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I have three means and standard deviations that represent subscales of a factor. Each subscale has a different number of items. I am summing the three subscale means together to form a factor, and am wondering if there is any way to obtain a standard deviation for the resulting factor score. Below is my example, with means listed in the first column and SDs in the second:

Sub1 (18 items) 49.80 8.16

Sub2 (13 items) 42.10 5.50

Sub3 (14 items) 36.27 7.91

Fact (45 items) 128.17 XXX

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  • $\begingroup$ Do you want to add the means or average them in some way? You have done the former, but I have my doubts that you want to. $\endgroup$
    – Dave
    May 13 at 19:28
  • $\begingroup$ Thank you so much for engaging me. I have been working on this for 6 hours. So the three subscales of this particular assessment instrument are traditionally added together to form a "factor" score that represents a latent variable of interest. $\endgroup$
    – John Engle
    May 13 at 19:29
  • $\begingroup$ So I do want to add them. $\endgroup$
    – John Engle
    May 13 at 19:30
  • $\begingroup$ 0. Welcome to CV.SE. 1. If we assume that each subscale is Gaussian and independent of the others, we can quickly derive the results with simple Algebra. Please see my answer below for more details. $\endgroup$
    – usεr11852
    May 16 at 1:52
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If the "factor" score of interest ($Z$) can be view as adding together the three subscales of this particular interest and each subscale is assumed to be normally distributed and independent of the others, then the we simply see $Z$ as being the linear combination of three independent Gaussian variables $A$, $B$ and $C$.

The general case is such that if our variable of interest $Z$ is the mixture of $K$ independent Gaussian variables, ie: $Z = \sum_k^{K} w_k N(\mu_{k}, \sigma_k^2)$, then $Z$ is itself normal-distributed as $Z\sim N(\sum_k^K w_k \mu_{k}, \sum_k^K (w_k \sigma_k)^2)$. The Wikipedia article on the sum of normally distributed random variables covers nicely if you want to read more details.

Now, using the above for the case of three Gaussian random variables $A$, $B$ and $C$, distributed as $N(\mu_A,\sigma_A^2)$, $N(\mu_B,\sigma_B^2)$ and $N(\mu_C,\sigma_C^2)$ respectively and assuming that $w_A = w_B = w_C = 1$, then we can immediately substitute $\mu_{A} = 49.80$, $\sigma_A = 8.16$, $\mu_{B} = 42.1$, $\sigma_b = 5.5$, $\mu_{C} = 36.27$ and $\sigma_C = 7.91$, to get:

$\mu_Z = w_A \mu_A + w_B \mu_B + w_C \mu_C = 128.17$ (as you have derived yourself) and $\sigma_Z^2 = w_A^2 \sigma_A^2 + w_B^2 \sigma_B^2 + w_C^2 \sigma_C^2 = 159.4037$ (or $\sigma_Z = 12.62552$)

Here is a quick R snippet showing this in action:

N = 10^6
mu = c(49.8, 42.1, 36.27)
sigma = c(8.16, 5.5, 7.91) 
set.seed(710)
A =  rnorm(N, mu[1], sd = sigma[1])
B =  rnorm(N, mu[2], sd = sigma[2])
C =  rnorm(N, mu[3], sd = sigma[3])
Z = A+B+C
round(c(mean(Z), sd(Z)),3)
# 128.170  12.625 as expected
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  • $\begingroup$ Thank you so much for helping! $\endgroup$
    – John Engle
    May 19 at 19:46
  • $\begingroup$ I am glad I could help. If this answer is helpful please consider upvoting it and if it resolves your question marking it as the accepted answer. $\endgroup$
    – usεr11852
    May 19 at 20:08

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