In doomlab/learnSEM: Learning Tutorials for Structural Equation Modeling
knitr::opts_chunk$set(
collapse = TRUE,
comment = "#>"
)
knitr::opts_chunk$set(echo = TRUE)
library(lavaan)
library(semPlot)
Latent Growth Modeling
- Measuring change over repeated time measurements
- Gives you more information than a traditional repeated measures ANOVA or regression line by itself.
- Advantages:
- Estimate means and covariances separately
- Estimating observed values and unobserved values separately
Assumptions
- Continuous measurement of the DVs
- This assumption is true for many structural models though!
- Time spacing is the same across people
- NOT across measurements, but people need to be spaced the same
- At least three time points per person (otherwise, you are doing a dependent t-test)
- Large samples!
Before you start
- You have to know the expected type of change before you start
- Generally it's linear, however, you can do curvilinear or power functions
Example model
- We will set the regression loading values to specific numbers to be able to estimate intercept and slope.
knitr::include_graphics("pictures/example_lgm.png")
Understand the idea
- Intercept
- You set these values to 1 indicating that you do NOT want to estimate them
- Basically, that gives you a starting value for the first time point, which is the average point where people start (y-intercept)
Understand the idea
- Slope – represents the change over time
- You can set these values to anything you want
- Usually the first time is indicated by a 0
- There's no slope for time 1, just an intercept
- Then the paths are set based on the time differences between them
Understand the idea
- Setting the parameters this way:
- Helps with identification
- Is theoretical to match the concept of slope and intercept estimation
- Allows you to not have to set the variances, so you can look at them
Questions
- What is the average change for each person?
- Where do they start?
- How does it change over time?
- What is the form of the change?
Questions
- Is the average slope and intercept a good fit for all participants?
- Or should we include a variance term to account for the differences between people?
- Models are called random effects if you add the random variances (similar to multilevel models)
- If we have a random effects model – is there another variable that explains those random effects?
Understanding the Output
- Intercept mean: the average starting point for time 1
- Slope mean: the average increment across time points
knitr::include_graphics("pictures/random_fixed.png")
Understanding the Output
- Intercept variance: the spread around the average start point
- Large scores indicate a lot of spread – meaning people start in a lot of different places
- Small scores indicate a small spread – everyone starts about the same place
knitr::include_graphics("pictures/random_fixed.png")
Understanding the Output
- Slope variance – the range of increments across time points
- Small variances mean that everyone is going up/down about the same amount
- Large variances mean that people scores are going up/down differently (almost like an interaction)
knitr::include_graphics("pictures/random_fixed.png")
Understanding the Output
- Factor covariance – examines the relationship between intercept and slope
- Positive covariance - positive slope
- People who start with higher intercepts have higher positive slopes
- Positive covariance - negative slope
- People who start higher go down faster (larger negative slopes)
- Negative covariance - positive slope
- People who start with higher intercepts go up slower (smaller slopes)
- Negative covariance - negative slope
- People who start with higher intercepts go down slower
Example: Crime Data across time
##load the data
crime.cov <- lav_matrix_lower2full(c(.63,
.50, .60,
.48, .48, .58,
.47, .48, .51, .67))
crime.mean <- c(5.17, 5.32, 5.40, 5.52)
names(crime.mean) <-
rownames(crime.cov) <-
colnames(crime.cov) <- c("Time1", "Time2", "Time3", "Time4")
New Functions
growth(): This function helps with the set up for a LCM, fixes the parameters correctly for you. - We will use the data as covariances and means, but you can import the data just like
cfa()
How to Test
- This procedure is very similar to multigroup model testing, only in reverse.
- We are going to start with the most constrained model and slowly let parameters free to see which is the best.
- We would prefer the unconstrained model be the best:
- The slope and intercept are useful
- But the variances to be low indicating everyone has the same pattern of slope and intercept
Intercept Only Model
- Use only the intercept (mean of the data) to estimate
- You want this model to be bad, otherwise you are saying that one average score is the best for all the time points
- Intercept variance is constrained to 0, so you only get a mean.
- Residuals are forced to be the same for each time point, so the variance is the same across all time points
crime.model1 <- '
# intercept
i =~ 1*Time1 + 1*Time2 + 1*Time3 + 1*Time4
i~~0*i
# residual variances
Time1~~r*Time1
Time2~~r*Time2
Time3~~r*Time3
Time4~~r*Time4
'
Intercept Only Model
crime.fit1 <- growth(crime.model1,
sample.cov=crime.cov,
sample.mean=crime.mean,
sample.nobs=952)
summary(crime.fit1,
standardized = TRUE,
fit.measures = TRUE,
rsquare = TRUE)
Intercept Only Model
semPaths(crime.fit1,
whatLabels = "par",
edge.label.cex = 1,
layout = "tree")
Comparison Table Fit Indices
library(knitr)
fit.table <- matrix(NA, nrow = 5, ncol = 6)
colnames(fit.table) <- c("Model", "X2", "df", "RMSEA", "SRMR", "CFI")
fit.table[1, ] <- c("Intercept Only", round(fitmeasures(crime.fit1, c("chisq", "df", "rmsea", "srmr", "cfi")),3))
kable(fit.table)
Comparison Table Parameters
#save the parameter estimates
crime.fit1.par <- parameterestimates(crime.fit1)
#make table
par.table <- matrix(NA, nrow = 5, ncol = 7)
colnames(par.table) <- c("Model", "Intercept Mean", "Intercept Variance", "Residual Variance", "Slope Mean", "Slope Variance", "Covariance")
#put data in table
par.table[1, ] <- c("Intercept Only",
round(crime.fit1.par$est[crime.fit1.par$lhs == "i" & crime.fit1.par$op == "~1"], 3),
"X",
round(crime.fit1.par$est[crime.fit1.par$lhs == "Time1" & crime.fit1.par$op == "~~"], 3),
"X",
"X",
"X")
kable(par.table)
Random Intercept Only Model
- We allow the intercept variance to be > 0
- People can start at different places
- Residual variances are still the same
- Still no slope
crime.model2 <- '
# intercept
i =~ 1*Time1 + 1*Time2 + 1*Time3 + 1*Time4
# residual variances
Time1~~r*Time1
Time2~~r*Time2
Time3~~r*Time3
Time4~~r*Time4
'
Random Intercept Only Model
crime.fit2 <- growth(crime.model2,
sample.cov=crime.cov,
sample.mean=crime.mean,
sample.nobs=952)
summary(crime.fit2,
standardized = TRUE,
fit.measures = TRUE,
rsquare = TRUE)
Comparison Table Fit Indices
- This model is much better than model 1
- Implies that people all start in different places
fit.table[2, ] <- c("Random Intercept", round(fitmeasures(crime.fit2, c("chisq", "df", "rmsea", "srmr", "cfi")),3))
kable(fit.table)
Comparison Table Parameters
#save the parameter estimates
crime.fit2.par <- parameterestimates(crime.fit2)
#put data in table
par.table[2, ] <- c("Random Intercept",
round(crime.fit2.par$est[crime.fit2.par$lhs == "i" & crime.fit2.par$op == "~1"], 3),
round(crime.fit2.par$est[crime.fit2.par$lhs == "i" & crime.fit2.par$op == "~~"], 3),
round(crime.fit2.par$est[crime.fit2.par$lhs == "Time1" & crime.fit2.par$op == "~~"], 3),
"X",
"X",
"X")
kable(par.table)
Random Slope + Intercepts
- Add in the random slope
s~0*1 (makes the average slope 0)s~~0*i (uncorrelated slope/intercept)- Leave the intercept and its variance in the model
- Keep the residuals constrained to be the same
crime.model3 <- '
# intercept
i =~ 1*Time1 + 1*Time2 + 1*Time3 + 1*Time4
# slope
s =~ 0*Time1 + 1*Time2 + 2*Time3 + 3*Time4
s~0*1
s~~0*i
# residual variances
Time1~~r*Time1
Time2~~r*Time2
Time3~~r*Time3
Time4~~r*Time4
'
Random Slope + Intercepts
crime.fit3 <- growth(crime.model3,
sample.cov=crime.cov,
sample.mean=crime.mean,
sample.nobs=952)
summary(crime.fit3,
standardized = TRUE,
fit.measures = TRUE,
rsquare = TRUE)
Comparison Table Fit Indices
- This model is much better than model 2
- Implies that having at least a random slope is better than no slope
fit.table[3, ] <- c("Random Slope", round(fitmeasures(crime.fit3, c("chisq", "df", "rmsea", "srmr", "cfi")),3))
kable(fit.table)
Comparison Table Parameters
#save the parameter estimates
crime.fit3.par <- parameterestimates(crime.fit3)
#put data in table
par.table[3, ] <- c("Random Slope",
round(crime.fit3.par$est[crime.fit3.par$lhs == "i" & crime.fit3.par$op == "~1"], 3),
round(crime.fit3.par$est[crime.fit3.par$lhs == "i" & crime.fit3.par$op == "~~" & crime.fit3.par$rhs == "i"], 3),
round(crime.fit3.par$est[crime.fit3.par$lhs == "Time1" & crime.fit3.par$op == "~~"], 3),
"X",
round(crime.fit3.par$est[crime.fit3.par$lhs == "s" & crime.fit3.par$op == "~~"], 3),
"X")
kable(par.table)
Full Slopes + Intercept
- Allow the slope and intercept to vary
- Constrain the residual variances
crime.model4 <- '
# intercept
i =~ 1*Time1 + 1*Time2 + 1*Time3 + 1*Time4
# slope
s =~ 0*Time1 + 1*Time2 + 2*Time3 + 3*Time4
# residual variances
Time1~~r*Time1
Time2~~r*Time2
Time3~~r*Time3
Time4~~r*Time4
'
Full Slopes + Intercept
- Significant covariance – negative covariance with positive slope
- People who start high go up slower than people who start lower.
crime.fit4 <- growth(crime.model4,
sample.cov=crime.cov,
sample.mean=crime.mean,
sample.nobs=952)
summary(crime.fit4,
standardized = TRUE,
fit.measures = TRUE,
rsquare = TRUE)
Comparison Table Fit Indices
- This model is much better than model 3
- Implies that having a single slope is better than just random slopes
fit.table[4, ] <- c("Full Slope", round(fitmeasures(crime.fit4, c("chisq", "df", "rmsea", "srmr", "cfi")),3))
kable(fit.table)
Comparison Table Parameters
#save the parameter estimates
crime.fit4.par <- parameterestimates(crime.fit4)
#put data in table
par.table[4, ] <- c("Full Slope",
round(crime.fit4.par$est[crime.fit4.par$lhs == "i" & crime.fit4.par$op == "~1"], 3),
round(crime.fit4.par$est[crime.fit4.par$lhs == "i" & crime.fit4.par$op == "~~" & crime.fit4.par$rhs == "i"], 3),
round(crime.fit4.par$est[crime.fit4.par$lhs == "Time1" & crime.fit4.par$op == "~~"], 3),
round(crime.fit4.par$est[crime.fit4.par$lhs == "s" & crime.fit4.par$op == "~1"], 3),
round(crime.fit4.par$est[crime.fit4.par$lhs == "s" & crime.fit4.par$op == "~~"], 3),
round(crime.fit4.par$est[crime.fit4.par$lhs == "i" & crime.fit4.par$op == "~~" & crime.fit4.par$rhs == "s"], 3))
kable(par.table)
Totally Unconstrained Model
- Now the residuals are free to vary
- You want the residuals to be small and roughly equal, so this model shouldn't be any different than the previous model
crime.model5 <- '
# intercept
i =~ 1*Time1 + 1*Time2 + 1*Time3 + 1*Time4
# slope
s =~ 0*Time1 + 1*Time2 + 2*Time3 + 3*Time4
'
Totally Unconstrained Model
crime.fit5 <- growth(crime.model5,
sample.cov=crime.cov,
sample.mean=crime.mean,
sample.nobs=952)
summary(crime.fit5,
standardized = TRUE,
fit.measures = TRUE,
rsquare = TRUE)
Comparison Table Fit Indices
- This model is equal to model 4
- This result implies that your residual variances are all approximately equal
fit.table[5, ] <- c("Unconstrained", round(fitmeasures(crime.fit5, c("chisq", "df", "rmsea", "srmr", "cfi")),3))
kable(fit.table)
Comparison Table Parameters
#save the parameter estimates
crime.fit5.par <- parameterestimates(crime.fit5)
residual_numbers <- paste(round(crime.fit5.par$est[crime.fit5.par$lhs == "Time1" & crime.fit5.par$op == "~~"], 3),
round(crime.fit5.par$est[crime.fit5.par$lhs == "Time2" & crime.fit5.par$op == "~~"], 3),
round(crime.fit5.par$est[crime.fit5.par$lhs == "Time3" & crime.fit5.par$op == "~~"], 3),
round(crime.fit5.par$est[crime.fit5.par$lhs == "Time4" & crime.fit5.par$op == "~~"], 3))
#put data in table
par.table[5, ] <- c("Unconstrained",
round(crime.fit5.par$est[crime.fit5.par$lhs == "i" & crime.fit5.par$op == "~1"], 3),
round(crime.fit5.par$est[crime.fit5.par$lhs == "i" & crime.fit5.par$op == "~~" & crime.fit5.par$rhs == "i"], 3),
residual_numbers,
round(crime.fit5.par$est[crime.fit5.par$lhs == "s" & crime.fit5.par$op == "~1"], 3),
round(crime.fit5.par$est[crime.fit5.par$lhs == "s" & crime.fit5.par$op == "~~"], 3),
round(crime.fit5.par$est[crime.fit5.par$lhs == "i" & crime.fit5.par$op == "~~" & crime.fit5.par$rhs == "s"], 3))
kable(par.table)
Summary
-
In this lecture you've learned:
-
Latent growth modeling
- The idea of random intercepts and slopes to account for participant variability
- How to test most to least restrictive models
- How to compare parameters and models
doomlab/learnSEM documentation built on Jan. 25, 2024, 2 p.m.
R Package Documentation
Browse R Packages
We want your feedback!
Note that we can't provide technical support on individual packages. You should contact the package authors for that.
knitr::opts_chunk$set( collapse = TRUE, comment = "#>" )
knitr::opts_chunk$set(echo = TRUE) library(lavaan) library(semPlot)
Latent Growth Modeling
- Measuring change over repeated time measurements
- Gives you more information than a traditional repeated measures ANOVA or regression line by itself.
- Advantages:
- Estimate means and covariances separately
- Estimating observed values and unobserved values separately
Assumptions
- Continuous measurement of the DVs
- This assumption is true for many structural models though!
- Time spacing is the same across people
- NOT across measurements, but people need to be spaced the same
- At least three time points per person (otherwise, you are doing a dependent t-test)
- Large samples!
Before you start
- You have to know the expected type of change before you start
- Generally it's linear, however, you can do curvilinear or power functions
Example model
- We will set the regression loading values to specific numbers to be able to estimate intercept and slope.
knitr::include_graphics("pictures/example_lgm.png")
Understand the idea
- Intercept
- You set these values to 1 indicating that you do NOT want to estimate them
- Basically, that gives you a starting value for the first time point, which is the average point where people start (y-intercept)
Understand the idea
- Slope – represents the change over time
- You can set these values to anything you want
- Usually the first time is indicated by a 0
- There's no slope for time 1, just an intercept
- Then the paths are set based on the time differences between them
Understand the idea
- Setting the parameters this way:
- Helps with identification
- Is theoretical to match the concept of slope and intercept estimation
- Allows you to not have to set the variances, so you can look at them
Questions
- What is the average change for each person?
- Where do they start?
- How does it change over time?
- What is the form of the change?
Questions
- Is the average slope and intercept a good fit for all participants?
- Or should we include a variance term to account for the differences between people?
- Models are called random effects if you add the random variances (similar to multilevel models)
- If we have a random effects model – is there another variable that explains those random effects?
Understanding the Output
- Intercept mean: the average starting point for time 1
- Slope mean: the average increment across time points
knitr::include_graphics("pictures/random_fixed.png")
Understanding the Output
- Intercept variance: the spread around the average start point
- Large scores indicate a lot of spread – meaning people start in a lot of different places
- Small scores indicate a small spread – everyone starts about the same place
knitr::include_graphics("pictures/random_fixed.png")
Understanding the Output
- Slope variance – the range of increments across time points
- Small variances mean that everyone is going up/down about the same amount
- Large variances mean that people scores are going up/down differently (almost like an interaction)
knitr::include_graphics("pictures/random_fixed.png")
Understanding the Output
- Factor covariance – examines the relationship between intercept and slope
- Positive covariance - positive slope
- People who start with higher intercepts have higher positive slopes
- Positive covariance - negative slope
- People who start higher go down faster (larger negative slopes)
- Negative covariance - positive slope
- People who start with higher intercepts go up slower (smaller slopes)
- Negative covariance - negative slope
- People who start with higher intercepts go down slower
Example: Crime Data across time
##load the data crime.cov <- lav_matrix_lower2full(c(.63, .50, .60, .48, .48, .58, .47, .48, .51, .67)) crime.mean <- c(5.17, 5.32, 5.40, 5.52) names(crime.mean) <- rownames(crime.cov) <- colnames(crime.cov) <- c("Time1", "Time2", "Time3", "Time4")
New Functions
growth(): This function helps with the set up for a LCM, fixes the parameters correctly for you.- We will use the data as covariances and means, but you can import the data just like
cfa()
How to Test
- This procedure is very similar to multigroup model testing, only in reverse.
- We are going to start with the most constrained model and slowly let parameters free to see which is the best.
- We would prefer the unconstrained model be the best:
- The slope and intercept are useful
- But the variances to be low indicating everyone has the same pattern of slope and intercept
Intercept Only Model
- Use only the intercept (mean of the data) to estimate
- You want this model to be bad, otherwise you are saying that one average score is the best for all the time points
- Intercept variance is constrained to 0, so you only get a mean.
- Residuals are forced to be the same for each time point, so the variance is the same across all time points
crime.model1 <- ' # intercept i =~ 1*Time1 + 1*Time2 + 1*Time3 + 1*Time4 i~~0*i # residual variances Time1~~r*Time1 Time2~~r*Time2 Time3~~r*Time3 Time4~~r*Time4 '
Intercept Only Model
crime.fit1 <- growth(crime.model1, sample.cov=crime.cov, sample.mean=crime.mean, sample.nobs=952) summary(crime.fit1, standardized = TRUE, fit.measures = TRUE, rsquare = TRUE)
Intercept Only Model
semPaths(crime.fit1, whatLabels = "par", edge.label.cex = 1, layout = "tree")
Comparison Table Fit Indices
library(knitr) fit.table <- matrix(NA, nrow = 5, ncol = 6) colnames(fit.table) <- c("Model", "X2", "df", "RMSEA", "SRMR", "CFI") fit.table[1, ] <- c("Intercept Only", round(fitmeasures(crime.fit1, c("chisq", "df", "rmsea", "srmr", "cfi")),3)) kable(fit.table)
Comparison Table Parameters
#save the parameter estimates crime.fit1.par <- parameterestimates(crime.fit1) #make table par.table <- matrix(NA, nrow = 5, ncol = 7) colnames(par.table) <- c("Model", "Intercept Mean", "Intercept Variance", "Residual Variance", "Slope Mean", "Slope Variance", "Covariance") #put data in table par.table[1, ] <- c("Intercept Only", round(crime.fit1.par$est[crime.fit1.par$lhs == "i" & crime.fit1.par$op == "~1"], 3), "X", round(crime.fit1.par$est[crime.fit1.par$lhs == "Time1" & crime.fit1.par$op == "~~"], 3), "X", "X", "X") kable(par.table)
Random Intercept Only Model
- We allow the intercept variance to be > 0
- People can start at different places
- Residual variances are still the same
- Still no slope
crime.model2 <- ' # intercept i =~ 1*Time1 + 1*Time2 + 1*Time3 + 1*Time4 # residual variances Time1~~r*Time1 Time2~~r*Time2 Time3~~r*Time3 Time4~~r*Time4 '
Random Intercept Only Model
crime.fit2 <- growth(crime.model2, sample.cov=crime.cov, sample.mean=crime.mean, sample.nobs=952) summary(crime.fit2, standardized = TRUE, fit.measures = TRUE, rsquare = TRUE)
Comparison Table Fit Indices
- This model is much better than model 1
- Implies that people all start in different places
fit.table[2, ] <- c("Random Intercept", round(fitmeasures(crime.fit2, c("chisq", "df", "rmsea", "srmr", "cfi")),3)) kable(fit.table)
Comparison Table Parameters
#save the parameter estimates crime.fit2.par <- parameterestimates(crime.fit2) #put data in table par.table[2, ] <- c("Random Intercept", round(crime.fit2.par$est[crime.fit2.par$lhs == "i" & crime.fit2.par$op == "~1"], 3), round(crime.fit2.par$est[crime.fit2.par$lhs == "i" & crime.fit2.par$op == "~~"], 3), round(crime.fit2.par$est[crime.fit2.par$lhs == "Time1" & crime.fit2.par$op == "~~"], 3), "X", "X", "X") kable(par.table)
Random Slope + Intercepts
- Add in the random slope
s~0*1(makes the average slope 0)s~~0*i(uncorrelated slope/intercept)- Leave the intercept and its variance in the model
- Keep the residuals constrained to be the same
crime.model3 <- ' # intercept i =~ 1*Time1 + 1*Time2 + 1*Time3 + 1*Time4 # slope s =~ 0*Time1 + 1*Time2 + 2*Time3 + 3*Time4 s~0*1 s~~0*i # residual variances Time1~~r*Time1 Time2~~r*Time2 Time3~~r*Time3 Time4~~r*Time4 '
Random Slope + Intercepts
crime.fit3 <- growth(crime.model3, sample.cov=crime.cov, sample.mean=crime.mean, sample.nobs=952) summary(crime.fit3, standardized = TRUE, fit.measures = TRUE, rsquare = TRUE)
Comparison Table Fit Indices
- This model is much better than model 2
- Implies that having at least a random slope is better than no slope
fit.table[3, ] <- c("Random Slope", round(fitmeasures(crime.fit3, c("chisq", "df", "rmsea", "srmr", "cfi")),3)) kable(fit.table)
Comparison Table Parameters
#save the parameter estimates crime.fit3.par <- parameterestimates(crime.fit3) #put data in table par.table[3, ] <- c("Random Slope", round(crime.fit3.par$est[crime.fit3.par$lhs == "i" & crime.fit3.par$op == "~1"], 3), round(crime.fit3.par$est[crime.fit3.par$lhs == "i" & crime.fit3.par$op == "~~" & crime.fit3.par$rhs == "i"], 3), round(crime.fit3.par$est[crime.fit3.par$lhs == "Time1" & crime.fit3.par$op == "~~"], 3), "X", round(crime.fit3.par$est[crime.fit3.par$lhs == "s" & crime.fit3.par$op == "~~"], 3), "X") kable(par.table)
Full Slopes + Intercept
- Allow the slope and intercept to vary
- Constrain the residual variances
crime.model4 <- ' # intercept i =~ 1*Time1 + 1*Time2 + 1*Time3 + 1*Time4 # slope s =~ 0*Time1 + 1*Time2 + 2*Time3 + 3*Time4 # residual variances Time1~~r*Time1 Time2~~r*Time2 Time3~~r*Time3 Time4~~r*Time4 '
Full Slopes + Intercept
- Significant covariance – negative covariance with positive slope
- People who start high go up slower than people who start lower.
crime.fit4 <- growth(crime.model4, sample.cov=crime.cov, sample.mean=crime.mean, sample.nobs=952) summary(crime.fit4, standardized = TRUE, fit.measures = TRUE, rsquare = TRUE)
Comparison Table Fit Indices
- This model is much better than model 3
- Implies that having a single slope is better than just random slopes
fit.table[4, ] <- c("Full Slope", round(fitmeasures(crime.fit4, c("chisq", "df", "rmsea", "srmr", "cfi")),3)) kable(fit.table)
Comparison Table Parameters
#save the parameter estimates crime.fit4.par <- parameterestimates(crime.fit4) #put data in table par.table[4, ] <- c("Full Slope", round(crime.fit4.par$est[crime.fit4.par$lhs == "i" & crime.fit4.par$op == "~1"], 3), round(crime.fit4.par$est[crime.fit4.par$lhs == "i" & crime.fit4.par$op == "~~" & crime.fit4.par$rhs == "i"], 3), round(crime.fit4.par$est[crime.fit4.par$lhs == "Time1" & crime.fit4.par$op == "~~"], 3), round(crime.fit4.par$est[crime.fit4.par$lhs == "s" & crime.fit4.par$op == "~1"], 3), round(crime.fit4.par$est[crime.fit4.par$lhs == "s" & crime.fit4.par$op == "~~"], 3), round(crime.fit4.par$est[crime.fit4.par$lhs == "i" & crime.fit4.par$op == "~~" & crime.fit4.par$rhs == "s"], 3)) kable(par.table)
Totally Unconstrained Model
- Now the residuals are free to vary
- You want the residuals to be small and roughly equal, so this model shouldn't be any different than the previous model
crime.model5 <- ' # intercept i =~ 1*Time1 + 1*Time2 + 1*Time3 + 1*Time4 # slope s =~ 0*Time1 + 1*Time2 + 2*Time3 + 3*Time4 '
Totally Unconstrained Model
crime.fit5 <- growth(crime.model5, sample.cov=crime.cov, sample.mean=crime.mean, sample.nobs=952) summary(crime.fit5, standardized = TRUE, fit.measures = TRUE, rsquare = TRUE)
Comparison Table Fit Indices
- This model is equal to model 4
- This result implies that your residual variances are all approximately equal
fit.table[5, ] <- c("Unconstrained", round(fitmeasures(crime.fit5, c("chisq", "df", "rmsea", "srmr", "cfi")),3)) kable(fit.table)
Comparison Table Parameters
#save the parameter estimates crime.fit5.par <- parameterestimates(crime.fit5) residual_numbers <- paste(round(crime.fit5.par$est[crime.fit5.par$lhs == "Time1" & crime.fit5.par$op == "~~"], 3), round(crime.fit5.par$est[crime.fit5.par$lhs == "Time2" & crime.fit5.par$op == "~~"], 3), round(crime.fit5.par$est[crime.fit5.par$lhs == "Time3" & crime.fit5.par$op == "~~"], 3), round(crime.fit5.par$est[crime.fit5.par$lhs == "Time4" & crime.fit5.par$op == "~~"], 3)) #put data in table par.table[5, ] <- c("Unconstrained", round(crime.fit5.par$est[crime.fit5.par$lhs == "i" & crime.fit5.par$op == "~1"], 3), round(crime.fit5.par$est[crime.fit5.par$lhs == "i" & crime.fit5.par$op == "~~" & crime.fit5.par$rhs == "i"], 3), residual_numbers, round(crime.fit5.par$est[crime.fit5.par$lhs == "s" & crime.fit5.par$op == "~1"], 3), round(crime.fit5.par$est[crime.fit5.par$lhs == "s" & crime.fit5.par$op == "~~"], 3), round(crime.fit5.par$est[crime.fit5.par$lhs == "i" & crime.fit5.par$op == "~~" & crime.fit5.par$rhs == "s"], 3)) kable(par.table)
Summary
-
In this lecture you've learned:
-
Latent growth modeling
- The idea of random intercepts and slopes to account for participant variability
- How to test most to least restrictive models
- How to compare parameters and models
R Package Documentation
Browse R Packages
We want your feedback!
Note that we can't provide technical support on individual packages. You should contact the package authors for that.
Embedding an R snippet on your website
Add the following code to your website.
For more information on customizing the embed code, read Embedding Snippets.