#### 1. Setup (You Try It) #### ## to start with, we will load a package for data management ## and load a package for fitting Linear Mixed effects Models (LMMs) ## recall that loading a package is like opening an app ## and you need to repeat this process each time you start up R ## if everything installed successfully already ## this code should work and return no errors ## if this does not work, try to install it first ## by uncommenting the install packages code # install.packages("data.table", dependencies = TRUE) # install.packages("lme4", dependencies = TRUE) # install.packages("ggplot2", dependencies = TRUE) library(data.table) library(lme4) library(ggplot2) ## load the data (note this is already an R dataset) ## so we use a new function readRDS() ## make sure that the data file is located in ## your project directory d <- readRDS("aces_daily_sim_processed.RDS") ## see the dimenions of the data (rows columns) ## should be: 6927 56 dim(d) ## see the names of the variables in the dataset names(d) ## linear regression with one continuous predictor ## using a dataset on 32 cars built into R m1 <- lm(hp ~ mpg, data = mtcars) summary(m1) ## linear regression with one binary (categorical) predictor m2 <- lm(hp ~ vs, data = mtcars) summary(m2) ## linear regression with two predictors m3 <- lm(hp ~ mpg + vs, data = mtcars) summary(m3) ## to visualize the multiple assessments per person ## and the variation in means we will focus ## one one variable, positive affect (PosAff) for just ## four individuals (IDs 1, 2, 6, and 9) ## first we subset the data to only rows ## where UserID is in IDs 1, 2, 6 or 9 ## now that we use quotes around the IDs ## because they are not stored as numbers but ## as discrete categorical values (i.e., IDs are not ## really a continuous "number", they represent discrete people). justfour <- d[UserID %in% c("1", "2", "6", "9")] ## now we can plot it ## (do not worry about understanding graphing, we will learn that later) ## Note the warnings about non-finite values are because of some ## missing data, you can ignore these safely, if you expect ## some missing data (we do) ggplot(justfour, aes(UserID, PosAff)) + geom_jitter() + stat_summary(fun.data = mean_cl_normal, colour = "blue", size = 1) #### 2. Individual, Random, Fixed Effects (Demonstration) #### ## individual ## fit a linear regression of PosAff on only an intercept and have R ## save the intercept BY UserID m.individual <- d[, as.list(coef(lm(PosAff ~ 1))), by = UserID] ## view the first few rows of this new individual mean dataset ## remember that with only an intercept in the model ## the intercept = individual mean head(m.individual) ## histogram of individual intercepts ## note a few near the maximum positive affect: 5 ggplot(m.individual, aes(`(Intercept)`)) + geom_histogram() ## fit an intercept only linear model (regression) ## and store the results in an object, m1lr m1lr <- lm(PosAff ~ 1, data = d) ## create a summary of our linear regression model summary(m1lr) ## the intercept is the overall mean of positive affect ## we can easily verify this mean(d$PosAff, na.rm = TRUE) ## the linear regression model does not allow any individual differences ## now we will fit a *l*inear *m*ixed *e*ffects *r*egression ## model (LMM) using the lmer() function ## again we only have an intercept, but this time we include ## the intercept as a fixed effect (which will capture the mean) ## and as a random effect (which will capture the standard deviation) m1lmm <- lmer(PosAff ~ 1 + (1 | UserID), data = d) ## generate a summary ## NOTE: in class walk through interpretation of output summary(m1lmm) ## this model shows the assumed normal distribution of the ## individual means of positive affect are summarized by ## M = 2.67866 and SD = 0.7930 ## we can compare these values to the values we would ## obtain if we simply calculated and summarized the individual means ## we can see the values are very close ## because of some differences in the estimation, these will not always ## match exactly, but they are typically close mean(m.individual$`(Intercept)`, na.rm = TRUE) sd(m.individual$`(Intercept)`, na.rm = TRUE) ## we also can use the variance (not standard deviation) ## estimates from the random effects section to estimate ## the intraclass correlation coefficient (ICC) for positive ## affect to provide a descriptive statistic for how much ## variance in positive affect occurs between people ## relative to its total variance ## ICC for positive affect .6289 / (.6289 + .5290) ## in this intercept only model, the fixed effects estimates ## are similar between the linear regression and the LMMs ## regression coefficients (fixed effects) from linear regression coef(m1lr) ## fixed effects only from LMM fixef(m1lmm) ## however, because the linear regression erroneously assumes that ## all observations are independent, the standard error is biased ## downwards resulting in biased, too narrow confidence intervals ## here we just compare the 95% confidence interval for the fixed effect ## intercept estimate, labelled "(Intercept)" from both models ## note that it is wider (appropriately) in the LMM confint(m1lr) confint(m1lmm, oldNames = FALSE) ## we can check some of the ranges of random effects ## to get random effects with the fixed effects already added (i.e., NOT deviations) ## use coef() instead of ranef() ranef.devs <- ranef(m1lmm)$UserID ranef.blups <- coef(m1lmm)$UserID View(ranef.blups) ## visualize the random intercept ggplot(ranef.devs, aes(`(Intercept)`)) + geom_histogram() + ggtitle("Deviations") ## same but this time instead of deviations from ## the fixed effect, add the fixed effect intercept in ggplot(ranef.blups, aes(`(Intercept)`)) + geom_histogram() + ggtitle("Random Intercept Est") ## summarise BLUPs for the random intercept summary(ranef.blups[["(Intercept)"]]) ## if we wanted, instead of using mean +/- SD to give a range of the estimates for ## the random intercept and slope, we could use the 25th and 75th percentiles ## (i.e., 1st and 3rd quartiles of the BLUPs) quantile(ranef.blups[["(Intercept)"]]) ## ## Sample Write up ## A total of 6,399 observations from 191 unique people were included ## in the analyses. Overall, the model estimated the average ## positive affect as b [95% CI] = 2.68 [2.56, 2.79]. ## However, the random effects revealed substantial differences between ## individual (Random Effect SD = 0.79) such that although the average was 2.68, most people ## fell between 1.89 and 3.47 (i.e., 2.67866 +/- 0.7930). ## The intraclass correlation coefficient was .63, indicating that 63% of ## the variance in positive affect exists between people and ## 37% of the variance in positive affect exists within people across days. #### 3. Diagnostics & Inference (Demonstration) #### ## make a dataset of the residuals and expected values ## to do this, we use the ## fitted() function for expected values ## resid() function for model residuals ## NOTE: these two functions work on linear regression models too d.residuals <- data.table( Yhat = fitted(m1lmm), Residuals = resid(m1lmm)) ## check for normality of the outcome ## by examining residuals ggplot(d.residuals, aes(Residuals)) + geom_histogram() ## check for normality of the outcome ## using QQ plot ggplot(d.residuals, aes(sample = Residuals)) + stat_qq() + stat_qq_line() ## check for homogeneity of variance assumption ## does it look rectangular with no clear pattern? ggplot(d.residuals, aes(Yhat, Residuals)) + geom_point(alpha = .2) ## make a dataset of the random effects by UserID d.random <- as.data.table(coef(m1lmm)$UserID) ## check whether the random effects are normally distributed ## note that these have mean 0 ggplot(d.random, aes(`(Intercept)`)) + geom_histogram() ## normality via QQ plot ggplot(d.random, aes(sample = `(Intercept)`)) + stat_qq() + stat_qq_line() ## fixed effects table fetable <- coef(summary(m1lmm)) ## view the table print(fetable) ## extract the t values (b / se) ## take their absolute value (as we typically do 2 sided hypothesis tests ## calculate the p-value using the normal distribution function, pnorm() ## and multiply by 2 so its a two-tailed test pnorm(abs(fetable[, "t value"]), lower.tail = FALSE) * 2 ## confidence intervals using the Wald method are based ## on assuming a normal distribution and are very fast and easy ## but do not give any confidence intervals for the random effects confint(m1lmm, method = "Wald") ## confidence intervals using the profile method are based ## on the change in model performance (the log likelihood) ## and are much slower, but generally a bit more precise and ## are appropriate for random effects confint(m1lmm, method = "profile") ## we can compare models using likelihood ratio tests ## which are OK (not dissimilar from assuming normal) for fixed effects ## and are about the best that can be done for random effects ## note that you always need at least one random effect in LMMs ## so you cannot simply remove the random intercept; generally this is ## used for testing more complex random effects we will learn in future weeks ## or testing several differences at once ## setup a new model with an additional predictor, SurveyInteger lmmalt <- lmer(PosAff ~ 1 + SurveyInteger + (1 | UserID), d) ## conduct a likelihood ratio test of these two, nested models anova(lmmalt, m1lmm, test = "LRT") ## assuming normal distribution, test the fixed effects in ## our alternate model, lmmalt fetablealt <- coef(summary(lmmalt)) pnorm(abs(fetablealt[, "t value"]), lower.tail = FALSE) * 2 #### 4. EXERCISE - Include in Workbook #### ## pick one of the other variables ## in the dataset (not positive affect) and fit an intercept only ## linear mixed model by completing the code below. ## if you need a refresher on what variables are available ## take a look at the table in the slides or look through the example ## above to see how we could find all the variable names in the dataset ## store the model results in an object called "m2lmm" m2lmm <- lmer() ## now make a summary of the model results summary( ) ## what is the intraclass correlation coefficient for this variable? ## if you try fitting an intercept only linear regression (not LMM) ## to the same variable, are the confidence intervals ## wider or narrower for the LMM or linear regression? m2lr <- lm( ) ## what is the confidence interval for the fixed effect intercept? confint( ) ## what variation may we expect in the intercept due to individual differences? quantile(coef( )$UserID)