The 95 Percent Confidence Interval for the Difference Between Two $\left( \bar { X } _ { 1 } - \bar { X } _ { 2 } \right) - 2 \left( s \bar { x } _ { 1 } - \bar { x } _ { 2 } \right)$

Multiple Choice

The 95 percent confidence interval for the difference between two independent population means is given by.

A) $\left( \bar { X } _ { 1 } - \bar { X } _ { 2 } \right) - 2 \left( s \bar { x } _ { 1 } - \bar { x } _ { 2 } \right)$ to $\left( \bar { X } _ { 1 } - \bar { X } _ { 2 } \right) + 2 \left( s _ { \bar { X } _ { 1 } - \bar { X } _ { 2 } } \right)$
B) $\left( \bar { X } _ { 1 } - \bar { X } _ { 2 } \right) - t .05 \left( s \bar { x } _ { 2 } = \bar { x } _ { 2 } \right)$ to $\left( \bar { X } _ { 1 } - \bar { X } _ { 2 } \right) + t _ { 05 } \left( s _ { \bar { x } _ { 1 } - \bar { x } _ { i } } \right)$
C) $\left( \bar { X } _ { 1 } - \bar { X } _ { 2 } \right) - t _ { .05 } \left( s _ { X ^ { \prime } } \right)$ to $\left( \bar { X } _ { 1 } - \bar { X } _ { 2 } \right) + t .05 \left( s _ { \bar { X } } \right)$
d $( \bar { X } - \bar { X } ) - t \cos \left( e _ { \bar { X } } \right)$ to $\left( \bar { X } _ { 1 } - \bar { X } _ { 2 } \right) + t _ { 95 } ( s \bar { X } )$