Underestimated Coverage probability Let $U0$ denotes intercept variance and $U1$ denotes slope variance.
Given that the coverage rate for the intercept variance is $91$% $(U0)$ , and the coverage rate for the slope variance is $91.2$% $(U1)$ . Also nominal coverage rate is $95$%.
Then it is written that

The amount of coverage here implies that the standard errors for the  variance components are estimated about  $15$% too small.

I have not understood from the information of "coverage rate for the intercept variance $91$% $(U0)$ and  coverage rate for the slope variance $91.2$% $(U1)$", how is the calculation  "the standard errors for the  variance components are estimated about $15$% too small"  derived ?
For double check , you can consider the following :
The coverage rate for slope variance is $91.2$%. This amount implies that the standard errors for the  slope variance is estimated about $3.1$% too small when the nominal coverage rate is $95$%.
Any help is appreciated. Thanks in advance.
 A: The nominal coverave probability is $95\%$, so your interval is estimated as $[\mu-1.96\hat{\sigma};[\mu-1.96\hat{\sigma}]$ because you find a $95\%$ value (in R) as 'qnorm(p=0.975) = qnorm(p=1.95/2)'. This value is 1.96 and is derived from tha standard normal distribution (note is $z$)
In one or another way (probably by simulation) the authros find that the true coverage probability is $91\%$.  So with this you have (qnorm(p=1.91/2)=1.695) , for this same sigma an true $91\%$ interval of $[\mu-1.695\hat{\sigma}_{real};\mu+1.695\hat{\sigma}_{real}]$, so $1.695\hat{\sigma}_{real}=1.96\hat{\sigma}]$ of $\hat{\sigma}=\frac{1.695}{1.96}\hat{\sigma_{real}}$, or an underestimation of $\frac{1.695}{1.96}-1=-13.5\%$, so more or less $15\%$ underestimated. 
Answers to the questions in the comments:
First, why subtract 1 ? If you want to have the relative difference between two numbers $a$ and $b$, so how much percent (of a) is the difference between a and b, then you compute $\frac{b-a}{a}=\frac{b}{a}-1$, e.g. the how much is 95 different from 100 in pct? $\frac{95}{100}-1=0.95-1=-0.05$. 
The relation between the two sigma's ? The nomilal coverage probability of $[\mu-1.96\hat{\sigma};\mu+1.96\hat{\sigma}]$ is $95\%$, but the authors know that this same interval has a true coverage probability of $92\%$, so it is the same interval as $[\mu-1.695\hat{\sigma}_{real};\mu+1.695\hat{\sigma}_{real}]$, but if these two intervals are the same, then it must hold that $1.695\hat{\sigma}_{real}=1.96\hat{\sigma}]$
