I'm using STATA 15.
My question: Would someone know how to report the 95% confidence interval for the random-intercept's standard deviation in ESTTAB?
My code is below -- please let me know if I can clarify. Thank you for your time!
Code:
. eststo: mixed absenty_p bmim || grade: Performing EM optimization: Performing gradient-based optimization: Iteration 0: log restricted-likelihood = 3883.39 Iteration 1: log restricted-likelihood = 3883.39 Computing standard errors: Mixed-effects REML regression Number of obs = 2,676 Group variable: grade Number of groups = 9 Obs per group: min = 225 avg = 297.3 max = 343 Wald chi2(1) = 0.11 Log restricted-likelihood = 3883.39 Prob > chi2 = 0.7432 ------------------------------------------------------------------------------ absenty_p | Coef. Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- bmim | .0000674 .0002057 0.33 0.743 -.0003357 .0004705 _cons | .0563113 .0051681 10.90 0.000 .046182 .0664406 ------------------------------------------------------------------------------ ------------------------------------------------------------------------------ Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval] -----------------------------+------------------------------------------------ grade: Identity | var(_cons) | .0000689 .0000403 .0000219 .0002167 -----------------------------+------------------------------------------------ var(Residual) | .0031651 .0000867 .0029996 .0033396 ------------------------------------------------------------------------------ LR test vs. linear model: chibar2(01) = 31.86 Prob >= chibar2 = 0.0000 (est5 stored) esttab using "/Users/maishahuq/Desktop/BMI-A-A/LDA_bmi_absenty_p.csv", replace se transform(ln*: exp(@) exp(@))
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