问题如下图：为什么置信区间越大，误差越大
选项：
A. 
B. 
C. 
D. 
解释：
NO.PZ2016062402000036问题如下 The measurement error in VAR, e to sampling variation, shoulgreater with More observations ana high confinlevel (e.g. , 99%) Fewer observations ana high confinlevel More observations ana low confinlevel (e.g. , 95%) Fewer observations ana low confinlevel Sampling variability (or imprecision) increases with (1) fewer observations an(2) greater confinlevels. To show (1), we crefer to the formula for the precision of the sample mean, whivaries inversely with the square root of the number of ta points. A similreasoning applies to (2). A greater confinlevel involves fewer observations in the left tails, from whiVis compute 是不是有个公式。，，
NO.PZ2016062402000036 The measurement error in VAR, e to sampling variation, shoulgreater with More observations ana high confinlevel (e.g. , 99%) Fewer observations ana high confinlevel More observations ana low confinlevel (e.g. , 95%) Fewer observations ana low confinlevel Sampling variability (or imprecision) increases with (1) fewer observations an(2) greater confinlevels. To show (1), we crefer to the formula for the precision of the sample mean, whivaries inversely with the square root of the number of ta points. A similreasoning applies to (2). A greater confinlevel involves fewer observations in the left tails, from whiVis compute
NO.PZ2016062402000036 The measurement error in VAR, e to sampling variation, shoulgreater with More observations ana high confinlevel (e.g. , 99%) Fewer observations ana high confinlevel More observations ana low confinlevel (e.g. , 95%) Fewer observations ana low confinlevel Sampling variability (or imprecision) increases with (1) fewer observations an(2) greater confinlevels. To show (1), we crefer to the formula for the precision of the sample mean, whivaries inversely with the square root of the number of ta points. A similreasoning applies to (2). A greater confinlevel involves fewer observations in the left tails, from whiVis compute 我看之前有一个老师的解答，说的是置信度越高，小于VAR的值就越少。 这句话是理解成这样么？因为VAR是损失的概念，所以在作为，当置信度越高的时候，VAR就越在左边，所以VAR左边的数值就越少，所以，VAR就越有可能是被正常估计的，所以measurement error就越小。 这样理解对么？
没看明白题目意思，能否一下
