When confidence intervals of two parameters do not overlap with the mean of the other, why is it considered to be statistically significant? Consider two parameters with means A and B with 95% confidence intervals. If the confidence intervals of the parameters do not overlap with the mean of the other, why is it considered to be statistically significant? Please give detailed explanation with examples.
 A: Two statistics are significantly different if the true difference between the two is different from 0.
$s_A$ the standard deviation of $A$ and $s_B$ the standard deviation of $B$, you can define the confidence interval for $A-B$ as 
$I_{A-B} = [(A-B) - z_{0.95}\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}, (A-B) + z_{0.95}\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}]$
And each CI for $A$ and $B$
$I_{A} = [A - z_{0.95}\sqrt{\frac{s_A^2}{n_A}}, A + z_{0.95}\sqrt{\frac{s_A^2}{n_A}}]$
$I_{B} = [B - z_{0.95}\sqrt{\frac{s_B^2}{n_B}}, B + z_{0.95}\sqrt{\frac{s_B^2}{n_B}}]$
When the confidence interval of $A$ do not overlap with the mean $B$ then the condition is not verified and $I_{A-B}$ does not contain 0. (True mean difference can then not be 0).
As an example, $A = 10, s_A = 5, B = 15, s_B = 7, n_A = n_B = 10$
$I_A = [7.763932, 12.23607]$
$I_B = [11.8695, 18.1305]$
$I_A$ and $I_B$ do overlap but no CI overlap the means $A$ or $B$.
$I_{A-B} = [-8.847077, -1.152923]$
So the true value of $A-B$ is lesser than 0, $A$ and $B$ are different.
If $B = 13$ then 
$I_B = [9.869505, 16.1305]$ ; $I_A = [7.763932, 12.23607]$
$I_A$ and $I_B$ now overlap so that $A$ is in $I_B$ but $B$ is not in $I_A$.
$I_{A-B} = [-6.847077, 0.8470768]$
So true value of $A-B$ may be 0 and thus you can not conclude that $A$ and $B$ are different.
