# Two Direct Effects

## Explanation

In a conceptual model, the concepts are normally placed in a rectangular. We have three concepts, $X_1$ , $X_2$ and Y.
The single headed arrow indicates that you assume causal (linear) relations from $X_1$ and $X_2$ (on the left) to Y on the right. Thus: if $X_1$ or $X_2$ increases, Y will increase as a result as well. In this example we have two direct effects. The implicit assumption of a conceptual model is that (both) concepts have an additive effect. Naturally, there would in principle be nothing wrong with a multiplicative effect/model.
It is not always necessary to label the paths but for this tutorial it will turn out to be handy. Normally, when there is no sign (or label) it is assumed that the path has a positive valence. It is, however, good practice to include the valence of the paths in your conceptual models, i.e. replace a with + or -.

Please note, I use the label concept and not variable. A variable is something that is part of your dataset, measured by, for example, a survey item. A concept is a theoretical construct, and the concept as intended may have relations - according to your theory - with other concepts. This concept is measured by one or more variables, the ‘concept as measured’.
It is generally a good idea to use concepts that are (very) close to your actual measurements. Thus, although you may use the concept social cohesion in your conceptual model, this concept is overly broad and there is a fierce debate on how it should be defined. If you have measured social cohesion with for example a single item on generalized trust (“Generally speaking would you say that most people can be trusted or that you can’t be too careful in dealing with people?”) why not use the concept generalized trust in your conceptual model?

## Abstract hypothesis/hypotheses

Hypo1: $X_1$ leads to Y.
Hypo2: $X_2$ leads to Y.

Normally we do not have an hypothesis on the association between $X_1$ and $X_2$. If we do, we could summarize our ideas with the following conceptual model:

Hypo3: $X_1$ and $X_2$ are positively associated.

## Real life example

$X_1$ is occupational success.
$X_2$ is being in a relationship (YES/NO) Y is health

Hypo1: Occupational success will lead to better health. (Money makes happy.) Hypo2: Being in a relationship will lead to better health. (Love makes happy.)

## Structural equations

• $Y = X_1$
• $Y = X_2$
• $X_1 = X_2$

or, following the syntax of the R package Lavaan

• Y~$$X_1$$
• Y~$$X_2$$
• $$X_1$$~~$$X_2$$

The single ~ indicates a direct effect (regression path).
The double ~~ indicates a (partial), (error)variance.

## Formal test of hypotheses

rm(list = ls())  #empty environment
require(haven)
nells <- read_dta("../static/NELLS panel nl v1_2.dta")  #change directory name to your working directory

Operationalize concepts.

# We will use the data of wave 2.
nellsw2 <- nells[nells$w2cpanel == 1, ] # As an indicator of occupational success we will use income in wave 2. table(nellsw2$w2fa61, useNA = "always")
attributes(nellsw2$w2fa61) # recode (I will start newly created variables with cm from conceptual models) nellsw2$cm_income <- nellsw2$w2fa61 nellsw2$cm_income[nellsw2$cm_income == 1] <- 100 nellsw2$cm_income[nellsw2$cm_income == 2] <- 225 nellsw2$cm_income[nellsw2$cm_income == 3] <- 400 nellsw2$cm_income[nellsw2$cm_income == 4] <- 750 nellsw2$cm_income[nellsw2$cm_income == 5] <- 1250 nellsw2$cm_income[nellsw2$cm_income == 6] <- 1750 nellsw2$cm_income[nellsw2$cm_income == 7] <- 2250 nellsw2$cm_income[nellsw2$cm_income == 8] <- 2750 nellsw2$cm_income[nellsw2$cm_income == 9] <- 3250 nellsw2$cm_income[nellsw2$cm_income == 10] <- 3750 nellsw2$cm_income[nellsw2$cm_income == 11] <- 4250 nellsw2$cm_income[nellsw2$cm_income == 12] <- 4750 nellsw2$cm_income[nellsw2$cm_income == 13] <- 5250 nellsw2$cm_income[nellsw2$cm_income == 14] <- 5750 nellsw2$cm_income[nellsw2$cm_income == 15] <- 6500 nellsw2$cm_income[nellsw2$cm_income == 16] <- 7500 nellsw2$cm_income[nellsw2$cm_income == 17] <- NA # let us scale the variable a bit and translate into income per 1000euro nellsw2$cm_income <- nellsw2$cm_income/1000 # from household income to personal income attributes(nellsw2$w2fa62)
table(nellsw2$w2fa62, useNA = "always") nellsw2$cm_income_per <- nellsw2$w2fa62 nellsw2$cm_income_per[nellsw2$cm_income_per == 1] <- 0 nellsw2$cm_income_per[nellsw2$cm_income_per == 2] <- 10 nellsw2$cm_income_per[nellsw2$cm_income_per == 3] <- 20 nellsw2$cm_income_per[nellsw2$cm_income_per == 4] <- 30 nellsw2$cm_income_per[nellsw2$cm_income_per == 5] <- 40 nellsw2$cm_income_per[nellsw2$cm_income_per == 6] <- 50 nellsw2$cm_income_per[nellsw2$cm_income_per == 7] <- 60 nellsw2$cm_income_per[nellsw2$cm_income_per == 8] <- 70 nellsw2$cm_income_per[nellsw2$cm_income_per == 9] <- 80 nellsw2$cm_income_per[nellsw2$cm_income_per == 10] <- 90 nellsw2$cm_income_per[nellsw2$cm_income_per == 11] <- 100 nellsw2$cm_income_per[nellsw2$cm_income_per == 12] <- NA nellsw2$cm_income_ind <- nellsw2$cm_income * nellsw2$cm_income_per/100

# habing a partner
attributes(nellsw2$w2fb1) table(nellsw2$w2fb1, useNA = "always")
nellsw2$cm_partner <- as.factor(ifelse(nellsw2$w2fb1 == 1, 0, 1))

# as an indicator of health we will use subjective well being from 5 (excellent) to 1 (bad) thus we
# have to reverse code original variable

attributes(nellsw2$w2scf1) table(nellsw2$w2scf1, useNA = "always")
nellsw2$cm_health <- 6 - nellsw2$w2scf1
##
##    1    2    3    4    5    6    7    8    9   10   11   12   13   14   15   16   17 <NA>
##   55   78  103  204  338  326  282  272  276  205  133   62   48   22   22   29  374    0
## $label ## [1] " wat is het netto inkomen per maand van u en uw partner samen?/van u?/ " ## ##$format.stata
## [1] "%8.0g"
##
## $labels ## Minder dan ¤150 per maand ¤150 - ¤299 per maand ¤300 - ¤499 per maand ## 1 2 3 ## ¤500 - ¤999 per maand ¤1.000 - ¤1.499 per maand ¤1.500 - ¤1.999 per maand ## 4 5 6 ## ¤2.000 - ¤2.499 per maand ¤2.500 - ¤2.999 per maand ¤3.000 - ¤3.499 per maand ## 7 8 9 ## ¤3.500 - ¤3.999 per maand ¤4.000 - ¤4.499 per maand ¤4.500 - ¤4.999 per maand ## 10 11 12 ## ¤5.000 - ¤5.499 per maand ¤5.500 - ¤5.999 per maand ¤6.000 - ¤6.999 per maand ## 13 14 15 ## ¤7.000 of meer per maand weet niet, wil niet zeggen ## 16 17 ## ##$class
## [1] "haven_labelled" "vctrs_vctr"     "double"
##
## $label ## [1] " hoe groot is uw bijdrage in dit inkomen ongeveer? kunt u een percentage noemen " ## ##$format.stata
## [1] "%8.0g"
##
## $labels ## vrijwel geen bijdrage ongeveer 10% ongeveer 20% ongeveer 30% ## 1 2 3 4 ## ongeveer 40% ongeveer 50% ongeveer 60% ongeveer 70% ## 5 6 7 8 ## ongeveer 80% ongeveer 90% ongeveer 100% weet niet ## 9 10 11 12 ## ##$class
## [1] "haven_labelled" "vctrs_vctr"     "double"
##
##
##    1    2    3    4    5    6    7    8    9   10   11   12 <NA>
##  253   48   89  259  233  242  183  229  114   63  887  229    0
## $label ## [1] " heeft u op dit moment een partner? " ## ##$format.stata
## [1] "%8.0g"
##
## $labels ## ik woon alleen en heb geen partner ik woon alleen en heb een partner ## 1 2 ## ik woon ongehuwd samen ik woon gehuwd samen ## 3 4 ## ##$class
## [1] "haven_labelled" "vctrs_vctr"     "double"
##
##
##    1    2    3    4 <NA>
##  665  264  475 1425    0
## $label ## [1] " wat vindt u, over het algemeen genomen, van uw gezondheid? " ## ##$format.stata
## [1] "%8.0g"
##
## $labels ## uitstekend zeer goed goed matig slecht ## 1 2 3 4 5 ## ##$class
## [1] "haven_labelled" "vctrs_vctr"     "double"
##
##
##    1    2    3    4    5 <NA>
##  438  853 1211  247   48   32

And test the direct effect. Naturally, there are many ways to test for a direct effects in R but in this tutorial I will try to do everything at least also in the package Lavaan.

But first plot the associations:

# I randomly select 200 respondents otherwise the plot will be too crowded
selection <- sample(1:length(nellsw2$cm_income_ind), 200, replace = FALSE) # because we are interested in a correlation, I plot the standardized variables par(mfrow = c(1, 2)) plot(nellsw2$cm_income_ind[selection], nellsw2$cm_health[selection], xlab = "income", ylab = "health", main = "Bivariate effect of income on health") abline(lm(nellsw2$cm_health ~ nellsw2$cm_income_ind), lwd = 4, col = "red") boxplot(cm_health ~ cm_partner, data = nellsw2, notch = TRUE, xlab = "having a partner", ylab = "health", main = "Bivariate effect partnership on health") I hope you observe that the regression line does not fit the data very well and that the means/median of health does not differ much whether or not you have a partner. And now,…estimate the direct effects via lm(): summary(lm(nellsw2$cm_health ~ nellsw2$cm_income_ind)) summary(lm(nellsw2$cm_health ~ nellsw2$cm_partner)) summary(lm(nellsw2$cm_health ~ nellsw2$cm_income_ind + nellsw2$cm_partner))
##
## Call:
## lm(formula = nellsw2$cm_health ~ nellsw2$cm_income_ind)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -2.5178 -0.5178 -0.3913  0.5382  1.6087
##
## Coefficients:
##                       Estimate Std. Error t value Pr(>|t|)
## (Intercept)            3.39132    0.03276 103.516  < 2e-16 ***
## nellsw2$cm_income_ind 0.07230 0.01860 3.886 0.000105 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 0.9132 on 2353 degrees of freedom ## (474 observations deleted due to missingness) ## Multiple R-squared: 0.006377, Adjusted R-squared: 0.005955 ## F-statistic: 15.1 on 1 and 2353 DF, p-value: 0.0001047 ## ## ## Call: ## lm(formula = nellsw2$cm_health ~ nellsw2$cm_partner) ## ## Residuals: ## Min 1Q Median 3Q Max ## -2.4985 -0.4946 -0.4946 0.5054 1.5054 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 3.498485 0.035747 97.869 <2e-16 *** ## nellsw2$cm_partner1 -0.003866   0.040896  -0.095    0.925
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.9183 on 2795 degrees of freedom
##   (32 observations deleted due to missingness)
## Multiple R-squared:  3.198e-06,  Adjusted R-squared:  -0.0003546
## F-statistic: 0.008938 on 1 and 2795 DF,  p-value: 0.9247
##
##
## Call:
## lm(formula = nellsw2$cm_health ~ nellsw2$cm_income_ind + nellsw2$cm_partner) ## ## Residuals: ## Min 1Q Median 3Q Max ## -2.5380 -0.5221 -0.3993 0.5364 1.6174 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 3.40777 0.04439 76.774 < 2e-16 *** ## nellsw2$cm_income_ind  0.07441    0.01900   3.916 9.25e-05 ***
## nellsw2$cm_partner1 -0.02521 0.04590 -0.549 0.583 ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 0.9133 on 2352 degrees of freedom ## (474 observations deleted due to missingness) ## Multiple R-squared: 0.006504, Adjusted R-squared: 0.00566 ## F-statistic: 7.699 on 2 and 2352 DF, p-value: 0.0004647 And with Lavaan. require(lavaan) model <- ' cm_health ~ cm_income_ind cm_health ~ cm_partner cm_income_ind ~~ cm_partner ' nellsw2$cm_partner <- as.numeric(nellsw2$cm_partner) fit <- sem(model, data = nellsw2) #I use sem instead of lavaan. The only advantage is that I don't have to tell lavaan that I also need the error variances. summary(fit, standardized = TRUE) fitMeasures(fit, fit.measures = "logl") inspect(fit, "r2") #to obtain r-squared # parameterEstimates(fit) ## lavaan 0.6-7 ended normally after 18 iterations ## ## Estimator ML ## Optimization method NLMINB ## Number of free parameters 6 ## ## Used Total ## Number of observations 2355 2829 ## ## Model Test User Model: ## ## Test statistic 0.000 ## Degrees of freedom 0 ## ## Parameter Estimates: ## ## Standard errors Standard ## Information Expected ## Information saturated (h1) model Structured ## ## Regressions: ## Estimate Std.Err z-value P(>|z|) Std.lv Std.all ## cm_health ~ ## cm_income_ind 0.074 0.019 3.919 0.000 0.074 0.082 ## cm_partner -0.025 0.046 -0.550 0.583 -0.025 -0.012 ## ## Covariances: ## Estimate Std.Err z-value P(>|z|) Std.lv Std.all ## cm_income_ind ~~ ## cm_partner 0.086 0.009 9.631 0.000 0.086 0.202 ## ## Variances: ## Estimate Std.Err z-value P(>|z|) Std.lv Std.all ## .cm_health 0.833 0.024 34.315 0.000 0.833 0.993 ## cm_income_ind 1.023 0.030 34.315 0.000 1.023 1.000 ## cm_partner 0.175 0.005 34.315 0.000 0.175 1.000 ## ## logl ## -7737.193 ## cm_health ## 0.007 Let us briefly discuss the results: • The direct effect for income on health does not change once we include our partner variable. • You will also observe that having a partner does not contribute to our SWB. • Money makes you happy but a partner does not. • According to the results, having a partner is positively related to income. Given the above, you may want to go back to your theoretical drawing board. And perhaps having done so, you now would like to estimate the following model: require(lavaan) model <- " cm_income_ind ~ cm_health cm_income_ind ~ cm_partner cm_income_ind ~ 1 cm_health ~~ cm_partner " nellsw2$cm_partner <- as.numeric(nellsw2\$cm_partner)

fit <- sem(model, data = nellsw2)  #I use sem instead of lavaan. The only advantage is that I don't have to tell lavaan that I also need the error variances.
summary(fit, standardized = TRUE)
fitMeasures(fit, fit.measures = "logl")
inspect(fit, "r2")  #to obtain r-squared
# parameterEstimates(fit)
## lavaan 0.6-7 ended normally after 22 iterations
##
##   Estimator                                         ML
##   Optimization method                           NLMINB
##   Number of free parameters                          9
##
##                                                   Used       Total
##   Number of observations                          2355        2829
##
## Model Test User Model:
##
##   Test statistic                                 0.000
##   Degrees of freedom                                 0
##
## Parameter Estimates:
##
##   Standard errors                             Standard
##   Information                                 Expected
##   Information saturated (h1) model          Structured
##
## Regressions:
##                    Estimate  Std.Err  z-value  P(>|z|)   Std.lv  Std.all
##   cm_income_ind ~
##     cm_health         0.087    0.022    3.919    0.000    0.087    0.079
##     cm_partner        0.488    0.049   10.047    0.000    0.488    0.202
##
## Covariances:
##                    Estimate  Std.Err  z-value  P(>|z|)   Std.lv  Std.all
##   cm_health ~~
##     cm_partner        0.002    0.008    0.248    0.804    0.002    0.005
##
## Intercepts:
##                    Estimate  Std.Err  z-value  P(>|z|)   Std.lv  Std.all
##    .cm_income_ind     0.272    0.117    2.312    0.021    0.272    0.269
##     cm_health         3.496    0.019  185.248    0.000    3.496    3.817
##     cm_partner        1.773    0.009  205.509    0.000    1.773    4.235
##
## Variances:
##                    Estimate  Std.Err  z-value  P(>|z|)   Std.lv  Std.all
##    .cm_income_ind     0.975    0.028   34.315    0.000    0.975    0.953
##     cm_health         0.839    0.024   34.315    0.000    0.839    1.000
##     cm_partner        0.175    0.005   34.315    0.000    0.175    1.000
##
##      logl
## -7737.193
## cm_income_ind
##         0.047

Take Home Messages

• It is easy to estimate equivalent models with similar fits (see the loglikelihood of the models above) but with a different interpretation.