slope variance is 8.8%. Although the coverage is not grotesquely wrong, the 95% confidence interval is clearly too short. The amount of non-coverage

here implies that the standard errors for the second-level variance components are estimated about 15% too small.

I have not understood how did they calculate that the second-level variance components are estimated about 15% too small?

I thought that the calculation might be \(2\times[(8.9-5)+(8.8-5)]=15.4\%\), but do not understanding the reasoning of multiplying by 2.