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Bayesian style dependent tests

Bayesian style dependent tests. Greg Francis. PSY 626: Bayesian Statistics for Psychological Science Fall 2018 Purdue University. Weapon priming. Anderson, Benjamin, & Bartholow (1998): 32 subjects read an aggression-related word (injure, shatter) out loud as fast as possible

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Bayesian style dependent tests

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  1. Bayesian style dependent tests Greg Francis PSY 626: Bayesian Statistics for Psychological Science Fall 2018 Purdue University

  2. Weapon priming • Anderson, Benjamin, & Bartholow (1998): 32 subjects read an aggression-related word (injure, shatter) out loud as fast as possible • After presentation of a weapon word (shotgun, grenade) • After presentation of a neutral word (rabbit fish) • There were additional trials with non-aggression-related words being read (we ignore them) • Our data is the mean response time for wording reading based on whether the first (prime) word is a weapon word or a neutral word • Follow along by downloading “WeaponPrime” and “WeaponPrime1.R” from the class web site

  3. Ignore dependencies? • WPdata<- read.csv(file="WeaponPrime.csv",header=TRUE,stringsAsFactors=TRUE) • traditional1 <- t.test(Time ~ Prime, data=WPdata, var.equal=TRUE) • print(traditional1) Two Sample t-test data: Time by Prime t = 0.54967, df = 62, p-value = 0.5845 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -18.86865 33.18115 sample estimates: mean in group Neutral mean in group Weapon 416.3438 409.1875

  4. Standard analysis • # Run traditional dependent t-test • traditional2 <- t.test(Time ~ Prime, data=WPdata, paired=TRUE) • print(traditional2) • Note: t-value and p-value change, but sample means do not • Difference is: 7.1563 • Note: confidence interval of difference changes a lot Paired t-test data: Time by Prime t = 2.2239, df = 31, p-value = 0.03358 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.593243 13.719257 sample estimates: mean of the differences 7.15625

  5. Regression analysis • # Compare to traditional independent linear regression • print(summary(lm(Time~Prime, data=WPdata))) • Prints out a table of estimates of Intercept (Neutral prime) and deviation for Weapon prime (difference of means) Call: lm(formula = Time ~ Prime, data = WPdata) Residuals: Min 1Q Median 3Q Max -123.187 -27.555 4.734 27.195 127.656 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 416.344 9.206 45.23 <2e-16 *** PrimeWeapon -7.156 13.019 -0.55 0.585 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘’ 1 Residual standard error: 52.08 on 62 degrees of freedom Multiple R-squared: 0.00485, Adjusted R-squared: -0.0112 F-statistic: 0.3021 on 1 and 62 DF, p-value: 0.5845

  6. Regression analysis • # Compare to traditional dependent linear regression • library(lme4) • print(lmer(Time ~ Prime + (1 | Subject), data = WPdata)) • Prints out a table of estimates of Intercept (Neutral prime) and deviation for Weapon prime (difference of means) Linear mixed model fit by REML ['lmerMod'] Formula: Time ~ Prime + (1 | Subject) Data: WPdata REML criterion at convergence: 606.886 Random effects: Groups Name Std.Dev. Subject (Intercept) 50.46 Residual 12.87 Number of obs: 64, groups: Subject, 32 Fixed Effects: (Intercept) PrimeWeapon 416.344 -7.156

  7. Bayesian variation of t-test Call: lm(formula = Time ~ Prime, data = WPdata) Residuals: Min 1Q Median 3Q Max -123.187 -27.555 4.734 27.195 127.656 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 416.344 9.206 45.23 <2e-16 *** PrimeWeapon -7.156 13.019 -0.55 0.585 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘’ 1 Residual standard error: 52.08 on 62 degrees of freedom Multiple R-squared: 0.00485, Adjusted R-squared: -0.0112 F-statistic: 0.3021 on 1 and 62 DF, p-value: 0.5845 • Treat data as independent • library(brms) • model1 = brm(Time ~ Prime, data = WPdata, iter = 2000, warmup = 200, chains = 3, thin = 2 ) Family: gaussian Links: mu = identity; sigma = identity Formula: Time ~ Prime Data: WPdata (Number of observations: 64) Samples: 3 chains, each with iter = 2000; warmup = 200; thin = 2; total post-warmup samples = 2700 Population-Level Effects: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat Intercept 416.68 9.12 398.96 434.34 2700 1.00 PrimeWeapon -7.65 13.40 -33.82 19.07 2700 1.00 Family Specific Parameters: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sigma 52.50 4.63 44.24 62.25 2495 1.00

  8. Bayesian variation of t-test Linear mixed model fit by REML ['lmerMod'] Formula: Time ~ Prime + (1 | Subject) Data: WPdata REML criterion at convergence: 606.886 Random effects: Groups Name Std.Dev. Subject (Intercept) 50.46 Residual 12.87 Number of obs: 64, groups: Subject, 32 Fixed Effects: (Intercept) PrimeWeapon 416.344 -7.156 • Treat data as dependent (different intercept for each subject) • model2 = brm(Time ~ Prime + (1 | Subject), data = WPdata, iter = 2000, warmup = 200, chains = 3, thin = 2 ) Family: gaussian Links: mu = identity; sigma = identity Formula: Time ~ Prime + (1 | Subject) Data: WPdata (Number of observations: 64) Samples: 3 chains, each with iter = 2000; warmup = 200; thin = 2; total post-warmup samples = 2700 Group-Level Effects: ~Subject (Number of levels: 32) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sd(Intercept) 51.18 6.63 39.92 66.44 488 1.00 Population-Level Effects: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat Intercept 416.92 8.82 399.12 434.25 272 1.00 PrimeWeapon -7.16 3.43 -13.75 -0.45 2429 1.00 Family Specific Parameters: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sigma 13.43 1.80 10.49 17.49 1612 1.00

  9. Model comparison • Does the independent or dependent model better fit the data? > loo(model1, model2) LOOIC SE model1 691.99 13.17 model2 554.79 9.50 model1 - model2 137.20 12.87 Warning messages: 1: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL' 2: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL' 3: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL' 4: Found 16 observations with a pareto_k > 0.7 in model 'model2'. With this many problematic observations, it may be more appropriate to use 'kfold' with argument 'K = 10' to perform 10-fold cross-validation rather than LOO.

  10. Model comparison • Does the independent or dependent model better fit the data? • Should be no surprise! • kfold(model1, model2) • …… • KFOLDIC SE • model1 691.78 12.91 • model2 563.56 10.25 • model1 - model2 128.22 13.42 • There were 50 or more warnings (use warnings() to see the first 50)

  11. Null model • No effect of prime, different intercepts for different subjects • model3 = brm(Time ~ 1 + (1 | Subject), data = WPdata, iter = 2000, warmup = 200, chains = 3, thin = 2 ) • > summary(model3) • Family: gaussian • Links: mu = identity; sigma = identity • Formula: Time ~ 1 + (1 | Subject) • Data: WPdata (Number of observations: 64) • Samples: 3 chains, each with iter = 2000; warmup = 200; thin = 2; • total post-warmup samples = 2700 • Group-Level Effects: • ~Subject (Number of levels: 32) • Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat • sd(Intercept) 51.62 6.77 39.98 66.46 539 1.01 • Population-Level Effects: • Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat • Intercept 413.64 9.26 396.23 432.35 355 1.01 • Family Specific Parameters: • Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat • sigma 14.15 1.85 10.98 18.07 1591 1.00

  12. Model comparison • Does the null (model3) or two-mean (model2) model better fit the data? • Try: • loo(model2, model3) • kfold(model2, model3) • …… • KFOLDIC SE • model2 571.58 11.75 • model3 574.79 13.02 • model2 - model3 -3.22 10.63

  13. Dependent ANOVA • ADHD Treatment effects • Children (n=24) given different dosages of drug MPH (methylphenidate) over different weeks • Measured ability to delay impulsive behavior responses (wait long enough to press a key to get a “star”) • Delay of Gratification task

  14. Dependent ANOVA • Standard dependent ANOVA

  15. Regression analysis Linear mixed model fit by REML ['lmerMod'] Formula: CorrectResponses ~ Dosage + (1 | SubjectID) Data: ATdata REML criterion at convergence: 658.3 Scaled residuals: Min 1Q Median 3Q Max -2.3690 -0.5831 0.1585 0.5923 1.9393 Random effects: Groups Name Variance Std.Dev. SubjectID (Intercept) 89.56 9.464 Residual 35.89 5.991 Number of obs: 96, groups: SubjectID, 24 Fixed effects: Estimate Std. Error t value (Intercept) 39.75000 2.28635 17.386 DosageD15 -0.08333 1.72948 -0.048 DosageD30 4.58333 1.72948 2.650 DosageD60 4.95833 1.72948 2.867 Correlation of Fixed Effects: (Intr) DsgD15 DsgD30 DosageD15 -0.378 DosageD30 -0.378 0.500 DosageD60 -0.378 0.500 0.500 • # Compare to traditional dependent linear regression • library(lme4) • print(summary( lmer(CorrectResponses ~ Dosage + (1 | SubjectID), data = ATdata)) ) • Prints out a table of estimates of Intercept (D0 dosage) and deviation for other Dosages • There is some debate about degrees of freedom and computing p-values

  16. Bayesian regression Linear mixed model fit by REML ['lmerMod'] Formula: CorrectResponses ~ Dosage + (1 | SubjectID) Data: ATdata REML criterion at convergence: 658.3 Scaled residuals: Min 1Q Median 3Q Max -2.3690 -0.5831 0.1585 0.5923 1.9393 Random effects: Groups Name Variance Std.Dev. SubjectID (Intercept) 89.56 9.464 Residual 35.89 5.991 Number of obs: 96, groups: SubjectID, 24 Fixed effects: Estimate Std. Error t value (Intercept) 39.75000 2.28635 17.386 DosageD15 -0.08333 1.72948 -0.048 DosageD30 4.58333 1.72948 2.650 DosageD60 4.95833 1.72948 2.867 Correlation of Fixed Effects: (Intr) DsgD15 DsgD30 DosageD15 -0.378 DosageD30 -0.378 0.500 DosageD60 -0.378 0.500 0.500 Family: gaussian Links: mu = identity; sigma = identity Formula: CorrectResponses ~ Dosage + (1 | SubjectID) Data: ATdata (Number of observations: 96) Samples: 3 chains, each with iter = 2000; warmup = 200; thin = 2; total post-warmup samples = 2700 Group-Level Effects: ~SubjectID (Number of levels: 24) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sd(Intercept) 9.73 1.63 7.02 13.49 1091 1.00 Population-Level Effects: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat Intercept 39.63 2.29 35.27 44.30 1213 1.00 DosageD15 -0.08 1.77 -3.44 3.25 2424 1.00 DosageD30 4.59 1.74 1.34 8.05 2576 1.00 DosageD60 4.93 1.76 1.48 8.46 2306 1.00 Family Specific Parameters: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sigma 6.09 0.53 5.19 7.19 1977 1.00 Samples were drawn using sampling(NUTS). For each parameter, Eff.Sample is a crude measure of effective sample size, and Rhat is the potential scale reduction factor on split chains (at convergence, Rhat = 1). • model1 = brm(CorrectResponses ~ Dosage + (1 |SubjectID), data = ATdata, iter = 2000, warmup = 200, chains = 3, thin = 2 ) • # print out summary of model • print(summary(model1))

  17. Bayesian regression • Check chains • plot(model1)

  18. Bayesian regression • Plot marginals • dev.new() • plot(marginal_effects(model1), points = TRUE)

  19. Bayesian regression • Consider a null model (no differences between means) • model2 = brm(CorrectResponses ~ 1 + (1 |SubjectID), data = ATdata, iter = 2000, warmup = 200, chains = 3, thin = 2 ) Family: gaussian Links: mu = identity; sigma = identity Formula: CorrectResponses ~ 1 + (1 | SubjectID) Data: ATdata (Number of observations: 96) Samples: 3 chains, each with iter = 2000; warmup = 200; thin = 2; total post-warmup samples = 2700 Group-Level Effects: ~SubjectID (Number of levels: 24) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sd(Intercept) 9.67 1.68 6.86 13.49 1288 1.00 Population-Level Effects: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat Intercept 41.92 2.10 37.90 46.12 1181 1.00 Family Specific Parameters: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sigma 6.61 0.56 5.63 7.76 2084 1.00 Samples were drawn using sampling(NUTS). For each parameter, Eff.Sample is a crude measure of effective sample size, and Rhat is the potential scale reduction factor on split chains (at convergence, Rhat = 1).

  20. Bayesian regression > loo(model1, model2, reloo=TRUE) No problematic observations found. Returning the original 'loo' object. 2 problematic observation(s) found. The model will be refit 2 times. Fitting model 1 out of 2 (leaving out observation 2) Start sampling Gradient evaluation took 2.7e-05 seconds 1000 transitions using 10 leapfrog steps per transition would take 0.27 seconds. Adjust your expectations accordingly! Elapsed Time: 0.365972 seconds (Warm-up) 0.257485 seconds (Sampling) 0.623457 seconds (Total) …….. LOOIC SE model1 647.94 13.74 model2 658.79 12.98 model1 - model2 -10.86 8.53 > loo(model1, model2) LOOIC SE model1 647.94 13.74 model2 659.03 13.07 model1 - model2 -11.10 8.56 Warning message: Found 2 observations with a pareto_k > 0.7 in model 'model2'. It is recommended to set 'reloo = TRUE' in order to calculate the ELPD without the assumption that these observations are negligible. This will refit the model 2 times to compute the ELPDs for the problematic observations directly. • Compare null (model2) and full (model1) models

  21. Standard ANOVA • Contrasts are t-tests (or F-tests, if you like)

  22. Bayesian Regression • Contrasts: Easiest to look at posteriors • Easy for comparisons to D0 (Intercept) > post<-posterior_samples(model1) > > # What is the probability that mean for D15 is larger than mean for D0? > cat("Probability CorrectResponses mean D15 is larger than mean D0 = ", length(post$b_DosageD15[post$b_DosageD15 >0])/length(post$b_DosageD15) ) Probability CorrectResponses mean D15 is larger than mean D0 = 0.4833333 > cat("Probability CorrectResponses mean D30 is larger than mean D0 = ", length(post$b_DosageD30[post$b_DosageD30 >0])/length(post$b_DosageD30) ) Probability CorrectResponses mean D30 is larger than mean D0 = 0.9937037 > cat("Probability CorrectResponses mean D60 is larger than mean D0 = ", length(post$b_DosageD60[post$b_DosageD60 >0])/length(post$b_DosageD60) ) Probability CorrectResponses mean D60 is larger than mean D0 = 0.9959259

  23. Bayesian Regression • Contrasts: Easiest to look at posteriors • A bit complicated for other comparisons # What is the probability that mean for D60 is larger than mean for D15? meanD15 <- post$b_Intercept + post$b_DosageD15 meanD60 <- post$b_Intercept + post$b_DosageD60 cat("Probability CorrectResponses mean D60 is larger than mean D15 = ", sum(meanD60 > meanD15)/length(meanD60) ) > Probability CorrectResponses mean D60 is larger than mean D15 = 0.9966667 meanD30 <- post$b_Intercept + post$b_DosageD30 > > cat("Probability CorrectResponses mean D60 is larger than mean D30 = ", sum(meanD60 > meanD30)/length(meanD60) ) Probability CorrectResponses mean D60 is larger than mean D30 = 0.5811111

  24. Setting priors • It is not always obvious what can be set • brms has a nice function get_prior() > get_prior(CorrectResponses ~ Dosage + (1 |SubjectID), data = ATdata) prior class coef group resp dpar nlpar bound 1 b 2 b DosageD15 3 b DosageD30 4 b DosageD60 5 student_t(3, 39, 10) Intercept 6 student_t(3, 0, 10) sd 7 sd SubjectID 8 sd Intercept SubjectID 9 student_t(3, 0, 10) sigma >

  25. Setting priors Family: gaussian Links: mu = identity; sigma = identity Formula: CorrectResponses ~ Dosage + (1 | SubjectID) Data: ATdata (Number of observations: 96) Samples: 3 chains, each with iter = 2000; warmup = 200; thin = 2; total post-warmup samples = 2700 Group-Level Effects: ~SubjectID (Number of levels: 24) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sd(Intercept) 9.73 1.63 7.02 13.49 1091 1.00 Population-Level Effects: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat Intercept 39.63 2.29 35.27 44.30 1213 1.00 DosageD15 -0.08 1.77 -3.44 3.25 2424 1.00 DosageD30 4.59 1.74 1.34 8.05 2576 1.00 DosageD60 4.93 1.76 1.48 8.46 2306 1.00 Family Specific Parameters: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sigma 6.09 0.53 5.19 7.19 1977 1.00 • Let’s set an informative (bad) prior on the slopes • model3 = brm(CorrectResponses ~ Dosage + (1 |SubjectID), data = ATdata, iter = 2000, warmup = 200, chains = 3, thin = 2, prior = c( prior(normal(10, 1), class = "b")) ) Family: gaussian Links: mu = identity; sigma = identity Formula: CorrectResponses ~ Dosage + (1 | SubjectID) Data: ATdata (Number of observations: 96) Samples: 3 chains, each with iter = 2000; warmup = 200; thin = 2; total post-warmup samples = 2700 Group-Level Effects: ~SubjectID (Number of levels: 24) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sd(Intercept) 9.56 1.64 6.90 13.35 1102 1.00 Population-Level Effects: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat Intercept 35.03 2.10 30.85 39.18 1223 1.00 DosageD15 8.13 0.90 6.40 9.91 2179 1.00 DosageD30 9.70 0.87 8.09 11.43 2210 1.00 DosageD60 9.81 0.87 8.15 11.52 1922 1.00 Family Specific Parameters: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sigma 6.90 0.61 5.83 8.19 2256 1.00

  26. Setting priors Family: gaussian Links: mu = identity; sigma = identity Formula: CorrectResponses ~ Dosage + (1 | SubjectID) Data: ATdata (Number of observations: 96) Samples: 3 chains, each with iter = 2000; warmup = 200; thin = 2; total post-warmup samples = 2700 Group-Level Effects: ~SubjectID (Number of levels: 24) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sd(Intercept) 9.73 1.63 7.02 13.49 1091 1.00 Population-Level Effects: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat Intercept 39.63 2.29 35.27 44.30 1213 1.00 DosageD15 -0.08 1.77 -3.44 3.25 2424 1.00 DosageD30 4.59 1.74 1.34 8.05 2576 1.00 DosageD60 4.93 1.76 1.48 8.46 2306 1.00 Family Specific Parameters: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sigma 6.09 0.53 5.19 7.19 1977 1.00 • Let’s set a crazy prior on just one slope • model4 = brm(CorrectResponses ~ Dosage + (1 |SubjectID), data = ATdata, iter = 2000, warmup = 200, chains = 3, thin = 2, prior = c( prior(normal(20, 1), class = "b", coef="DosageD60")) ) Family: gaussian Links: mu = identity; sigma = identity Formula: CorrectResponses ~ Dosage + (1 | SubjectID) Data: ATdata (Number of observations: 96) Samples: 3 chains, each with iter = 2000; warmup = 200; thin = 2; total post-warmup samples = 2700 Group-Level Effects: ~SubjectID (Number of levels: 24) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sd(Intercept) 9.26 1.63 6.59 13.01 1553 1.00 Population-Level Effects: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat Intercept 33.21 2.23 28.72 37.62 1665 1.00 DosageD15 6.21 2.08 2.30 10.35 2327 1.00 DosageD30 10.90 2.05 6.94 14.93 2176 1.00 DosageD60 17.63 0.98 15.68 19.53 2448 1.00 Family Specific Parameters: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sigma 8.05 0.76 6.71 9.64 2059 1.00

  27. Bayesian regression > loo(model1, model2, model3, model4, reloo=TRUE) No problematic observations found. Returning the original 'loo' object. 2 problematic observation(s) found. The model will be refit 2 times. Fitting model 1 out of 2 (leaving out observation 2) Start sampling Gradient evaluation took 9.5e-05 seconds 1000 transitions using 10 leapfrog steps per transition would take 0.95 seconds. Adjust your expectations accordingly! Elapsed Time: 0.285336 seconds (Warm-up) 0.248522 seconds (Sampling) 0.533858 seconds (Total) …….. LOOIC SE model1 647.94 13.74 model2 658.78 12.98 model3 669.23 11.38 model4 698.55 13.77 model1 - model2 -10.84 8.53 model1 - model3 -21.29 9.67 model1 - model4 -50.61 12.17 model2 - model3 -10.45 12.02 model2 - model4 -39.77 17.23 model3 - model4 -29.32 9.55 > loo(model1, model2, model3, model4) LOOIC SE model1 647.94 13.74 model2 659.03 13.07 model3 669.11 11.33 model4 698.55 13.77 model1 - model2 -11.10 8.56 model1 - model3 -21.18 9.69 model1 - model4 -50.61 12.17 model2 - model3 -10.08 12.07 model2 - model4 -39.52 17.28 model3 - model4 -29.43 9.58 Warning messages: 1: Found 2 observations with a pareto_k > 0.7 in model 'model2'. It is recommended to set 'reloo = TRUE' in order to calculate the ELPD without the assumption that these observations are negligible. This will refit the model 2 times to compute the ELPDs for the problematic observations directly. 2: Found 1 observations with a pareto_k > 0.7 in model 'model3'. It is recommended to set 'reloo = TRUE' in order to calculate the ELPD without the assumption that these observations are negligible. This will refit the model 1 times to compute the ELPDs for the problematic observations directly. • Compare all our models • Different priors are different models!

  28. Regression • We may doing something fundamentally wrong here • We are treating dosage as a categorical variable, but it is really a quantitative variable • Maybe we should really be doing a proper regression instead of coercing it into an ANOVA design

  29. Frequentist regression • library(lme4) • print(lmer(CorrectResponses ~ DosageNumber + (1 | SubjectID), data = ATdata)) Linear mixed model fit by REML ['lmerMod'] Formula: CorrectResponses ~ DosageNumber + (1 | SubjectID) Data: ATdata REML criterion at convergence: 675.6007 Random effects: Groups Name Std.Dev. SubjectID (Intercept) 9.451 Residual 6.069 Number of obs: 96, groups: SubjectID, 24 Fixed Effects: (Intercept) DosageNumber 39.64167 0.09421

  30. Bayesian regression • library(brms) • model5 = brm(CorrectResponses ~ DosageNumber + (1 |SubjectID), data = ATdata, iter = 2000, warmup = 200, chains = 3, thin = 2 ) Family: gaussian Links: mu = identity; sigma = identity Formula: CorrectResponses ~ DosageNumber + (1 | SubjectID) Data: ATdata (Number of observations: 96) Samples: 3 chains, each with iter = 2000; warmup = 200; thin = 2; total post-warmup samples = 2700 Group-Level Effects: ~SubjectID (Number of levels: 24) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sd(Intercept) 9.69 1.59 7.11 13.25 1193 1.00 Population-Level Effects: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat Intercept 39.38 2.18 35.04 43.62 902 1.00 DosageNumber 0.09 0.03 0.04 0.15 2394 1.00 Family Specific Parameters: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sigma 6.16 0.54 5.20 7.32 2010 1.00 Linear mixed model fit by REML ['lmerMod'] Formula: CorrectResponses ~ DosageNumber + (1 | SubjectID) Data: ATdata REML criterion at convergence: 675.6007 Random effects: Groups Name Std.Dev. SubjectID (Intercept) 9.451 Residual 6.069 Number of obs: 96, groups: SubjectID, 24 Fixed Effects: (Intercept) DosageNumber 39.64167 0.09421

  31. Bayesian regression • library(brms) • model1 = brm(CorrectResponses ~ DosageNumber + (1 |SubjectID), data = ATdata, iter = 2000, warmup = 200, chains = 3, thin = 2 )

  32. Bayesian ANOVA vs. regression • loo(model1, model5, reloo=TRUE) • I re-ran model1, so the LOO value is a bit different from what we had previously • I am not sure this kind of comparison is appropriate because we are technically using different independent variables (it feels the same, though) • Hardly any difference between the two models • Which makes sense given the data LOOIC SE model1 648.57 13.57 model5 648.24 14.02 model1 - model5 0.33 4.50

  33. Activity • Find the possible priors for the linear regression model • What would be reasonable priors? • Implement them and run the model • Discuss

  34. Conclusions • Dependent tests • Pretty straightforward once you get the notation straight • Which is really just the notation of regression • Something similar to contrasts is done by looking at the posteriors • No messy hypothesis testing

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