Effect Size Estimation Why and How An Overview. Statistical Significance Only tells you sample results unlikely were the null true. Null is usually that.

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Effect Size Estimation Why and How An Overview

Statistical Significance Only tells you sample results unlikely were the null true. Null is usually that the effect size is absolutely zero. If power is high, the size of a significant effect could be trivial. If power is low, a big effect could fail to be detected

Nonsignificant Results Effect size estimates should be reported here too, especially when power was low. Will help you and others determine whether or not it is worth the effort to repeat the research under conditions providing more power.

Comparing Means Student’s T Tests Even with complex research, the most important questions can often be addressed by simple contrasts between means or sets of means. Reporting strength of effect estimates for such contrasts can be very helpful.

Symbols Different folks use different symbols. Here are those I shall use  – the parameter, Cohen’s. d – the sample statistic, There is much variation with respect to choice of symbols. Some use d to stand for the parameter, for example.

One Sample On SAT-Q, is µ for my students same as national average? Point estimate does not indicate precision of estimation. We need a confidence interval.

Constructing the Confidence Interval Approximate method – find unstandardized CI, divide endpoints by sample SD. OK with large sample sizes. With small sample sizes should use an exact method. Computer-intensive, iterative procedure, must estimate µ and σ.

Programs to Do It SAS SPSS The mean math SAT of my undergraduate statistics students (M = 535, SD = 93.4) was significantly greater than the national norm (516), t(113) = 2.147, p =.034, d =.20. A 95% confidence interval for the mean runs from 517 to 552. A 95% confidence interval for  runs from.015 to.386.

Benchmarks for  What would be a small effect in one context might be a large effect in another. Cohen reluctantly provided these benchmarks for behavioral research.2 = small, not trivial.5 = medium.8 = large

Reducing Error Not satisfied with the width of the CI,.015 to.386 (trivial to small/medium)? Get more data, or Do any of the other things that increase power.

Why Standardize? Statisticians argue about this. If the unit of measure is meaningful (cm, \$, ml), do not need to standardize. Weight reduction intervention produced average loss of 17.3 pounds. Residents of Mississippi average 17.3 points higher than national norm on measure of neo-fascist attitudes.

Bias in Effect Size Estimation Lab research may result in over-estimation of the size of the effect in the natural world. Sample Homogeneity Extraneous Variable Control Mean difference = 25 Lab SD = 15, d = 1.67, whopper effect Field SD = 100, d =.25, small effect

Two Independent Means

Programs Will do all this for you and give you a CI. Conf_Interval-d2.sas CI-d-SPSS.zip Confidence Intervals, Pooled and Separate Variances TConfidence Intervals, Pooled and Separate Variances T

Example Pooled t(86) = 3.267 t = 3.267 ; df = 86 ; n 1 = 33 ; n 2 = 55 ; d = t/sqrt(n1*n2/(n1+n2)); ncp_lower = TNONCT(t,df,.975); ncp_upper = TNONCT(t,df,.025); d_lower = ncp_lower*sqrt((n1+n2)/(n1*n2)); d_upper = ncp_upper*sqrt((n1+n2)/(n1*n2)); output; run; proc print; var d d_lower d_upper; run ; Obs d d_lower d_upper 1 0.71937 0.27268 1.16212 Among Vermont school-children, girls’ GPA (M = 2.82, SD =.83, N = 33) was significantly higher than boys’ GPA (M = 2.24, SD =.81, N = 55), t(65.9) = 3.24, p =.002, d =.72. A 95% confidence interval for the difference between girls’ and boys’ mean GPA runs from.23 to.95 in raw score units and from.27 to 1.16 in standardized units. This is an almost large effect by Cohen’s guidelines.

Glass’ Delta Use the control group SD rather than pooled SD as the standardizer. When the control group SD is a better estimate of SD in the population of interest.

Point Biserial r Simply correlate group membership with the scores on the outcome variable. Or compute For the regression Score = a + b  Group, b = difference in group means =.588. standardized slope = This is a medium-sized effect by Cohen’s benchmarks. Hmmmm. It was large when we used g.

Eta-Squared For two mean comparisons, this is simply the squared point biserial r. Can be interpreted as a proportion of variance. CI: Conf-Interval-R2-Regr.sas or CI-R2-SPSS.zipConf-Interval-R2-Regr.sasCI-R2-SPSS.zip For our data,  2 =.11, CI.95 =.017,.240. Again, overestimation results from EV control.  2

Cohen’s Benchmarks for  and  2  –.1 is small but not trivial (r 2 = 1%) –.3 is medium (9%) –.5 is large (25%)  2 –.01 (1%) is small but not trivial –.06 is medium –.14 is large Note the inconsistency between these two sets of benchmarks.

Effect of n 1 /n 2 on d and r pb n 1 /n 2 = 1 M 1 = 5.5, SD 1 = 2.306, n 1 = 20, M 2 = 7.8, SD 2 = 2.306, n 2 = 20 –t(38) = 3.155, p =.003 –M 2 -M 1 = 2.30 d = 1.00 r pb =.456 Large effect

Effect of n 1 /n 2 on d and r pb n 1 /n 2 = 25 M 1 = 5.500, SD 1 = 2.259, n 1 = 100, M 2 = 7.775, SD 2 = 2.241, n 2 = 4 –t(102) = 1.976, p =.051 –M 2 -M 1 = 2.30 d = 1.01 r pb =.192 Large or (Small to Medium) Effect?

How does n 1 /n 2 affect r pb ? The point biserial r is the standardized slope for predicting the outcome variable from the grouping variable (coded 1,2). The unstandardized slope is the simple difference between group means. Standardize by multiplying by the SD of the grouping variable and dividing by the SD of the outcome variable. The SD of the grouping variable is a function of the sample sizes. For example, for N = 100, the SD of the grouping variable is –.503 when n1, n2 = 50, 50 –.473 when n1, n2 = 67, 33 –.302 when n1, n2 = 90, 10

Common Language Effect Size Statistic Find the lower-tailed p for For our data, p =.5, If you were to randomly select one boy & one girl. P(Girl GPA > Boy GPA) =.69. Odds =.69/(1-.69) = 2.23.

Two Related Samples Treat the data as if they were from independent samples when calculating d. If you standardize with the SD of the difference scores, you will overestimate . There is not available software to get an exact CI, and approximation procedures are only good with large data sets.

Correlation/Regression Even in complex research, many questions of great interest are addressed by zero-order correlation coefficients. Pearson r,  are already standardized. Cohen’s Benchmarks: –.1 = small, not trivial –.3 = medium –.5 = large

CI for , Correlation Model All variables random rather than fixed. Use R2 program to obtain CI for ρ 2.

R2 Program (Correlation Model)

Oh my, p <.05, but the 95% CI includes zero.

That’s better. The 90% CI does NOT include zero. Do note that the “lower bound” from the 95% CI is identical to the “lower limit” of the 90% CI.

CI for , Regression Model Y random, X fixed. Tedious by-hand method: See handout. SPSS and SAS programs for comparing Pearson correlations and OLS regression coefficients.SPSS and SAS programs for comparing Pearson correlations and OLS regression coefficients Web calculator at VassarVassar

Vassar Web App.

More Apps. R2 will not handle N > 5,000. Use this approximation instead: Conf-Interval-R2-Regr-LargeN.sas Conf-Interval-R2-Regr-LargeN.sas For Regression analysis (predictors are fixed, not random), use this: Conf-Interval-R2-Regr (SAS) or CI-R2-SPSS.zip (SPSS) Conf-Interval-R2-Regr CI-R2-SPSS.zip

What Confidence Coefficient Should I Use? For R 2, if you want the CI to be concordant with a test of the null that ρ 2 = 0, Use a CC of (1 - 2α), not (1 - α). Suppose you obtain r =.26 from n = 62 pairs of scores. F(1, 60) = 4.35. The p value is.041, significant with the usual.05 criterion.

Bias in Sample R 2 Sample R 2 overestimates population ρ 2. With large df numerator this can result in the CI excluding the point estimate. This should not happen if you use the shrunken R 2 as your point estimate.

Common Language Statistic Sample two cases (A & B) from paired X,Y. CL=P(Y A > Y B | X A > X B ) For one case, CL = P(Y > M y | X > M x )

r to CL r.00.10.30.50.70.90.99 CL50%53%60%67%75%86%96% Odds11.131.5236.124

Multiple R 2 Cohen:.02 = small (2% of variance).15 = medium (13% of variance).35 = large (26% of variance)

Partial and Semipartial

Example Grad GPA = GRE-Q, GRE-V, MAT, AR R 2 =.6405 For GRE-Q, pr 2 =.16023, sr 2 =.06860

One-Way ANOVA sdsdfsdsd

CI for  2 Conf-Interval-R2-Regr.sas CI-R2-SPSS at my SPSS Programs PageSPSS Programs Page CI.95 =.84,.96

22 Sample  2 overestimates population  2  2 is less biased For our data,  2 =.93.

Misinterpretation of Estimates of Proportion of Variance Explained 6% (Cohen’s benchmark for medium  2 sounds small. Aspirin study: Outcome = Heart Attack? –Preliminary results so dramatic study was stopped, placebo group told to take aspirin –Odds ratio = 1.83 –r 2 =.0011 Report r instead of r 2 ? r =.033

Extraneous Variable Control May artificially inflate strength of effect estimates (including d, r, , , etc.). Effect estimate from lab research >> that from field research. A variable that explains a large % of variance in highly controlled lab research may explain little out in the natural world.

Standardized Differences Between Means When k > 2 Plan focused contrasts between means or sets of means. Chose contrasts that best address the research questions posed. Do not need to do ANOVA. Report d for each contrast.

Standardized Differences Among Means in ANOVA Find an average value of d across pairs of means. Or the average standardized difference between group mean and grand mean. Steiger has proposed the RMSSE as the estimator.

Root Mean Square Standardized Effect k is the number of groups, M j is group mean, GM is grand mean. Standardizer is pooled SD, SQRT(MSE) For our data, RMSSE = 4.16. Godzilla. The population parameter is .

Place a CI on RMSSE http://www.statpower.net/Content/NDC/NDC.exe

Click Compute Get CI for lambda, the noncentrality parameter.

Transform CI to RMSSE The CI for lambda = 102.646, 480.288 CI for  = 2.616, 5.659.

CI → Hypothesis Test H 0 :  = 0.  cannot be less than 0, so a one-tailed p would be appropriate. Accordingly we find a 100(1-2α)% CI. For the usual.05 test, that is a 90% CI. If the CI excludes 0, then the ANOVA is significant.

Factorial Analysis of Variance For each effect,  2 = SS effect /SS total  2 : as before, use SS effect in place of SS AmongGroups Now suppose that one of the factors is experimental (present in the lab but not in the natural world). And the other is variable in both lab and the natural world.

Modify the Denominator of  2 Sex x Experimental Therapy ANOVA Sex is variable in lab and natural world Experimental Therapy only in lab Estimate effect of sex with variance due to Therapy and Interaction excluded from denominator. The resulting statistic is called partial eta- squared.

Partial  2 When estimating Therapy and Interaction, one should not remove effect of Sex from the denominator.

Explaining More Than 100% of the Variance Pierce, Block, and Aguinis (2004) Many articles, in good journals, where partial  2 was wrongly identified as  2 Even when total variance explained exceeded 100%. In one case, 204%. Why don’t authors, reviewers, and editors notice such foolishness?

CI for  2 or Partial  2 Use Conf-Interval-R2-Regr.sasConf-Interval-R2-Regr.sas Use the ANOVA F to get CI for partial  2 To get CI for  2 will need compute a modified F. See Two-Way Independent Samples ANOVA on SASTwo-Way Independent Samples ANOVA on SAS

Contingency Table Analysis 2 x 2 table: Phi = Pearson r between dichotomous variables. Cramér’s φ = similar, for a x b tables where a and/or b > 2. Odds ratio: (odds of A|B)/(odds of A| not B)

Small Effect Phi =.1 Odds ratio = (55/45)  (45/55) = 1.49

Medium Effect Phi =.3 Odds ratio = (65/35)  (35/65) = 3.45

Large Effect Phi =.5 Odds ratio = (75/25)  (25/75) = 9.00

Phi and Odds Ratios The marginals were uniform in the contingency tables above. For a fixed odds ratio, phi decreases as the marginals deviate from uniform. See http://core.ecu.edu/psyc/wuenschk/StatHel p/Phi-OddsRatio.doc http://core.ecu.edu/psyc/wuenschk/StatHel p/Phi-OddsRatio.doc

CI for Odds Ratio Conduct a binary logistic regression and ask for confidence intervals for the odds ratios.

Multivariate Analysis Most provide statistics similar to r 2 and  2 Canonical correlation/regression –For each root get a canonical r –Is the corr between a weighted combination of the Xs and a weighted combination of the Ys Other analyses are just simplifications or special cases of canonical corr/regr.

MANOVA and DFA: Canonical r For each root get a squared canonical r. There will be one root for each treatment df. If you were to use ANOVA to compare the groups on that root, this canonical r 2 would be

MANOVA and DFA: 1 -  For each effect, Wilks  is, basically, Accordingly, you can compute a multivariate  2 as 1 - . If k = 2, 1 -  is the canonical r 2.

Binary Logistic Regression Cox & Snell R 2 –Has an upper boundary less than 1. Nagelkerke R 2 –Has an upper boundary of 1. Classification results speak to magnitude of omnibus effect. Odds ratios speak to magnitude of partial effects.

Comparing Predictors’ Contributions It may help to standardize continuous predictors prior to computing odds ratios Consider these results

Relative Contributions of the Predictors The event being predicted is retention in ECU’s engineering program. Each one point increase in HS GPA multiplies the odds of retention by 3.656. A one point inrease in Quantitative SAT increases the odds by only 1.006 But a one point increase in GPA is a helluva lot larger than a one point increase in SAT.

Standardized Predictors Here we see that the relative contributions of the three predictors do not differ much.

Why Confidence Intervals? They are not often reported. So why do I preach their usefulness? IMHO, they give one everything given by a hypothesis test p AND MORE. Let me illustrate, using confidence intervals for 

Significant Results, CI =.01,.03 We can be confident of the direction of the effect. We can also be confident that the size of the effect is so small that it might as well be zero. “Significant” in this case is a very poor descriptor of the effect.

Significant Results, CI =.02,.84 We can be confident of the direction of the effect & it is probably not trivial in magnitude, But it is estimated with little precision. Could be trivial, could be humongous. Need more data to get more precise estimation of size of effect.

Significant Results, CI =.51,.55 We can be confident of the direction of the effect & that it is large in magnitude (in most contexts). We have great precision.

Not Significant, CI = -.46, +.43 Effect could be anywhere from large in one direction to large in the other direction. This tells us we need more data (or other power-enhancing characteristics).

Not Significant, CI = -.74, +.02 Cannot be very confident about the direction of the effect, but It is likely that is negative. Need more data/power.

Not Significant, CI = -.02, +.01 A very impressive result. Tells us that the effect is of trivial magnitude. Suppose X = generic vs. brand-name drug Y = response to drug. We have established bioequivalence.

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