correlation between slopes, Latent growth modeling

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Nicolas Barreto

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Apr 30, 2018, 5:07:41 PM4/30/18

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Hello again all,

First off, I apologize for the long post. Any help or direction anyone can give me would be of great help.

I just want to confirm that I am using the appropriate method (Latent growth curve modeling) and Lavaan properly. i have a few questions And i will try to ask them as I go.

My PI is interested in "if changes in PH are associated with changes in TNF". Additionally, there are two groups (G1 = BC and G2 = HC) which should have different trajectories. Of note BC has an n = 106 and HC has an n = 56.

The expected trajectory for both PH for BC is to deviate from baseline at time 2, then come closer at time 3, with no differences at time 4, while the trajectory from HC is simply linear. (see the code for PH below)

PH <-'

#intercept

i.PH =~ 1*sq_phT1 + 1*sq_phT2 + 1*sq_phT3 + 1*sq_phT4

#slope

s.PH =~ 0*sq_phT1 + 1*sq_phT2 + c(.5,2)*sq_phT3 + c(.5,3)*sq_phT4

#fixed efects

i.PH ~ 1

s.PH ~ 1

#random effects

i.PH ~~ i.PH

s.PH ~~ s.PH

i.PH ~~ s.PH

summary(lavaan(PH,data=dat, auto.var=T,estimator="MLR", group = "HCvBC"))

Q1) I interpreted my PI's question to be asking about the correlation between slopes of PH and TNF. Does that make sense?

Q2) Is multivariate latent growth curve modeling the most appropriate analytic strategy? I know there are other methods (repeated measure anova, panel models, etc.) to do longitudinal data, but do they answer if the longitudinal change in one variable is associated with the longitudinal change in the other?

Q3) have I done the model specification properly?

below is the model specification for assessing both variables together

PH.TNF <- '

#intercept

i.PH =~ 1*sq_phT1 + 1*sq_phT2 + 1*sq_phT3 + 1*sq_phT4

i.TNF =~ 1*Log10T1sTNFRII + 1*Log10T2sTNFRII + 1*Log10sTNFRII_T3 + 1*Log10sTNFRII_T4

#slope

s.PH =~ 0*sq_phT1 + 1*sq_phT2 + c(.5,2)*sq_phT3 + c(.5,3)*sq_phT4

#s.TNF =~ 0*Log10T1sTNFRII + 1*Log10T2sTNFRII + c(.5,2)*Log10sTNFRII_T3 + c(.5,3)*Log10sTNFRII_T4

s.TNF =~ 0*Log10T1sTNFRII + 1*Log10T2sTNFRII + 2*Log10sTNFRII_T3 + 3*Log10sTNFRII_T4

#fixed effects

i.PH ~ 1

s.PH ~ 1

i.TNF ~ 1

s.TNF ~ 1

#random effects (intercept and slope)

i.PH ~~ i.PH

i.TNF ~~ i.TNF

s.PH ~~ s.PH

s.TNF ~~ s.TNF

i.PH ~~ s.PH

i.TNF ~~ s.TNF

#regressions

s.PH ~ s.TNF

i.PH ~ i.TNF

fit.PHTNF <- lavaan(PH.TNF, data = dat, auto.var = T, estimator = "MLR", group = "HCvBC")

summary(fit.PHTNF, standardized = T, ci = TRUE)

the problem is that when i run it i get a warning. I believe because in the HC group the variance of the slope of PH is estimated as negative, but it isnt when run on its own.

Q4) have I done something wrong?

Warning message:

In lav_object_post_check(object) :

lavaan WARNING: some estimated lv variances are negative

> summary(fit.PHTNF, standardized = T, ci = TRUE)

lavaan (0.5-23.1097) converged normally after 391 iterations

Used Total

Number of observations per group

BC 94 106

HC 53 56

Estimator ML Robust

Minimum Function Test Statistic 111.383 121.369

Degrees of freedom 48 48

P-value (Chi-square) 0.000 0.000

Scaling correction factor 0.918

for the Yuan-Bentler correction

Chi-square for each group:

BC 66.682 72.660

HC 44.701 48.709

Parameter Estimates:

Information Observed

Standard Errors Robust.huber.white

Group 1 [BC]:

Latent Variables:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

i.PH =~

sq_phT1 1.000 1.000 1.000 2.166 0.839

sq_phT2 1.000 1.000 1.000 2.166 0.846

sq_phT3 1.000 1.000 1.000 2.166 0.872

sq_phT4 1.000 1.000 1.000 2.166 0.832

i.TNF =~

Log10T1sTNFRII 1.000 1.000 1.000 0.131 0.963

Log10T2sTNFRII 1.000 1.000 1.000 0.131 0.834

Lg10sTNFRII_T3 1.000 1.000 1.000 0.131 0.857

Lg10sTNFRII_T4 1.000 1.000 1.000 0.131 0.859

s.PH =~

sq_phT1 0.000 0.000 0.000 0.000 0.000

sq_phT2 1.000 1.000 1.000 0.786 0.307

sq_phT3 0.500 0.500 0.500 0.393 0.158

sq_phT4 0.500 0.500 0.500 0.393 0.151

s.TNF =~

Log10T1sTNFRII 0.000 0.000 0.000 0.000 0.000

Log10T2sTNFRII 1.000 1.000 1.000 0.023 0.146

Lg10sTNFRII_T3 2.000 2.000 2.000 0.046 0.300

Lg10sTNFRII_T4 3.000 3.000 3.000 0.069 0.451

Regressions:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

s.PH ~

s.TNF -30.093 11.547 -2.606 0.009 -52.725 -7.461 -0.875 -0.875

i.PH ~

i.TNF -8.884 1.310 -6.779 0.000 -11.452 -6.316 -0.536 -0.536

Covariances:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

.i.PH ~~

.s.PH -0.881 0.832 -1.059 0.290 -2.510 0.749 -1.267 -1.267

i.TNF ~~

s.TNF -0.000 0.000 -0.003 0.998 -0.001 0.001 -0.000 -0.000

Intercepts:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

.i.PH 36.628 4.399 8.326 0.000 28.005 45.250 16.909 16.909

.s.PH -1.340 0.291 -4.608 0.000 -1.910 -0.770 -1.705 -1.705

i.TNF 3.362 0.015 230.834 0.000 3.333 3.390 25.738 25.738

s.TNF 0.014 0.004 3.672 0.000 0.006 0.021 0.605 0.605

.sq_phT1 0.000 0.000 0.000 0.000 0.000

.sq_phT2 0.000 0.000 0.000 0.000 0.000

.sq_phT3 0.000 0.000 0.000 0.000 0.000

.sq_phT4 0.000 0.000 0.000 0.000 0.000

.Log10T1sTNFRII 0.000 0.000 0.000 0.000 0.000

.Log10T2sTNFRII 0.000 0.000 0.000 0.000 0.000

.Lg10sTNFRII_T3 0.000 0.000 0.000 0.000 0.000

.Lg10sTNFRII_T4 0.000 0.000 0.000 0.000 0.000

Variances:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

.i.PH 3.346 0.660 5.067 0.000 2.052 4.640 0.713 0.713

i.TNF 0.017 0.005 3.419 0.001 0.007 0.027 1.000 1.000

.s.PH 0.144 1.352 0.107 0.915 -2.506 2.795 0.234 0.234

s.TNF 0.001 0.000 3.406 0.001 0.000 0.001 1.000 1.000

.sq_phT1 1.972 0.664 2.972 0.003 0.672 3.273 1.972 0.296

.sq_phT2 3.010 0.793 3.794 0.000 1.455 4.565 3.010 0.459

.sq_phT3 2.206 0.467 4.725 0.000 1.291 3.121 2.206 0.357

.sq_phT4 2.807 0.649 4.323 0.000 1.535 4.080 2.807 0.414

.Log10T1sTNFRII 0.001 0.001 1.597 0.110 -0.000 0.003 0.001 0.072

.Log10T2sTNFRII 0.007 0.002 3.274 0.001 0.003 0.011 0.007 0.284

.Lg10sTNFRII_T3 0.004 0.001 3.895 0.000 0.002 0.006 0.004 0.176

.Lg10sTNFRII_T4 0.001 0.001 1.583 0.113 -0.000 0.003 0.001 0.059

Group 2 [HC]:

Latent Variables:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

i.PH =~

sq_phT1 1.000 1.000 1.000 1.773 0.888

sq_phT2 1.000 1.000 1.000 1.773 0.851

sq_phT3 1.000 1.000 1.000 1.773 0.887

sq_phT4 1.000 1.000 1.000 1.773 0.793

i.TNF =~

Log10T1sTNFRII 1.000 1.000 1.000 0.109 0.974

Log10T2sTNFRII 1.000 1.000 1.000 0.109 0.957

Lg10sTNFRII_T3 1.000 1.000 1.000 0.109 0.837

Lg10sTNFRII_T4 1.000 1.000 1.000 0.109 0.941

s.PH =~

sq_phT1 0.000 0.000 0.000 NaN NaN

sq_phT2 1.000 1.000 1.000 NaN NaN

sq_phT3 2.000 2.000 2.000 NaN NaN

sq_phT4 3.000 3.000 3.000 NaN NaN

s.TNF =~

Log10T1sTNFRII 0.000 0.000 0.000 0.000 0.000

Log10T2sTNFRII 1.000 1.000 1.000 0.012 0.104

Lg10sTNFRII_T3 2.000 2.000 2.000 0.024 0.182

Lg10sTNFRII_T4 3.000 3.000 3.000 0.036 0.306

Regressions:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

s.PH ~

s.TNF -9.705 10.200 -0.952 0.341 -29.697 10.286 NaN NaN

i.PH ~

i.TNF -3.570 2.234 -1.598 0.110 -7.948 0.808 -0.220 -0.220

Covariances:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

.i.PH ~~

.s.PH 0.017 0.160 0.104 0.917 -0.297 0.331 0.078 0.078

i.TNF ~~

s.TNF -0.000 0.000 -0.189 0.850 -0.001 0.000 -0.041 -0.041

Intercepts:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

.i.PH 18.934 7.375 2.567 0.010 4.479 33.388 10.680 10.680

.s.PH 0.295 0.148 1.993 0.046 0.005 0.585 NaN NaN

i.TNF 3.327 0.016 211.691 0.000 3.296 3.358 30.412 30.412

s.TNF 0.012 0.002 5.130 0.000 0.008 0.017 1.048 1.048

.sq_phT1 0.000 0.000 0.000 0.000 0.000

.sq_phT2 0.000 0.000 0.000 0.000 0.000

.sq_phT3 0.000 0.000 0.000 0.000 0.000

.sq_phT4 0.000 0.000 0.000 0.000 0.000

.Log10T1sTNFRII 0.000 0.000 0.000 0.000 0.000

.Log10T2sTNFRII 0.000 0.000 0.000 0.000 0.000

.Lg10sTNFRII_T3 0.000 0.000 0.000 0.000 0.000

.Lg10sTNFRII_T4 0.000 0.000 0.000 0.000 0.000

Variances:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

.i.PH 2.991 0.560 5.343 0.000 1.894 4.088 0.951 0.951

i.TNF 0.012 0.002 6.040 0.000 0.008 0.016 1.000 1.000

.s.PH -0.015 0.098 -0.156 0.876 -0.208 0.177 NaN NaN

s.TNF 0.000 0.000 1.016 0.310 -0.000 0.000 1.000 1.000

.sq_phT1 0.843 0.416 2.027 0.043 0.028 1.657 0.843 0.211

.sq_phT2 1.166 0.597 1.952 0.051 -0.005 2.337 1.166 0.269

.sq_phT3 0.804 0.257 3.128 0.002 0.300 1.308 0.804 0.201

.sq_phT4 1.786 0.597 2.990 0.003 0.615 2.956 1.786 0.357

.Log10T1sTNFRII 0.001 0.000 1.420 0.156 -0.000 0.002 0.001 0.052

.Log10T2sTNFRII 0.001 0.000 2.461 0.014 0.000 0.002 0.001 0.082

.Lg10sTNFRII_T3 0.005 0.002 1.965 0.049 0.000 0.010 0.005 0.279

.Lg10sTNFRII_T4 0.001 0.001 0.661 0.509 -0.001 0.002 0.001 0.045

Warning messages:

1: In sqrt(ETA2) : NaNs produced

2: In sqrt(ETA2) : NaNs produced

3: In sqrt(ETA2) : NaNs produced

Terrence Jorgensen

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May 4, 2018, 6:49:50 AM5/4/18

to lavaan

Q1) I interpreted my PI's question to be asking about the correlation between slopes of PH and TNF. Does that make sense?

Sounds reasonable, that's how I would interpret it, too.

Q2) Is multivariate latent growth curve modeling the most appropriate analytic strategy? I know there are other methods (repeated measure anova, panel models, etc.) to do longitudinal data, but do they answer if the longitudinal change in one variable is associated with the longitudinal change in the other?

Since you have nonlinear change of PH in one group, I think it is more complicated. In the case of linear change, a single slope describes the change, so you can look at a single correlation between the latent slopes of PH and TNF. But to model nonlinear change in the other group, it sounds like your expectations would be better suited to using latent difference scores, which are equivalent to having a latent "slope" at each time after baseline that merely represents change from baseline (you could also specify it to be change from the previous time, which is the usual application of latent difference scores). Here is a good book:

http://www.longitudinalsem.com/

Anyway, in that case, you can't boil the "relationship" between the 2 change process down to a single correlation. The change (linear slope?) in TNF will have a correlation between each latent-change factor used to represent nonlinear change in PH. I suppose you could report each one, but talk to your PI about it.

Q3) have I done the model specification properly?

You have 2 groups, so you need to specify both fixed values, even if it is the same fixed value:

i.PH =~ c(1, 1)*sq_phT1 + c(1, 1)*sq_phT2 + ...

And make sure to free parameters using NA, to make sure they are estimated, for example:

i.PH ~ c(NA, NA)*1

#regressions
s.PH ~ s.TNF
i.PH ~ i.TNF
'

Why as regressions? You said you wanted the correlation, so just specify them as covariances (~~)

the problem is that when i run it i get a warning. I believe because in the HC group the variance of the slope of PH is estimated as negative,

Correct

Q4) have I done something wrong?

You have very small samples, so there is a lot of sampling variability in these estimates. As you can see, the 95% CI includes positive values, so this negative-variance estimate is consistent with sampling error.

http://journals.sagepub.com/doi/10.1177/0049124112442138

Variances:
Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

.s.PH -0.015 0.098 -0.156 0.876 -0.208 0.177 NaN NaN

Terrence D. Jorgensen

Postdoctoral Researcher, Methods and Statistics

Research Institute for Child Development and Education, the University of Amsterdam

UvA web page: http://www.uva.nl/profile/t.d.jorgensen

Nicolas Barreto

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May 21, 2018, 6:31:25 PM5/21/18

to lav...@googlegroups.com

Hello,

First, thank you so much for the advice and direction Terrence, it was of great help. I got the Newsome book and the latent difference score models work great.

I do have some follow up questions though,

I still want to compare to groups, and I can write the code to run it, but I am curious how to apply some of the difference score analyses the book suggets. For example, setting the means of the latent change scores to be the same will allow the use of the chi-square test statistic to serve as something of a test for if the change is the same across between all time points (i.e. linear change). My question is how to apply this to multigroup analysis.

below is my code and output for an analysis.

I have set the means to be the same within each subgroup (a for group1 and b for group2), so that i can determine if there is an argument for assuming linearity or not. I can then run a dual latent difference score model with two variables that change over time, and correlate their difference scores at each time point to assess their relationship across time, within each group.

However, I am not sure now what/how i should be interpreting the chi-squares for the model information.

Q1) is the minimum test statistic being non-significant now suggest that the model fits better keeping the groups separated?

Q2) would i look at the individual group chi-square and as the measure of if there is a difference in the differences across time? i.e. for group1 chi-square = 5.061. if that is over the critical value (see Q3) it would be significant, which would suggest that the difference scores are not equivalent across time.

Q3) how do i go about determining the df for that chi-square? my guess is # difference scores - 1...but i am not sure

Any additional help would be wonderful. Thank you, and sorry again for the long post.

il6.multi <- ' #defin latent difference factors;

ch.il2 =~ 1*Log10T2IL6

ch.il3 =~ 1*Log10T3IL6

ch.il4 =~ 1*Log10T4IL6

#autoregressive paths set to 1;

Log10T2IL6 ~ 1*Log10T1IL6

Log10T3IL6 ~ 1*Log10T2IL6

Log10T4IL6 ~ 1*Log10T3IL6

#means and intercepts;

ch.il2~c(a,b)*1

ch.il3~c(a,b)*1

ch.il4~c(a,b)*1

Log10T1IL6~1

Log10T2IL6~0

Log10T3IL6~0

Log10T4IL6~0

#estimate exogenous covariances;

Log10T1IL6 ~~ ch.il2 + ch.il3 + ch.il4

ch.il2 ~~ ch.il3 + ch.il4

ch.il3 ~~ ch.il4

#estimate exogenous variances;

Log10T1IL6 ~~ Log10T1IL6

ch.il2 ~~ ch.il2

ch.il3 ~~ ch.il3

ch.il4 ~~ ch.il4

#disturbances set to 0;

Log10T2IL6 ~~ 0*Log10T2IL6

Log10T3IL6 ~~ 0*Log10T3IL6

Log10T4IL6 ~~ 0*Log10T4IL6

dif.il6.multi <- lavaan(il6.multi, data = dat, auto.var=TRUE, estimator = "MLR",

group = "HCvBC", group.equal = "loadings")

> summary(dif.il6.multi, standardized = T, ci = T)

lavaan (0.5-23.1097) converged normally after 114 iterations

Used Total

Number of observations per group

Breast Cancer 92 106

Healthy Control 52 56

Estimator ML Robust

Minimum Function Test Statistic 6.068 6.683

Degrees of freedom 4 4

P-value (Chi-square) 0.194 0.154

Scaling correction factor 0.908

for the Yuan-Bentler correction

Chi-square for each group:

Breast Cancer 5.061 5.574

Healthy Control 1.007 1.109

Parameter Estimates:

Information Observed

Standard Errors Robust.huber.white

Group 1 [Breast Cancer]:

Latent Variables:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

ch.il2 =~

Log10T2IL6 1.000 1.000 1.000 0.278 0.961

ch.il3 =~

Log10T3IL6 1.000 1.000 1.000 0.248 0.834

ch.il4 =~

Log10T4IL6 1.000 1.000 1.000 0.214 0.769

Regressions:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

Log10T2IL6 ~

Log10T1IL6 1.000 1.000 1.000 1.000 1.032

Log10T3IL6 ~

Log10T2IL6 1.000 1.000 1.000 1.000 0.970

Log10T4IL6 ~

Log10T3IL6 1.000 1.000 1.000 1.000 1.068

Covariances:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

ch.il2 ~~

Log10T1IL6 -0.041 0.017 -2.497 0.013 -0.074 -0.009 -0.149 -0.499

ch.il3 ~~

Log10T1IL6 0.005 0.009 0.525 0.599 -0.013 0.022 0.019 0.062

ch.il4 ~~

Log10T1IL6 -0.007 0.007 -1.070 0.284 -0.020 0.006 -0.033 -0.110

ch.il2 ~~

ch.il3 -0.033 0.009 -3.531 0.000 -0.051 -0.015 -0.476 -0.476

ch.il4 0.000 0.006 0.066 0.947 -0.012 0.013 0.007 0.007

ch.il3 ~~

ch.il4 -0.022 0.007 -3.011 0.003 -0.036 -0.008 -0.410 -0.410

Intercepts:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

ch.il2 (a) -0.003 0.011 -0.258 0.796 -0.024 0.018 -0.010 -0.010

ch.il3 (a) -0.003 0.011 -0.258 0.796 -0.024 0.018 -0.011 -0.011

ch.il4 (a) -0.003 0.011 -0.258 0.796 -0.024 0.018 -0.013 -0.013

Log10T1IL6 0.319 0.031 10.130 0.000 0.257 0.380 0.319 1.067

.Log10T2IL6 0.000 0.000 0.000 0.000 0.000

.Log10T3IL6 0.000 0.000 0.000 0.000 0.000

.Log10T4IL6 0.000 0.000 0.000 0.000 0.000

Variances:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

Log10T1IL6 0.089 0.019 4.694 0.000 0.052 0.126 0.089 1.000

ch.il2 0.077 0.015 5.014 0.000 0.047 0.107 1.000 1.000

ch.il3 0.062 0.014 4.277 0.000 0.033 0.090 1.000 1.000

ch.il4 0.046 0.008 5.668 0.000 0.030 0.062 1.000 1.000

.Log10T2IL6 0.000 0.000 0.000 0.000 0.000

.Log10T3IL6 0.000 0.000 0.000 0.000 0.000

.Log10T4IL6 0.000 0.000 0.000 0.000 0.000

Group 2 [Healthy Control]:

Latent Variables:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

ch.il2 =~

Log10T2IL6 1.000 1.000 1.000 0.139 0.661

ch.il3 =~

Log10T3IL6 1.000 1.000 1.000 0.166 0.802

ch.il4 =~

Log10T4IL6 1.000 1.000 1.000 0.195 0.843

Regressions:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

Log10T2IL6 ~

Log10T1IL6 1.000 1.000 1.000 1.000 1.137

Log10T3IL6 ~

Log10T2IL6 1.000 1.000 1.000 1.000 1.019

Log10T4IL6 ~

Log10T3IL6 1.000 1.000 1.000 1.000 0.892

Covariances:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

ch.il2 ~~

Log10T1IL6 -0.016 0.006 -2.605 0.009 -0.028 -0.004 -0.116 -0.486

ch.il3 ~~

Log10T1IL6 -0.009 0.006 -1.598 0.110 -0.020 0.002 -0.054 -0.226

ch.il4 ~~

Log10T1IL6 0.008 0.007 1.205 0.228 -0.005 0.021 0.041 0.172

ch.il2 ~~

ch.il3 -0.006 0.004 -1.573 0.116 -0.013 0.001 -0.242 -0.242

ch.il4 -0.002 0.003 -0.743 0.458 -0.009 0.004 -0.091 -0.091

ch.il3 ~~

ch.il4 -0.019 0.004 -4.586 0.000 -0.027 -0.011 -0.592 -0.592

Intercepts:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

ch.il2 (b) -0.012 0.008 -1.549 0.121 -0.028 0.003 -0.089 -0.089

ch.il3 (b) -0.012 0.008 -1.549 0.121 -0.028 0.003 -0.075 -0.075

ch.il4 (b) -0.012 0.008 -1.549 0.121 -0.028 0.003 -0.063 -0.063

Log10T1IL6 0.176 0.031 5.739 0.000 0.116 0.236 0.176 0.735

.Log10T2IL6 0.000 0.000 0.000 0.000 0.000

.Log10T3IL6 0.000 0.000 0.000 0.000 0.000

.Log10T4IL6 0.000 0.000 0.000 0.000 0.000

Variances:

Estimate Std.Err z-value P(>|z|) ci.lower ci.upper Std.lv Std.all

Log10T1IL6 0.057 0.012 4.976 0.000 0.035 0.080 0.057 1.000

ch.il2 0.019 0.004 5.006 0.000 0.012 0.027 1.000 1.000

ch.il3 0.027 0.005 5.453 0.000 0.018 0.037 1.000 1.000

ch.il4 0.038 0.007 5.331 0.000 0.024 0.052 1.000 1.000

.Log10T2IL6 0.000 0.000 0.000 0.000 0.000

.Log10T3IL6 0.000 0.000 0.000 0.000 0.000

.Log10T4IL6 0.000 0.000 0.000 0.000 0.000

Nicolas Barreto, MPH

Doctoral Student, Applied Social Psychology

Department of Behavioral and Organizational Science

Claremont Graduate University

nclsb...@gmail.com

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Terrence Jorgensen

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May 22, 2018, 12:23:23 AM5/22/18

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Q1) is the minimum test statistic being non-significant now suggest that the model fits better keeping the groups separated?

The test statistic for a model is testing whether that model fits as well as a saturated model (i.e., freely estimating all possible means and (co)variances). It is not comparing that model to any other restricted/hypothesized model.

Q2) would i look at the individual group chi-square and as the measure of if there is a difference in the differences across time? i.e. for group1 chi-square = 5.061. if that is over the critical value (see Q3) it would be significant, which would suggest that the difference scores are not equivalent across time.

The group contributions to the chi-squared are not test statistics, so you cannot use them to test any hypothesis.

Q3) how do i go about determining the df for that chi-square? my guess is # difference scores - 1...but i am not sure

The df for the test is listed with the test statistic. In order to test a particular hypothesis, you need to specify a model under which the null hypothesis is true (e.g., constrain differences to be equal, implying linearity) and a model under which the null is false (i.e., without the constraint), then compare those models using the anova() function, which returns a chi-squared difference test (along with its df).

Nicolas Barreto

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May 23, 2018, 7:28:00 PM5/23/18

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Terrence, thank you so much for all your help, i am sorry for all the questions.

Q1) is the minimum test statistic being non-significant now suggest that the model fits better keeping the groups separated?

Q3) how do i go about determining the df for that chi-square? my guess is # difference scores - 1...but i am not sure

Ah yes, that makes sense. My goal at this stage is to determine if the mean difference score at each time point is different between group 1 and group 2. So just to confirm that i have the logic right there are four possible versions of the constraints on the means (see below) and i would use the anova() command to compare between then (though the minimum test statistic would compare all of the models to free estimation). So my question is;

Q1) a significant minimum test statistic means that a given model is significantly different from free estimation of everything, which means that there is evidence that that constraint is not accurate (so if i constrained between the groups and it was significant then the groups should be different). Is that correct?

free estimation

#means;

ch.tnf2~1

ch.tnf3~1

ch.tnf4~1

Constrianed by time (T1 for group1 and group2 forced to be the same etc.)

#means;

ch.tnf2~a*1

ch.tnf3~b*1

ch.tnf4~c*1

constrained by group

#means;

ch.tnf2~ c(a,b)*1

ch.tnf3~ c(a,b)*1

ch.tnf4~ c(a,b)*1

Complete constraint

#means;

ch.tnf2~ m*1

ch.tnf3~ m*1

ch.tnf4~ m*1

Nicolas Barreto, MPH

Doctoral Student, Applied Social Psychology

Department of Behavioral and Organizational Science

Claremont Graduate University

nclsb...@gmail.com

Terrence Jorgensen

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Jun 2, 2018, 9:49:17 AM6/2/18

to lavaan

Q1) a significant minimum test statistic means that a given model is significantly different from free estimation of everything, which means that there is evidence that that constraint is not accurate (so if i constrained between the groups and it was significant then the groups should be different). Is that correct?

If the models are nested, the null hypothesis is that the more restricted model is correct, so the less restricted model estimates additional parameters that can be fixed/equated. As in any other NHST scenario, a significant test statistic implies you should reject the null.

Nicolas Barreto

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Jun 4, 2018, 11:33:39 AM6/4/18

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Excellent.

Thank you again for all the help. You have been extremely helpful.

Best,

Nic

Nicolas Barreto, MPH

Doctoral Student, Applied Social Psychology

Department of Behavioral and Organizational Science

Claremont Graduate University

nclsb...@gmail.com

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