#::::::::::::::::: PSY9140 SEM course Day 4 :::::::::::::::::::::::::::::### ################### (0) Preparations ####################################### # installing packages; has just to be done once. # install.packages("lavaan") # install.packages("psych") # install.packages("lavaanPlot") # install.packages("semPlot") # install.packages("semTable") # install.packages("ggcorrplot") # load packages. Has to be done each time R is started. library(lavaan) library(psych) library(lavaanPlot) library(semPlot) library(semTable) library(ggcorrplot) ################### (1) Exercise 1 ####################################### # load the datafile "data3.csv" as a dataframe with the name "df1" by using the "read.csv" command df1 <- read.csv("https://www.sv.uio.no/psi/personer/vit/tilmanns/psy9140/data3.csv") # descriptive statistics for all variables in "df1" by using "describe" from the "psych" package describe(df1) # (1) Specify the model shown in the figure # Model specification model <- "diet3 ~ diet2 + body2 body3 ~ body2 + diet2 diet2 ~ diet1 + body1 body2 ~ body1 + diet1 diet3 ~~ body3 diet2 ~~ body2 diet1 ~~ body1" # Running the model fit <- sem(model, missing = "ML", meanstructure = TRUE, data = df1) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE, modindices = TRUE) # (2) Check model fit and re-specify the model, according to modification indices and conceptual considerations. # Answer: Model fit shows that at least the autorgressions from T1 to T3 should be estimated. They are estimated here. model <- "diet3 ~ diet2 + body2 + diet1 body3 ~ body2 + diet2 + body1 diet2 ~ diet1 + body1 body2 ~ body1 + diet1 diet3 ~~ body3 diet2 ~~ body2 diet1 ~~ body1" # Running the model fit <- sem(model, missing = "ML", meanstructure = TRUE, data = df1) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) # (3) Both girls and boys are included in the sample. Specify a model with adjustment for gender. model <- "diet3 ~ diet2 + body2 + diet1 + girl body3 ~ body2 + diet2 + body1 + girl diet2 ~ diet1 + body1 + girl body2 ~ body1 + diet1 + girl diet3 ~~ body3 diet2 ~~ body2 diet1 ~~ body1 + girl body1 ~~ girl" # Running the model fit <- sem(model, missing = "ML", meanstructure = TRUE, data = df1) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) ################### (2) Exercise 2 ####################################### # load the datafile "data4.csv" as a dataframe with the name "df1" by using the "read.csv" command df2 <- read.csv("https://www.sv.uio.no/psi/personer/vit/tilmanns/psy9140/data4.csv") # descriptive statistics for all variables in "df2" by using "describe" from the "psych" package describe(df2) # (1) Use the input file day4.R, Exercise 2. Run the model and draw a path diagram of the model that is estimated. # Specify the model model <- "# Latent factors diet diet1 =~ diet11 + diet12 + diet13 + diet14 diet2 =~ diet21 + diet22 + diet23 + diet24 # Latent factors body body1 =~ body11 + body12 + body13 + body14 + body15 body2 =~ body21 + body22 + body23 + body24 + body25 # Covariances diet1 + diet2 ~~ body1 + body2 diet1 ~~ diet2 body1 ~~ body2" # Run the model fit <- sem(model, missing = "ML", meanstructure = TRUE, data = df2) # Increase the number of rows printed in the output options(max.print=10000) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE, modindices = TRUE) # I often prefer to draw the diagram with paper and pencil, to get a better overview over the model, # but if you want use semPlot: # Path diagram with a design with good readability. Intercepts are not depicted to increase readability # This is one of many possible ways to draw the model. semPaths(fit, whatLabels = "std", layout = "spring", layoutSplit = TRUE, curve = .3, nCharNodes = 0, measurementLayout = "tree3", intercepts = FALSE, subScale = .5, subScale2 = .5) # (2) The fit indices show that the model does not fit the data well. How could you improve the model in a way # that is conceptually meaningful? # Specify the model model <- "# Latent factors diet diet1 =~ diet11 + diet12 + diet13 + diet14 diet2 =~ diet21 + diet22 + diet23 + diet24 # Latent factors body body1 =~ body11 + body12 + body13 + body14 + body15 body2 =~ body21 + body22 + body23 + body24 + body25 # Covariances diet1 + diet2 ~~ body1 + body2 diet1 ~~ diet2 body1 ~~ body2 # Covariances between residuals diet11 ~~ diet21 diet12 ~~ diet22 diet13 ~~ diet23 diet14 ~~ diet24 body11 ~~ body21 body12 ~~ body22 body13 ~~ body23 body14 ~~ body24 body15 ~~ body25" # Run the model fit <- sem(model, missing = "ML", meanstructure = TRUE, data = df2) # Increase the number of rows printed in the output options(max.print=10000) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE, modindices = TRUE) # (3) Re-specify the model to obtain measurement invariance across time. # Specify the model model <- "# Latent factors diet diet1 =~ diet11 + d2*diet12 + d3*diet13 + d4*diet14 diet2 =~ diet21 + d2*diet22 + d3*diet23 + d4*diet24 # Latent factors body body1 =~ body11 + b2*body12 + b3*body13 + b4*body14 + b5*body15 body2 =~ body21 + b2*body22 + b3*body23 + b4*body24 + b5*body25 # Covariances diet1 + diet2 ~~ body1 + body2 diet1 ~~ diet2 body1 ~~ body2 # Covariances between residuals diet11 ~~ diet21 diet12 ~~ diet22 diet13 ~~ diet23 diet14 ~~ diet24 body11 ~~ body21 body12 ~~ body22 body13 ~~ body23 body14 ~~ body24 body15 ~~ body25" # Run the model fit <- sem(model, missing = "ML", meanstructure = TRUE, data = df2) # Increase the number of rows printed in the output options(max.print=10000) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE, modindices = TRUE) # (4) Re-specify the model to obtain a cross- lagged, autoregressive model. # Specify the model model <- "# Latent factors diet diet1 =~ diet11 + d2*diet12 + d3*diet13 + d4*diet14 diet2 =~ diet21 + d2*diet22 + d3*diet23 + d4*diet24 # Latent factors body body1 =~ body11 + b2*body12 + b3*body13 + b4*body14 + b5*body15 body2 =~ body21 + b2*body22 + b3*body23 + b4*body24 + b5*body25 # Covariances diet2 ~ diet1 + body1 body2 ~ body1 + diet1 diet1 ~~ body1 diet2 ~~ body2 # Covariances between residuals diet11 ~~ diet21 diet12 ~~ diet22 diet13 ~~ diet23 diet14 ~~ diet24 body11 ~~ body21 body12 ~~ body22 body13 ~~ body23 body14 ~~ body24 body15 ~~ body25" # Run the model fit <- sem(model, missing = "ML", meanstructure = TRUE, data = df2) # Increase the number of rows printed in the output options(max.print=10000) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) ################### (3) Exercise 3 ####################################### # (1) Use the script day4.R, Exercise 3 . Estimate a growth curve based on the mean # scores of body image (variables body1 to body3 ) at three time points. # Specify the model in the next line model <- "i =~ 1*body1 + 1*body2 + 1*body3 s =~ 0*body1 + 1*body2 + 2*body3 i ~~ s" # Running the model fit <- growth(model, missing = "ML", meanstructure = TRUE, data = df1) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) # (2) What kind of information do the results provide about the mean and # variance of the intercept and the slope of body image? # Answer: Mean intercept = 3.43, meaning that the estimated mean at T1 is 3.43. # Mean slope = 0.03, meaning that body image is increasing with 0.03 from one time point # to the next with on average 0.03 units. The variance for the intercept and slope is significant. # (3) The second data collection was two years after the first data collection. The third data collection was seven # years after the first data collection. Estimate a new growth curve model that takes into account the uneven # intervals between time points. How can we interpret the mean of the slope? # Specify the model in the next line model <- "i =~ 1*body1 + 1*body2 + 1*body3 s =~ 0*body1 + 2*body2 + 7*body3 i ~~ s" # Running the model fit <- growth(model, missing = "ML", meanstructure = TRUE, data = df1) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) # Answer: The mean of the slope indicates that body image increases on average with .008 units per year. ################### (4) Exercise 4 ##################################### # (1) Run the code in day4.R, Exercise 4. # Model specification model <- "# LGC specification i =~ 1*body1 + 1*body2 + 1*body3 s =~ 0*body1 + 2*body2 + 7*body3 # Covariance between i and s i ~~ s # Regression i + s ~ girl" # Running the model fit <- growth(model, missing = "ML", meanstructure = TRUE, data = df1) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) # (2) How is gender related to the intercept and slope of body image, and what do such relationships # tell us about how girls' and boys' body image is developing over time? # Answer: The regression coefficient for girl --> i is -0.431 (p<.05) and for girl --> s is 0.004 (p>.05) # The results indicate that the estimated level of body image at T1 for boys (who are coded 0) is 3.683 # (the intercept of the intercept), and the estimated level for girls is 3.252 (i.e., 3.683 - 0.431) # Moreover, the estimated growth in body image from one year to the next for boys is 0.006 (intercept of the slope), # and the estimated growth for girls is 0.010 (i.e., .006 + 0.004). However, the gender difference in the slope is not significant. # (3) Model a growth curve for dieting and interpret the estimates of the slope and the intercept. # Model specification model <- "# LGC specification i =~ 1*diet1 + 1*diet2 + 1*diet3 s =~ 0*diet1 + 2*diet2 + 7*diet3 # Covariance between i and s i ~~ s" # Running the model fit <- growth(model, missing = "ML", meanstructure = TRUE, data = df1) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) # (4) Include gender as time invariant covariate, predicting the intercept and the slope of dieting. # Is there any problem with this model? # Model specification model <- "# LGC specification i =~ 1*diet1 + 1*diet2 + 1*diet3 s =~ 0*diet1 + 2*diet2 + 7*diet3 # Covariance between i and s i ~~ s # Regression i + s ~ girl" # Running the model fit <- growth(model, missing = "ML", meanstructure = TRUE, data = df1) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) # Answer: The RMSEA indicates low model fit. However, CFI and TLI are ok. ################### (5) Exercise 5 ####################################### # (1) Run the code. Two growth curves are estimated in one model: One for dieting and one for body image. # Examine how intercepts and slopes are correlated with each other. # Model specification model <- "# LGC specification i_b =~ 1*body1 + 1*body2 + 1*body3 s_b =~ 0*body1 + 2*body2 + 7*body3 i_d =~ 1*diet1 + 1*diet2 + 1*diet3 s_d =~ 0*diet1 + 2*diet2 + 7*diet3 # Covariance between i and s i_b ~~ s_b i_d ~~ s_d i_d ~~ i_b s_d ~~ s_b i_d ~~ s_b i_b ~~ s_d" # Running the model fit <- growth(model, missing = "ML", meanstructure = TRUE, data = df1) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) # (2) Re-specify the model such that the intercept of dieting predicts the slope of body image. # The intercept of body image should predict the slope of dieting. # Model specification model <- "# LGC specification i_b =~ 1*body1 + 1*body2 + 1*body3 s_b =~ 0*body1 + 2*body2 + 7*body3 i_d =~ 1*diet1 + 1*diet2 + 1*diet3 s_d =~ 0*diet1 + 2*diet2 + 7*diet3 # Covariance between i and s i_b ~~ s_b i_d ~~ s_d i_d ~~ i_b s_d ~~ s_b s_b ~ i_d s_d ~ i_b" # Running the model fit <- growth(model, missing = "ML", meanstructure = TRUE, data = df1) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) # (3) Draw a diagram of the model you have just estimated. # Path diagram with a design with good readability. There are other solutions possible. semPaths(fit, whatLabels = "par", layout = "tree") ################### (6) Exercise 6 ####################################### # (1) Run the code and interpret the means of the intercept and slope. # load the datafile "data5.csv" as a dataframe with the name "df3" by using the "read.csv" command df3 <- read.csv("https://www.sv.uio.no/psi/personer/vit/tilmanns/psy9140/data5.csv") # descriptive statistics for all variables in "df1" by using "describe" from the "psych" package describe(df3) # Specify the model in the next line model <- "# LGC specification i =~ 1*body1 + 1*body2 + 1*body3 + 1*body4 s =~ 0*body1 + 2*body2 + 7*body3 + 13*body4" # Running the model fit <- growth(model, missing = "ML", meanstructure = TRUE, data = df3) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) # Answer: The estimated mean level of body image at T1 is 3.439. # The estimated growth in body image from one year to the next is 0.005. # (2) Include a quadratic slope in the model. # Model specification model <- "# LGC specification i =~ 1*body1 + 1*body2 + 1*body3 + 1*body4 s =~ 0*body1 + .2*body2 + .7*body3 + 1.3*body4 q =~ 0*body1 + .04*body2 + .49*body3 + 1.69*body4" # Running the model fit <- growth(model, missing = "ML", meanstructure = TRUE, data = df3) # Providing output for the model summary(fit, fit.measures = TRUE, standardized=TRUE) # (3) Try to draw a graph representing the development of body image according to the new, non-linear trajectory. # Answer: I would recommend to draw the graph by using paper and pencil to get familiar # with how such models represent growth. The model can also be drawn by for example using the ggplot2 package, see below. # If not installed yet, the ggplot2 package has to be installed first. # install.packages("ggplot2") library(ggplot2) ggplot(data.frame(x=c(0, 1.3)), aes(x)) + theme_classic() + theme(axis.text = element_text(color = "black", size = 14)) + # value label size theme(axis.title = element_text(size = 16)) + # axis title size theme(axis.ticks = element_line(size = 1)) + # thickness of axis ticks theme(axis.ticks.length = unit(.3,"cm")) + # length of axis ticks theme(axis.line = element_line(color = "black", size = 1)) + # size of x and y axis stat_function(fun=function(x) 3.431 + .125*x - .059*x^2, size = 1) + # provides the growth curve function coord_cartesian(xlim = c(0, 1.3), ylim = c(3.3, 3.6)) + # range of graph scale_x_continuous(breaks=seq(0, 1.3, .1), expand = c(0,0)) + #Labels specifies x-axis value labels xlab("Time in decades (0.1 = 1 year)") + ylab("Body image")