Common Issues in Structural Equation Modelling (SEM)
This post presents some very common issues we face when doing Structural Equation Modelling (SEM). Most of the problems and their remedies are either taken from a book or academic social interactions, e.g. Research Gate. Courtesy to all sources are mentioned along with the respective post/suggestion.
Problem 1: What are the goodness-of-fit criteria in structural equation model (SEM)?
Source: Hair JF, Black WC, Babin BJ, Anderson RE, Tatham RL (2006). Multivariate data analysis. Pearson Prentice Hall Upper Saddle River
Problem 2: What is the acceptable range for factor loading in SEM?
In CFA models there are some displays concerning the fitness level of your model. If these displays are in the suitable ranges which are widely known in the literature, do not worry about the factor loadings. But I strongly recommend you to conduct a EFA first to assess your variables then go on with CFA. In EFA it is widely accepted that items with factor loadings less than 0.5, and items having high factor loadings more than one factor are discarded from the model. You can filter your model via EFA.
Courtesy: Ufuk Turen
Problem 3: Any Citation for Cronbach alpha more than 0.6 is acceptable?
-Define reliability using Cronbach’s Alpha > 0.7 , (Hair, 2005).
-According to Ramayah (2011), Cronbach’s alpha coefficient values of more than 0.7 are considered good but values of more than 0.5 are acceptable.
-According to Sekaran and Bougie (2013), reliabilities less than 0.6 are considered to be poor, those in the range of 0.7 – 0.79 are said to be acceptable, and those above 0.8 are said to be good.
Courtesy: Qais Almaamari
Problem 4: What can we do when Cronbach’s alpha is below 0.7?
“Nunnally (1978) recommended calculation of coefficient alpha (also known as Cronbach alpha) in order to assess the reliability of a multiple-item variable. Churchill and Peter (1984) suggested an accepted level for the alpha coefficient. According to them a value of alpha below 0.60 is undesirable. Nunnally (1978; 1988) indicated that new developed measures can be accepted with an alpha value of 0.60, otherwise, 0.70 should be the threshold. However, considering the use of these scales for the first time in a new culture, the cut off value for the alpha coefficient was set up for 0.60 for all the scales (self-developed scales).”
Courtesy: Nishad Nawaz
Another comment on problem 4:
Thus if you have an alpha of 1 then scale is useless, the items are identical and not capturing the breadth of the construct, so too perfect an alpha is not acceptable.
If you have an alpha of 0.5 (you think it is a unidimensional scale) then the items are not hanging together very well and you probably do not have a unidimensional scale (back to factor analysis)
Practical issues about alpha.
Alpha is affected by the intercorrelations of items and importantly the number of items. low correlations reduce alpha
An Alpha calculated on Pearson correlations for dichotomous items will be an underestimate, similarly for item responses with less than 7 points
An alpha calculated on 20 items will be an overestimate etc.
Thus with 5 to 10 items with 7 or higher point scales, an alpha below 0.7 suggests you have a poor scale, check which items, one at a time, improve the alpha by removal. an alpha of 0.95 says you have too many items and you may want to remove the items that do not affect the size of alpha.
Courtesy: Robert T Brooks
Problem 5: What is the acceptable value of Average Variance Extracted (AVE)?
In this article: http://www.ijiet.org/papers/267-IT0040.pdf found the following:
Average Variance Extracted (AVE) is higher than 0.5 but we can accept 0.4. Because Fornell and Larcker said that if AVE is less than 0.5, but composite reliability is higher than 0.6, the convergent validity of the construct is still adequate (Fornell & Larcker, 1981)
- Fornell and D. F. Larcker, “Evaluating structural equation models with unobservable variables and measurement error,” Journal of marketing research, pp. 39-50, 1981.
Courtesy: Han Ping Fung
Another comment on problem 5:
I tend to agree with Marcie. AVE < .5 does not convey sufficient variance for the variables (items/questions) to converge into a single construct, which means items are less-than-effective measure of the latent construct. There’s more error variance than explained variance.
The logic is that:
when an item loading is .71, then it’s communality is .50 because communality is the loading square.
Thus, AVE < .50 means, on average item loading is less than .70 (Hair et al., 2010, p.777)
This also explains while your loadings are good (.60), AVE is less than perfect (< .50.).
Check variance inflation factor (VIF) to see if multicollinearity is an issue in your dataset.
if VIF > 5, then check your construct correlation. If any constructs’ correlation is more than .80 or .85 then you can combine these constructs as one.
Note that, this is subject to your underpinning theory.
Hair, J. F., Black, W. C., Babin, B. J., & Anderson, R. E. (2010). Multivariate data analysis (7 ed.). Upper Saddle River, NJ, USA: Prentice-Hall, Inc.
Courtesy: Saiyidi Mat Roni
Problem 6: How to include control variables in a structural equation model (SEM)?
This post is compiled by Ziaul H. Munim, the founder of the ResearchHUB.
Ziaul Haque Munim is an Associate Professor at the School of Business and Law of University of Agder, Norway. His main research interests include maritime logistics, transport economics, supply chain management and international business. Dynamics of container shipping freight rates is one of his special interests. He received the Best Paper Award at the IAME 2016 Conference in Hamburg, and was nominated for the Best Young Researcher Award at the IAME 2017 Conference in Kyoto. Follow his research on ResearchGate and Google Scholar.