Why is RMSEA typically reported with a 90% confidence interval, and not 95%?

Question

Kline (2016) writes that

[RMSEA] is usually reported in computer output with the 90% confidence interval $[\hat{\epsilon}_{L},\hat{\epsilon}_{U}]$ where $\hat{\epsilon}_{L}$ is the lower-bound estimate of $\epsilon$, the parameter estimated by $\hat{\epsilon}$, and $\hat{\epsilon}_{U}$ is the upper-bound estimate. If $\hat{\epsilon} = 0$, then $\hat{\epsilon}_{L}$ and the whole interval is a one-sided confidence interval where $\hat{\epsilon}_{U} > \hat{\epsilon}$. This explains why the confidence level is 90% instead of the more typical 95%, the conventional level for two-sided confidence intervals.

I am having trouble following Kline's logic here. I understand that a RMSEA below 0 is nonsensical, but how does that argue in favor of a 90% CI?

Kline, R. B. (2016). Principles and practice of structural equation modeling. Guilford publications.

Answer 1

Curran et al. (2003) write that:

It is common to report 90 percent confidence intervals for the RMSEA, primarily because of the resulting direct link to hypothesis testing based on the model test statistic.

Three hypothesis tests sometimes reported in the SEM literature are.

The test of exact fit, $H_{0}: \epsilon = 0$
and
The test of close fit, $H_{0}: \epsilon \leq .05$
and
The test of not-close fit, $H_{0}: \epsilon \geq .05$

Thus the ultimate rationale for using a 90% CI is that if you do that you can infer the results of those hypothesis tests from the CI. The relationship is illustrated in a table in MacCullum et al. (1996) p. 137.

Curran, P. J., Bollen, K. A., Chen, F., Paxton, P., & Kirby, J. B. (2003). Finite sampling properties of the point estimates and confidence intervals of the RMSEA. Sociological Methods & Research, 32, 208-252.

MacCallum, R. C., Browne, M. W., & Sugawara, H. M. (1996). Power analysis and determination of sample size for covariance structure modeling. Psychological methods, 1, 130.