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[Q] How to interpret mixed effects multilevel logistic regression with random slope and interaction?

I am trying to understand mixed effects multilevel regression. At the moment, I am struggling with the interpretation. First, let me present an example to make it a bit easier. Data is found here.
Suppose I want to investigate the following: Does extraversion predict popularity among pupils and is this effect interacted by gender?
Why include random slopes in this example? Well, Heising & Schaffer recommends to "always Include a Random Slope for the Lower-Level Variable Involved in a Cross-Level Interaction". I guess extraversion is the lower-level in this example, and gender is the upper-level?
Let me build the models:
library(lme4) library(haven) library(tidyverse) library(texreg) # Load df df <- read_sav(file ="https://github.com/MultiLevelAnalysis/Datasets-third-edition-Multilevel-book/blob/mastechapter%202/popularity/SPSS/popular2.sav?raw=true") # Make popular-variable binary for logistic analysis df <- df %>% mutate( popular = case_when( popular < 5 ~ 0, # Not popular popular > 4 ~ 1, # Popular ), sex = as_factor(sex)) # Without interaction or random slope model1 <- glmer(formula = popular ~ extrav + sex + (1|class), data = df, family = binomial(link = "logit")) # Adding random slope model2 <- glmer(formula = popular ~ extrav + sex + (1 + sex |class), data = df, family = binomial(link = "logit")) # Adding interaction model3 <- glmer(formula = popular ~ extrav + sex + sex:extrav + (1 + sex |class), data = df, family = binomial(link = "logit")) 
And here are the results (texreg doesn't seem to work with random slopes?):
Model 1: Without interaction or random slope
> summary(model1) Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) [glmerMod] Family: binomial ( logit ) Formula: popular ~ extrav + sex + (1 | class) Data: df AIC BIC logLik deviance df.resid 1733.6 1756.0 -862.8 1725.6 1996 Scaled residuals: Min 1Q Median 3Q Max -4.9153 -0.4036 0.0819 0.4061 6.0866 Random effects: Groups Name Variance Std.Dev. class (Intercept) 3.497 1.87 Number of obs: 2000, groups: class, 100 Fixed effects: Estimate Std. Error z value Pr(>|z|) (Intercept) -6.27476 0.45151 -13.90 <2e-16 *** extrav 0.98324 0.07305 13.46 <2e-16 *** sexgirl 2.93735 0.15760 18.64 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Correlation of Fixed Effects: (Intr) extrav extrav -0.883 sexgirl -0.452 0.346 
Model 2: Adding random slope
> summary(model2) Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) [glmerMod] Family: binomial ( logit ) Formula: popular ~ extrav + sex + (1 + sex | class) Data: df AIC BIC logLik deviance df.resid 1736.5 1770.1 -862.2 1724.5 1994 Scaled residuals: Min 1Q Median 3Q Max -4.7925 -0.3991 0.0857 0.4060 6.3603 Random effects: Groups Name Variance Std.Dev. Corr class (Intercept) 3.95039 1.9876 sexgirl 0.04961 0.2227 -1.00 Number of obs: 2000, groups: class, 100 Fixed effects: Estimate Std. Error z value Pr(>|z|) (Intercept) -6.313119 0.001122 -5628.2 <2e-16 *** extrav 0.985692 0.001122 878.5 <2e-16 *** sexgirl 2.928904 0.001122 2610.4 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Correlation of Fixed Effects: (Intr) extrav extrav 0.000 sexgirl 0.000 0.000 convergence code: 0 Model failed to converge with max|grad| = 0.0630261 (tol = 0.002, component 1) Model is nearly unidentifiable: very large eigenvalue - Rescale variables? 
Model 3: Adding interaction
> summary(model3) Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) [glmerMod] Family: binomial ( logit ) Formula: popular ~ extrav + sex + extrav:sex + (1 + sex | class) Data: df AIC BIC logLik deviance df.resid 1737.1 1776.3 -861.6 1723.1 1993 Scaled residuals: Min 1Q Median 3Q Max -5.1437 -0.4070 0.0772 0.3897 5.8930 Random effects: Groups Name Variance Std.Dev. Corr class (Intercept) 3.603215 1.89821 sexgirl 0.003442 0.05867 -0.99 Number of obs: 2000, groups: class, 100 Fixed effects: Estimate Std. Error z value Pr(>|z|) (Intercept) -5.82587 0.57189 -10.187 < 2e-16 *** extrav 0.89746 0.09709 9.243 < 2e-16 *** sexgirl 2.02635 0.72615 2.791 0.00526 ** extrav:sexgirl 0.17737 0.14192 1.250 0.21140 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Correlation of Fixed Effects: (Intr) extrav sexgrl extrav -0.926 sexgirl -0.672 0.691 extrv:sxgrl 0.605 -0.648 -0.975 convergence code: 0 Model failed to converge with max|grad| = 0.204295 (tol = 0.002, component 1) 
I know there are some problems with model 2 and 3, but I guess they will do fine as example.
  • Are the coefficients presented as usual logits/log-odds? And will exponentiating them reveal the odds-ratio as usual?
  • How do I interpret the intercepts and coefficients of the random effect differently from the fixed effect (I guess the fixed effect is the one that is reported most commonly?)
  • Do I interpret the (fixed) interaction term as usual (model 3): The effect of extrav increases by 0.18 leaving the final coefficient: 0.9 + 0.18 = 1.17.
  • In their article, Heising & Schaffer argues that the random slope for the lower-level variable of the cross-level interaction captures heteroscedasticity and cluster-correlated errors. Heteroscedasticity is not important in logistic regression, but I guess cluster-correlated is?
I know this is quite a post. I appreciate all inputs!
submitted by Kaka95 to statistics

Creating new variable based on sum of multiple variables.

Hey, so here's the situation I got myself into, I'll try to be brief and concise.
In my experiment I created 20 different "scenerios" (short stories basically). Each participant only encounters 3 of them (randomly). After each scenerio is presented, a 19-item Likart scale questionnaire is filled, which means each participant answered 57 questions with 1 to 7.
I took a look at the output of the data extracted from the experiment platform I used, and basically what happened is I now have 380 (19*20) variables, each variable representing an item. I basically need the sum of the questionnaire for each scenerio (so 3 scores for each participant), and I could not find an elegant way to approach the problem.
I tried using the SUM function under 'compute variable' or the syntax version, but selecting each and every variable manually is quite a tedious task. For example:
COMPUTE Q4 = SUM(Q4_1,Q4_2,Q4_3,Q4_4,Q4_5,Q4_6,Q4_7,Q4_8,Q4_9,Q4_10,Q4_11,Q4_12,Q4__13,Q4_14,Q4_15,Q4_16,Q4_17,Q4_18,Q4_19).
You get the idea I think. I'm pretty inexperienced with SPSS and was hoping some veterans could help me out.
Thank you!
submitted by lazania901 to spss

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