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For a power analysis of a structural equation model, I'm trying to follow the approach outlined in in the following article:

Wang, Y. A., & Rhemtulla, M. (2021). Power Analysis for Parameter Estimation in Structural Equation 
Modeling: A Discussion and Tutorial. Advances in Methods and Practices in Psychological Science, 
4(1), 1–17. https://doi.org/10.1177/2515245920918253

In the paper the authors also give a guide for their online calculator (yilinandrewang.shinyapps.io/pwrSEM/) which is based on R's lavaan packages. In order for it to work it requires the user to input certain parameters such as the factor loadings for the indicator variables (which could be based on theory or previous validation studies).

On page 10 however, the authors mention the following:

"In our scenario, the researcher has a good idea of the likely population parameter values in the
model and inputs those values into the parameter table. The researcher sets the factor loading of 
each indicator of X and Y to .70 (corresponding to a scale reliability of .74 for X and Y), the 
factor loading of each indicator of M to .80 (corresponding to a scale reliability of .84 for M), and 
the a, b, and c paths to 0.30, 0.20, and 0.10, respectively."(p.10)

In this example, the model is a basic mediation model where each of the latent variables (X, Y, M) has three indicator variables (fig. 4).

I'm trying to understand how the factor loadings correspond to the scale reliability. E.g. how do the authors convert the set of three factor loadings of .70 to a scale reliability of .74, or vice versa? In previous post here I did not find any equation that represents this relation.

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I believe Wang and Rhemtulla (2021) are using a formula for reliability that corresponds to the $\omega$ family of coefficients (see McNeish, 2018, for a nice review of $\omega$ vs. alternatives like $\alpha$). $\omega_T$ is a ratio of (summed and squared) standardized loadings ($\lambda$) for each manifest variable over the (summed and squared) standardized loadings + the uniqueness (i.e., standardized error variance ($\sigma$), given as 1 - the manifest variable's communality ($\lambda^2$, or $h^2$)).

$(\sum_{i=1}^{k}\lambda _i)^2/(\sum_{I=1}^{k}\lambda _i)^2+\sum_{i=1}^{k}\sigma^{2}_{ii}$

So from their path diagram in Fig 7:

Reliability for X or Y (3 variables w/ standardized loadings of .70):

$(\sum_{i=1}^{k}\lambda _i)^2 = (.70 + .70 + .70)^2 = 4.41$

$h^2 = .7^2, .7^2, .7^2$ so $.49, .49, .49$

and so $\sigma = .51, .51, .51$

$\sum_{i=1}^{k}\sigma^{2}_{ii} = 1.53$

Finally: $4.41/(4.41 = 1.53) = .74 $

Again for reliability for M (3 variables w/ standardized loadings of .80):

$(\sum_{i=1}^{k}\lambda _i)^2 = (.80 + .80 + .80)^2 = 5.76$

$h^2 = .8^2, .8^2, .8^2$ so $.64, .64, .64$

and so $\sigma = .36, .36, .36$

$\sum_{i=1}^{k}\sigma^{2}_{ii} = 1.08$

Finally: $5.76/(5.76 = 1.08) = .84 $

It's worth clarifying that the title of your question specifically asks about $\alpha$, and this is not the reliability metric that Wang and Rhemtulla (2021) use. Indeed, although one can calculate $\alpha$ in a latent variable framework, one must jump through some (questionable, IMO) hoops to get there, including the specification of a model in which there is no latent variable underlying a set of indicators (see Geldof et al., 2014, for a review).

References

Geldhof, G. J., Preacher, K. J., & Zyphur, M. J. (2014). Reliability estimation in a multilevel confirmatory factor analysis framework. Psychological Methods, 19(1), 72-91. doi: 10.1037/a0032138

McNeish, D. (2018). Thanks coefficient alpha, we’ll take it from here. Psychological Methods, 23(3), 412-433. doi: 10.1037/met0000144

Wang, Y. A., & Rhemtulla, M. (2021). Power analysis for parameter estimation in structural equation modeling: A discussion and tutorial. Advances in Methods and Practices in Psychological Science, 4(1), 1-17. doi: 10.1177/2515245920918253

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  • $\begingroup$ This is a helpful answer, but noting that you've got an equals where I think you meant to put a plus in 4.41/(4.41=1.53)=.74. Also I think you need an extra set of parentheses in the denomonator in your first formula $\endgroup$
    – Ben Ogorek
    Commented May 26, 2023 at 19:24
  • $\begingroup$ I'm also looking at Cho's paper Cronbach’s Coefficient Alpha: Well Known but Poorly Understood, and on p 33 he's got the unidimension version of omega with that includes the latent factor variance. If it's set to 1 during estimation, then the formula reduces to yours (with the parentheses added). I'm thinking at if you set the first loadign to 1 and let the factor variance float, you'd have to use (sum of loadings) ^2 * latent factor variance in the numerator and denominator $\endgroup$
    – Ben Ogorek
    Commented May 26, 2023 at 19:43

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