CFA warning message

Gouvidé Jean GBAGUIDI

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Feb 3, 2023, 2:21:12 PM2/3/23

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Hello!,I'm running CFA . lavaan ends nrmally and I can get all the information with poor fit of the model due to the warning message.

Warning message:
In lav_object_post_check(object) :
lavaan WARNING: the covariance matrix of the residuals of the observed
variables (theta) is not positive definite;
use lavInspect(fit, "theta") to investigate.

Can someone please give guidance on where I should go with this error?

See below my code and the syntax of the model.

setwd("~/My topic/My topic/DATA/VALID DATA/Data_final")
Mal<-read.csv("CC_MalariaS.csv",header = TRUE,sep = ",")
str('Mal')
head(Mal)
names(Mal)
attach(Mal)
summary(is.na(Mal))
str(Mal)
#### Convert to numerical#############
Mal$Min_Relative.Humidity<-as.numeric(as.factor(Mal$Min_Relative.Humidity))
Mal$Max_Relative.Humidity<-as.numeric(as.factor(Mal$Max_Relative.Humidity))#######
str(Mal)

####Smooth the data(non positive matrix)###########
d<-cor.smooth(Mal,eig.tol = 10^-4)### Smooth the matrix(non positive)

cfaMod2<-"F1=~Rainfall+Min_Relative.Humidity+Mean_Relative.Humidity
+Max_Relative.Humidity
F2=~a*Mean_Temperature+a*Min_Temperature
Malaria_Incidence~F1+F2

Rainfall~~Max_Temperature+Min_Relative.Humidity+Mean_Relative.Humidity
+Max_Relative.Humidity
Mean_Temperature~~Max_Temperature
Max_Temperature~~Min_Relative.Humidity+Max_Relative.Humidity+Mean_Relative.Humidity
Mean_Relative.Humidity~~Min_Relative.Humidity+Max_Relative.Humidity
Min_Relative.Humidity~~Max_Relative.Humidity
"
fit.cfaMod<-cfa(cfaMod2,sample.cov=d,sample.nobs=1193, estimator = "ML",bounds=F,std.lv=T,
orthogonal=T)
summary(fit.cfaMod,fit.measures=TRUE,standardized=TRUE,rsq=TRUE)

> lavInspect(fit.cfaMod, "theta")
Ranfll Min_Rltv.Hmdty Men_Rltv.Hmdty Mx_R.H
Rainfall 0.684
Min_Relative.Humidity 0.323 0.265
Mean_Relative.Humidity 0.234 0.168 0.118
Max_Relative.Humidity 0.114 0.033 0.037 0.022
Mean_Temperature 0.000 0.000 0.000 0.000
Min_Temperature 0.000 0.000 0.000 0.000
Malaria_Incidence 0.000 0.000 0.000 0.000
Max_Temperature -0.082 -0.087 -0.083 -0.075
Men_Tmprtr Min_Tmprtr Mlr_In Mx_Tmp
Rainfall
Min_Relative.Humidity
Mean_Relative.Humidity
Max_Relative.Humidity
Mean_Temperature 0.369
Min_Temperature 0.000 0.388
Malaria_Incidence 0.000 0.000 0.929
Max_Temperature 0.708 0.000 0.000 0.867

Yves Rosseel

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Feb 5, 2023, 11:10:03 AM2/5/23

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The residual variance for Max_Relative.Humidity is 0.022, but some other
values on that row are larger!

It looks like you have too many residual covariances in your model. You
could perhaps have one or two (if you have a good rationale for it), but
not that many.

Yves.

Gouvidé Jean GBAGUIDI

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Feb 5, 2023, 1:26:04 PM2/5/23

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Than you for your answer, please I'm sorry to ask you once again how I can have one or two residuals. Should I remove some variables like max_relative humidity and others? Please . Kindly tell me what I can do to solve this warning ⚠️.

My Best regards,

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Gouvidé Jean GBAGUIDI

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Feb 6, 2023, 6:42:25 AM2/6/23

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Hello Yves, thank you for feed back. I try to reduce the residuals covariance by my understanding through the syntaxbelow but I got the same warning message:

Warning message:
In lav_object_post_check(object) :
lavaan WARNING: the covariance matrix of the residuals of the observed
variables (theta) is not positive definite;
use lavInspect(fit, "theta") to investigate.

I will be very grateful for your deep help to solve this warning message

cfaMod2<-"F1=~Rainfall+Min_Relative.Humidity+Mean_Relative.Humidity
+Max_Relative.Humidity
F2=~Mean_Temperature+b1*Min_Temperature
Malaria_Incidence~F1+F2

Mean_Temperature~~Max_Temperature
Min_Relative.Humidity~~Max_Relative.Humidity
"
fit.cfaMod<-cfa(cfaMod2,sample.cov=d,sample.nobs=1193, estimator = "ML",orthogonal=T,std.lv=T,)

Le dim. 5 févr. 2023 à 17:10, Yves Rosseel <yros...@gmail.com> a écrit :

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Gouvidé Jean GBAGUIDI
WASCAL PhD-Candidate
Climate Change and Disaster Risks Management
MSc Ecohydrologist

MSc: Natural Sciences

Secretary of Benin Jeunesse Elite NGO

Benin National Focal Point: United International Federation

of Youth for Water and Climate(UN1FY)

Infolines:be...@un1fy.org

E-mail: gouvid...@gmail.com/gbaguid...@yahoo.fr
Phone:(00229)96258002/95036686 (Benin)

(00228)93649692 (Togo-LOME)
Phillipians: 4:13: "I can do all things through him who gives"

Yves Rosseel

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Feb 6, 2023, 7:18:04 AM2/6/23

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> cfaMod2<-"F1=~Rainfall+Min_Relative.Humidity+Mean_Relative.Humidity
> +Max_Relative.Humidity
> F2=~Mean_Temperature+b1*Min_Temperature
> Malaria_Incidence~F1+F2
>
> Mean_Temperature~~Max_Temperature
> Min_Relative.Humidity~~Max_Relative.Humidity
> "

You need to remove the last two lines of your syntax (the the two lines
with ~~ operator)

Yves.

Gouvidé Jean GBAGUIDI

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Feb 6, 2023, 7:31:44 AM2/6/23

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Thank you Yves for your assistance, please Yves, when I remove

Mean_Temperature~~Max_Temperature
Min_Relative.Humidity~~Max_Relative.Humidity

from my syntaxe I got the message:

Warning message:
In lav_object_post_check(object) :
lavaan WARNING: some estimated ov variances are negative

My question is how can I deal with the negative variance to reduce the residuals

This the output of my summary:

> summary(fit.cfaMod,fit.measures=TRUE,standardized=TRUE,rsq=TRUE)
lavaan 0.6.13 ended normally after 83 iterations

Estimator ML
Optimization method NLMINB
Number of model parameters 15

Number of observations 1193

Model Test User Model:

Test statistic 3103.257
Degrees of freedom 13
P-value (Chi-square) 0.000

Model Test Baseline Model:

Test statistic 14022.412
Degrees of freedom 21
P-value 0.000

User Model versus Baseline Model:

Comparative Fit Index (CFI) 0.779
Tucker-Lewis Index (TLI) 0.643

Loglikelihood and Information Criteria:

Loglikelihood user model (H0) -6386.477
Loglikelihood unrestricted model (H1) -4834.848

Akaike (AIC) 12802.953
Bayesian (BIC) 12879.217
Sample-size adjusted Bayesian (SABIC) 12831.571

Root Mean Square Error of Approximation:

RMSEA 0.446
90 Percent confidence interval - lower 0.433
90 Percent confidence interval - upper 0.460
P-value H_0: RMSEA <= 0.050 0.000
P-value H_0: RMSEA >= 0.080 1.000

Standardized Root Mean Square Residual:

SRMR 0.234

Parameter Estimates:

Standard errors Standard
Information Expected
Information saturated (h1) model Structured

Latent Variables:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
F1 =~
Rainfall 0.424 0.022 18.928 0.000 0.424 0.424
Mn_Rltv.H 0.957 0.021 44.969 0.000 0.957 0.958
Mn_Rltv.H 1.022 0.020 51.107 0.000 1.022 1.022
Mx_Rltv.H 0.954 0.021 44.679 0.000 0.954 0.954
F2 =~
Mn_Tmprtr 1.260 0.122 10.302 0.000 1.260 1.260
Mn_Tmprtr (b1) 0.498 0.055 9.025 0.000 0.498 0.498

Regressions:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
Malaria_Incidence ~
F1 0.069 0.018 3.858 0.000 0.069 0.069
F2 -0.233 0.036 -6.447 0.000 -0.233 -0.234

Covariances:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
F1 ~~
F2 0.000 0.000 0.000

Variances:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
.Rainfall 0.819 0.031 26.748 0.000 0.819 0.820
.Min_Rltv.Hmdty 0.083 0.003 26.405 0.000 0.083 0.083
.Men_Rltv.Hmdty -0.045 0.002 -25.157 0.000 -0.045 -0.045
.Max_Rltv.Hmdty 0.089 0.003 26.466 0.000 0.089 0.090
.Mean_Temperatr -0.587 0.307 -1.912 0.056 -0.587 -0.588
.Min_Temperatur 0.752 0.057 13.225 0.000 0.752 0.752
.Malaria_Incdnc 0.931 0.039 23.625 0.000 0.931 0.941
F1 1.000 1.000 1.000
F2 1.000 1.000 1.000

R-Square:
Estimate
Rainfall 0.180
Min_Rltv.Hmdty 0.917
Men_Rltv.Hmdty NA
Max_Rltv.Hmdty 0.910
Mean_Temperatr NA
Min_Temperatur 0.248
Malaria_Incdnc 0.059

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Keith Markus

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Feb 6, 2023, 11:15:35 PM2/6/23

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Gouvide,

You can think of the broader conceptual issue with residual covariances in this way: You are fitting a model to a covariance matrix. If you free all the residual covariances between the variables, this is sufficient to trivially reproduce their covariance matrix and any additional free parameters are redundant. As a general rule, the more residual covariances you free, the less work there is for the rest of your model parameters to do.

So, it is actually progress that you now have negative variances. This probably indicates that you now have enough constraints in your model to make it testable. Quite possibly what this indicates is that your observed covariance matrix cannot be reproduced by your model structure. Some variables may covary more than the model predicts and some parameters may therefore be estimated in a biased manner in an attempt to accommodate that.

You have two variables with negative disturbance variances, suggesting that the model is trying to predict more variance in those variables than they have. One possibility is that, because they are both means, your mean variables covary with one another more than they covary with other measures that are not means. You could conceptualize that as shared method variance. If you have enough observed variables, you can model that using multi-trait, multi-method models. Otherwise, you might try simply removing the mean indicator for F1 from the model.

Keith

------------------------
Keith A. Markus
John Jay College of Criminal Justice, CUNY
http://jjcweb.jjay.cuny.edu/kmarkus
Frontiers of Test Validity Theory: Measurement, Causation and Meaning.
http://www.routledge.com/books/details/9781841692203/

Gouvidé Jean GBAGUIDI

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Feb 7, 2023, 5:10:12 AM2/7/23

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Dear Keith,

Thank you for the assistance, after reading your comments, I tried to remove the variable Mean_relative Humidity from but I got the same warning message.

Warning message:
In lav_object_post_check(object) :
lavaan WARNING: some estimated ov variances are negative

Here is my new syntaxe

d<-cor.smooth(Mal,eig.tol = 10^-4)

cfaMod2<-"F1=~Rainfall+Min_Relative.Humidity
+Max_Relative.Humidity
F2=~Mean_Temperature+Min_Temperature
Malaria_Incidence~F1+F2

"
fit.cfaMod<-cfa(cfaMod2,sample.cov=d,sample.nobs=1193, estimator = "ML",

orthogonal=T,std.lv=T,bounds=F)
summary(fit.cfaMod,fit.measures=TRUE,standardized=TRUE,rsq=TRUE)

This is the output of my summary.

> summary(fit.cfaMod,fit.measures=TRUE,standardized=TRUE,rsq=TRUE)
lavaan 0.6.13 ended normally after 24 iterations

Estimator ML
Optimization method NLMINB

Number of model parameters 13

Number of observations 1193

Model Test User Model:

Test statistic 1870.534
Degrees of freedom 8

P-value (Chi-square) 0.000

Model Test Baseline Model:

Test statistic 5682.608
Degrees of freedom 15

P-value 0.000

User Model versus Baseline Model:

Comparative Fit Index (CFI) 0.671
Tucker-Lewis Index (TLI) 0.384

Loglikelihood and Information Criteria:

Loglikelihood user model (H0) -8247.724
Loglikelihood unrestricted model (H1) -7312.457

Akaike (AIC) 16521.447
Bayesian (BIC) 16587.542
Sample-size adjusted Bayesian (SABIC) 16546.249

Root Mean Square Error of Approximation:

RMSEA 0.442
90 Percent confidence interval - lower 0.425
90 Percent confidence interval - upper 0.459

P-value H_0: RMSEA <= 0.050 0.000
P-value H_0: RMSEA >= 0.080 1.000

Standardized Root Mean Square Residual:

SRMR 0.192

Parameter Estimates:

Standard errors Standard
Information Expected
Information saturated (h1) model Structured

Latent Variables:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
F1 =~

Rainfall 0.789 0.024 32.295 0.000 0.789 0.789
Min_Rltv.Hmdty 1.022 0.021 49.101 0.000 1.022 1.023
Max_Rltv.Hmdty 0.861 0.023 36.703 0.000 0.861 0.862
F2 =~
Mean_Temperatr 0.838 0.055 15.288 0.000 0.838 0.838
Min_Temperatur 0.748 0.051 14.772 0.000 0.748 0.748

Regressions:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
Malaria_Incidence ~

F1 0.248 0.027 9.179 0.000 0.248 0.249
F2 -0.252 0.031 -8.088 0.000 -0.252 -0.253

Covariances:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
F1 ~~
F2 0.000 0.000 0.000

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

.Rainfall 0.377 0.017 22.254 0.000 0.377 0.377
.Min_Rltv.Hmdty -0.046 0.012 -3.811 0.000 -0.046 -0.046
.Max_Rltv.Hmdty 0.257 0.013 19.081 0.000 0.257 0.258
.Mean_Temperatr 0.297 0.084 3.538 0.000 0.297 0.297
.Min_Temperatur 0.440 0.069 6.405 0.000 0.440 0.440
.Malaria_Incdnc 0.866 0.036 23.969 0.000 0.866 0.874

F1 1.000 1.000 1.000
F2 1.000 1.000 1.000

R-Square:
Estimate

Rainfall 0.623
Min_Rltv.Hmdty NA
Max_Rltv.Hmdty 0.742
Mean_Temperatr 0.703
Min_Temperatur 0.560
Malaria_Incdnc 0.126

I would like to let you know also that I don't know how to test the first possibility,which is the use of multi-trait, multi-method models. I'm new in lavaan and I'm not quite good in lavaan.

I can notice that all my eigen values is positive.

> eigen(d)
eigen() decomposition
$values
[1] 4.6794554401 2.0082591013 0.9422425056 0.8286588823 0.3271748668
[6] 0.1326205058 0.0710471903 0.0099625567 0.0005789511

$vectors
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 0.3953031 -0.05058885 -0.14584043 0.1573405672 0.77828175 -0.43097320
[2,] -0.2307815 -0.56345073 0.32597342 -0.0576799319 0.11081673 -0.08695177
[3,] 0.1263033 -0.65886525 0.06176549 -0.0750538018 0.15566622 0.46782981
[4,] -0.4043429 -0.17441095 0.36931407 -0.0114638993 0.01239381 -0.50547677
[5,] 0.4435527 -0.13343348 0.11857213 0.0470538894 -0.26665050 -0.11217114
[6,] 0.4441680 -0.11190549 0.04693173 0.0770007862 -0.06238938 0.21854322
[7,] 0.4169275 -0.14813454 0.17020707 -0.0009070617 -0.47547487 -0.45471791
[8,] -0.1029920 -0.34572211 -0.78230345 -0.3703817083 -0.17026329 -0.25489301
[9,] 0.1804700 0.20891744 0.27392146 -0.9059944012 0.16598412 0.01573764
[,7] [,8] [,9]
[1,] -0.06074803 -2.230407e-06 7.929241e-05
[2,] 0.07214553 -7.032796e-01 5.424215e-03
[3,] -0.29679119 4.573540e-01 -1.190007e-02
[4,] 0.34262031 5.442657e-01 1.700349e-02
[5,] 0.21315943 -3.120347e-07 -8.000383e-01
[6,] 0.72111921 2.270963e-06 4.586870e-01
[7,] -0.43969563 4.080371e-06 3.861080e-01
[8,] 0.16348174 4.610788e-06 -1.123261e-03
[9,] 0.01107399 -2.591047e-06 -2.055218e-03

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Shu Fai Cheung (張樹輝)

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Feb 7, 2023, 7:00:58 AM2/7/23

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Dear Gouvide,

I do not have a solution but I am thinking about what else you can explore.

I tried to recover the covariance matrix from the eigenvalues and vectors, and then examine the correlations. Some of them are .80 or even .90. I don't know what the columns represent but maybe you can check whether some indicators are so strongly correlated that you may consider using only one of them.

You can also try fitting a CFA model with only F1 (Rainfall, Min_Relative.Humidity, Max_Relative.Humidity) because it is Min_Relative.Humidity that has a negative error variance:

cfaMod2_F1 <- "F1 =~ Rainfall + Min_Relative.Humidity + Max_Relative.Humidity"

The 3-indicator-1-factor model is a saturated model. We usually don't fit this model. However, I came across cases in which the error variance of an indicator can be negative in this saturated model, regardless of sample size. Maybe your case has a similar problem. If yes, then the solution may be different, although I do not yet know what the solution is, other than removing one of the indicators.

-- Shu Fai

Gouvidé Jean GBAGUIDI

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Feb 7, 2023, 7:32:50 AM2/7/23

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Shu,

Thank you so much for your response and interest. The goal of this work is to model the link between climate variables(Rainfall, temperature, relative humidity, wind speed) and the incidence of malaria. I have 1193 observations after cleaning my data.

Here is the correlation between the variables.

Correlation
Rainfall Mean_Temperature Min_Temperature
Rainfall 1
Mean_Temperature -0.39 1
Min_Temperature 0.3 0.63 1
Max_Temperature -0.75 0.76 -0.02
Mean_Relative.Humidity 0.76 -0.3 0.42
Min_Relative.Humidity 0.81 -0.34 0.4
Max_Relative.Humidity 0.67 -0.25 0.41
Wind.Speed -0.13 0.28 0.35
Malaria_Incidence 0.2 -0.3 -0.09
Max_Temperature Mean_Relative.Humidity
Rainfall
Mean_Temperature
Min_Temperature
Max_Temperature 1
Mean_Relative.Humidity -0.74 1
Min_Relative.Humidity -0.78 0.97
Max_Relative.Humidity -0.66 0.97
Wind.Speed 0.07 -0.2
Malaria_Incidence -0.31 0.3
Min_Relative.Humidity Max_Relative.Humidity Wind.Speed
Rainfall
Mean_Temperature
Min_Temperature
Max_Temperature
Mean_Relative.Humidity
Min_Relative.Humidity 1
Max_Relative.Humidity 0.88 1
Wind.Speed -0.19 -0.19 1
Malaria_Incidence 0.28 0.31 -0.17
Malaria_Incidence
Rainfall
Mean_Temperature
Min_Temperature
Max_Temperature
Mean_Relative.Humidity
Min_Relative.Humidity
Max_Relative.Humidity
Wind.Speed
Malaria_Incidence 1

These are the output of EFA

-Parallel analysis suggests that the number of factors = 3 and the number of components = 2

> parallel$fa.values
[1] 4.43593272 1.20387208 0.24724435 0.02491603 -0.02509572 -0.07242975
[7] -0.13741032 -0.46940710 -0.80233980

so with the Kaiser values, I should have at least 2 factors. kaiser greater than 1 indicates the number of factors.

Factors analysis

> print(Twofactor,cut = 0.3,digits = 2)
Factor Analysis using method = ml
Call: fa(r = data.frame(subsetMal1), nfactors = 2, rotate = "oblimin",
SMC = F, covar = T, fm = "ML")
Standardized loadings (pattern matrix) based upon correlation matrix
ML1 ML2 h2 u2 com
Rainfall 0.76 0.616 0.3836 1.1
Mean_Temperature 0.94 0.995 0.0049 1.1
Min_Temperature 0.48 0.82 0.804 0.1957 1.6
Max_Temperature -0.71 0.52 0.860 0.1400 1.8
Mean_Relative.Humidity 1.00 0.995 0.0046 1.0
Min_Relative.Humidity 0.97 0.950 0.0497 1.0
Max_Relative.Humidity 0.97 0.926 0.0743 1.0
Wind.Speed 0.091 0.9091 1.9
Malaria_Incidence 0.137 0.8633 1.8

ML1 ML2
SS loadings 4.40 1.97
Proportion Var 0.49 0.22
Cumulative Var 0.49 0.71
Proportion Explained 0.69 0.31
Cumulative Proportion 0.69 1.00

With factor correlations of
ML1 ML2
ML1 1.00 -0.12
ML2 -0.12 1.00

Mean item complexity = 1.4
Test of the hypothesis that 2 factors are sufficient.

The degrees of freedom for the null model are 36 and the objective function was 15.85
The degrees of freedom for the model are 19 and the objective function was 5.44

The root mean square of the residuals (RMSR) is 0.07
The df corrected root mean square of the residuals is 0.09

Fit based upon off diagonal values = 0.98
Measures of factor score adequacy
ML1 ML2
Correlation of (regression) scores with factors 1.00 1.00
Multiple R square of scores with factors 1.00 0.99
Minimum correlation of possible factor scores 0.99 0.99

The different syntax of my model

Model1: One factor

cfaMod1<-'F=~Rainfall+Min_Relative.Humidity+Mean_Relative.Humidity
+Max_Relative.Humidity
Malaria_Incidence~F
Malaria_Incidence~~Malaria_Incidence'

Model2: Two factors

cfaMod2<-"F1=~Rainfall+Min_Relative.Humidity+Mean_Relative.Humidity
+Max_Relative.Humidity
F2=~Mean_Temperature+Min_Temperature
Malaria_Incidence~F1+F2
"

fit.cfaMod<-cfa(cfaMod2,sample.cov=d,sample.nobs=1193, estimator = "ML",orthogonal=T,std.lv=T,)

summary(fit.cfaMod,fit.measures=TRUE,standardized=TRUE,rsq=TRUE)

I got the same warning message when I run the two different model.

Warning message:
In lav_object_post_check(object) :
lavaan WARNING: some estimated ov variances are negative

Thank you for more attention on the data ,Shu

I attach my data

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CC_MalariaS.csv

Prof. Gavin Brown

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Feb 7, 2023, 2:51:18 PM2/7/23

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Hi Gouvidé

Your negative variance result happens for a number of reasons.

One reason is randomness. It is possible to set a negative value to a small positive value if the standard error range crosses zero into positive territory and if there is evidence from previous studies that the true value is positive.

In your case

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

.Min_Rltv.Hmdty -0.046 0.012 -3.811 0.000 -0.046 -0.046
the value for this variable is clearly much larger than the standard error (see the z value) so you should not force it to become .005

So this speaks to a serious mis-specification of your model.

I suspect this variable is highly correlated with .Max_Rltv.Hmdty (I bet these are statistically and logically linearly dependent). In that case I would delete one of these two variables (preferably the one with the negative error value).

See

Chen, F., Bollen, K. A., Paxton, P., Curran, P. J., & Kirby, J. B. (2001). Improper solutions in structural equation models: Causes, consequences, and strategies. Sociological Methods & Research, 29(4), 468-508. https://doi.org/10.1177/0049124101029004003

Gouvidé Jean GBAGUIDI

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Feb 7, 2023, 4:33:33 PM2/7/23

to lav...@googlegroups.com

Thank you so much Prof Gavin. Yes exactly, I have tried to remove Min_Relative Humidity from the syntax but the issu is that lavan indicates that Pearson correlations of the raw data were found when I compute the code: > d<-cor.smooth(Mal,eig.tol = 10^-4)### Smooth the matrix(non positive).

I can compute the number of factors

frr<-char2numeric(subsetMal1)
> parallel <-fa.parallel(frr,fm="ML",fa='both',
+ main="Parallel Analysis Scree Plots",SMC=T,n.obs = 1193)
In smc, smcs < 0 were set to .0
Parallel analysis suggests that the number of factors = 4 and the number of components = 2

> parallel$fa.values
[1] 3.54167884 1.45748687 0.46106118 0.12922021 -0.02409378 -0.03378708
[7] -0.18159654 -0.72522599

But I got error when I compute this code:

twofactor<-fa(data.frame(subsetMal1),nfactors =2,rotate ="oblimin",fm="ML",SMC = T,covar = T)
summary(threefactor)

> threefactor<-fa(data.frame(subsetMal1),nfactors =2,rotate ="oblimin",fm="ML",SMC = T,covar = T)
In smc, smcs < 0 were set to .0
In smc, smcs < 0 were set to .0
Error in optim(start, FAfn, FAgr, method = "L-BFGS-B", lower = 0.005, :
L-BFGS-B needs finite values of 'fn'
In addition: Warning message:
In log(e) : NaNs produced

So I was unable to determine the variables of each factor before defining the syntax of the CFA

Thank you for your attention.

I will be grateful to solve these issues

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