Cynderella

New Member
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

New Member
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
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?