groups… If n 1 =n 2, the sums of T 1 and T 2 should be equal. But if n 1 ≠n 2, then T 2 (n 2= bigger group) should automatically be bigger. But how much bigger under the null? For example, if team 1 has 3 people and team 2 has 10, we could rank all 13 participants from 1 to 13 on individual performance. If team1 (X) and team2 don’t differ in talent, the ranks ought to be spread evenly among the two groups, e.g.… 1 2 X 4 5 6 X 8 9 10 X 12 13 (exactly even distribution if team1 ranks 3 rd, 7 th, and 11 th )

Remember this? sum of within-group ranks for smaller group. sum of within-group ranks for larger group. Take-home point:

It turns out that, if the null hypothesis is true, the difference between the larger-group sum of ranks and the smaller-group sum of ranks is exactly equal to the difference between T 1 and T 2

From slide 23 From slide 24 Define new statistics Here, under null: U2=55+30-70 U1=6+30-21 U2+U1=30

 under null hypothesis, U 1 should equal U 2 : The U’s should be equal to each other and will equal n 1 n 2 /2 : U 1 + U 2 = n 1 n 2 Under null hypothesis, U 1 = U 2 = U 0  E(U 1 + U 2 ) = 2E( U 0 ) = n 1 n 2 E(U 1 = U 2 = U 0 ) = n 1 n 2 /2 So, the test statistic here is not quite the difference in the sum-of-ranks of the 2 groups  It’s the smaller observed U value: U 0 For small n’s, take U 0, and get p-value directly from a U table.

For large enough n’s (>10 per group)…

Add observed data to the example… Example: If the girls on the two gymnastics teams were ranked as follows: Team 1: 1, 5, 7 Observed T 1 = 13 Team 2: 2,3,4,6,8,9,10,11,12,13 Observed T 2 = 78 Are the teams significantly different? Total sum of ranks = 13*14/2 = 91 n 1 n 2 =3*10 = 30 Under the null hypothesis: expect U 1 - U 2 = 0 and U 1 + U 2 = 30 (each should equal about 15 under the null) and U 0 = 15 U 1 =30 + 6 – 13 = 23 U 2 = 30 + 55 – 78 = 7  U 0 = 7 Not quite statistically significant in U table…p=.1084 (see attached) x2 for two-tailed test

Example problem 2 A study was done to compare the Atkins Diet (low-carb) vs. Jenny Craig (low-cal, low-fat). The following weight changes were obtained; note they are very skewed because someone lost 100 pounds; the mean loss for Atkins is going to look higher because of the bozo, but does that mean the diet is better overall? Conduct a Mann-Whitney U test to compare ranks. AtkinsJenny Craig -100-11 -8-15 -4-5 +5+6 +8-20 +2

Answer Corresponding Ranks (lower is more weight loss!): AtkinsJenny Craig 14 53 76 910 112 8 Sum of ranks for JC = 25 (n=5) Sum of ranks for Atkins=41 (n=6) n 1 n 2 =5*6 = 30 under the null hypothesis: expect U 1 - U 2 = 0 and U 1 + U 2 = 30 and U 0 = 15 U 1 =30 + 15 – 25 = 20 U 2 = 30 + 21 – 41 = 10 U 0 = 10; n 1 =5, n 2 =6 Go to Mann-Whitney chart….p=.2143x 2 =.42

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