Simple Algebraic Calculation about underestimation

[Cynderella]

In a findings, it is found that the non-coverage rate for the second-level intercept variance is 8.9%, and the non-coverage rate for the second-level
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.

 

[Dason]

Let's assume normal distributions for the confidence intervals. What would be the z* value you would use to make a 91.2% confidence interval? How does it compare to the z* value needed to make a 95% confidence interval?

 

[Cynderella]

Let's assume normal distributions for the confidence intervals. What would be the z* value you would use to make a 91.2% confidence interval?

the z* value used to make a 91.2% confidence interval is 1.706043. R code

qnorm(.088/2,lower.tail=FALSE)

And z* value used to make a 95% confidence interval is 1.96. That is, The standard error for 91.2% CI is estimated \(\frac{ 1.706043-1.96}{1.706043}\times 100\%=-14.88\%\), or 15% too small.

Isn't it?