Dear Srinidhi, You are trying to fit 1 random intercept and 2 random slopes per individual, while you have at most 3 observations per individual. You simply don't have enough data to fit the random slopes. Reduce the random part to (1|ID).
Best regards, Thierry ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE AND FOREST Team Biometrie & Kwaliteitszorg / Team Biometrics & Quality Assurance thierry.onkel...@inbo.be Havenlaan 88 bus 73, 1000 Brussel *Postadres:* Koning Albert II-laan 15 bus 186, 1210 Brussel *Poststukken die naar dit adres worden gestuurd, worden ingescand en digitaal aan de geadresseerde bezorgd. Zo kan de Vlaamse overheid haar dossiers volledig digitaal behandelen. Poststukken met de vermelding ‘vertrouwelijk’ worden niet ingescand, maar ongeopend aan de geadresseerde bezorgd.* www.inbo.be /////////////////////////////////////////////////////////////////////////////////////////// To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey /////////////////////////////////////////////////////////////////////////////////////////// <https://www.inbo.be> Op ma 6 mei 2024 om 01:59 schreef Srinidhi Jayakumar via R-sig-mixed-models <r-sig-mixed-mod...@r-project.org>: > I am running a multilevel growth curve model to examine predictors of > social anhedonia (SA) trajectory through ages 12, 15 and 18. SA is a > continuous numeric variable. The age variable (Index1) has been coded as 0 > for age 12, 1 for age 15 and 2 for age 18. I am currently using a time > varying predictor, stress (LSI), which was measured at ages 12, 15 and 18, > to examine whether trajectory/variation in LSI predicts difference in SA > trajectory. LSI is a continuous numeric variable and was grand-mean > centered before using in the models. The data has been converted to long > format with SA in 1 column, LSI in the other, ID in another, and age in > another column. I used the code below to run my model using lmer. However, > I get the following error. Please let me know how I can solve this error. > Please note that I have 50% missing data in SA at age 12. > modelLSI_maineff_RE <- lmer(SA ~ Index1* LSI+ (1 + Index1+LSI |ID), data = > LSIDATA, control = lmerControl(optimizer ="bobyqa"), REML=TRUE) > summary(modelLSI_maineff_RE) > Error: number of observations (=1080) <= number of random effects (=1479) > for term (1 + Index1 + LSI | ID); the random-effects parameters and the > residual variance (or scale parameter) are probably unidentifiable > > I did test the within-person variance for the LSI variable and the > within-person variance is significant from the Greenhouse-Geisser, > Hyunh-Feidt tests. > > I also tried control = lmerControl(check.nobs.vs.nRE = "ignore") which gave > me the following output. modelLSI_maineff_RE <- lmer(SA ~ Index1* LSI+ (1 + > Index1+LSI |ID), data = LSIDATA, control = lmerControl(check.nobs.vs.nRE = > "ignore", optimizer ="bobyqa", check.conv.singular = .makeCC(action = > "ignore", tol = 1e-4)), REML=TRUE) > > summary(modelLSI_maineff_RE) > Linear mixed model fit by REML. t-tests use Satterthwaite's method > ['lmerModLmerTest'] > Formula: SA ~ Index1 * LSI + (1 + Index1 + LSI | ID) > Data: LSIDATA > Control: lmerControl(check.nobs.vs.nRE = "ignore", optimizer = "bobyqa", > check.conv.singular = .makeCC(action = "ignore", tol = 1e-04)) > > REML criterion at convergence: 7299.6 > > Scaled residuals: > Min 1Q Median 3Q Max > -2.7289 -0.4832 -0.1449 0.3604 4.5715 > > Random effects: > Groups Name Variance Std.Dev. Corr > ID (Intercept) 30.2919 5.5038 > Index1 2.4765 1.5737 -0.15 > LSI 0.1669 0.4085 -0.23 0.70 > Residual 24.1793 4.9172 > Number of obs: 1080, groups: ID, 493 > > Fixed effects: > Estimate Std. Error df t value Pr(>|t|) > (Intercept) 24.68016 0.39722 313.43436 62.133 < 2e-16 *** > Index1 0.98495 0.23626 362.75018 4.169 3.83e-05 *** > LSI -0.05197 0.06226 273.85575 -0.835 0.4046 > Index1:LSI 0.09797 0.04506 426.01185 2.174 0.0302 * > Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 > > Correlation of Fixed Effects: > (Intr) Index1 LSI > Index1 -0.645 > LSI -0.032 0.057 > Index1:LSI 0.015 0.037 -0.695 > > I am a little vary of the output still as the error states that I have > equal observations as the number of random effects (i.e., 3 observations > per ID and 3 random effects). Hence, I am wondering whether I can simplify > the model as either of the below models and choose the one with the > best-fit statistics: > > modelLSI2 <- lmer(SA ~ Index1* LSI+ (1 |ID)+ (Index1+LSI -1|ID),data = > LSIDATA, control = lmerControl(optimizer ="bobyqa"), REML=TRUE) *OR* > > modelLSI3 <- lmer(SA ~ Index1* LSI+ (1+LSI |ID),data = LSIDATA, control = > lmerControl(optimizer ="bobyqa"), REML=TRUE) [image: example of dataset] > <https://i.sstatic.net/JcRKS2C9.png> > > [[alternative HTML version deleted]] > > _______________________________________________ > r-sig-mixed-mod...@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models > [[alternative HTML version deleted]] ______________________________________________ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
