Suppose you want to build a 95% confidence interval for the ATE when the average outcome for the treated (55 subjects) is 12, with a sample standard deviation of 4, while the average outcome for the control (65 subjects) is 10, with a sample standard deviation of 6. Which of the following would be the proper confidence interval?
A) 0 ± 1.96 $\left( \sqrt { \frac { 4 ^ { 2 } } { 55 } + \frac { 6 ^ { 2 } } { 65 } } \right)$
B) 0 ± 2.58 $\left( \sqrt { \frac { 4 ^ { 2 } } { 55 } + \frac { 6 ^ { 2 } } { 65 } } \right)$
C) (12 - 10) ± 1.96 $\left( \sqrt { \frac { 4 ^ { 2 } } { 55 } + \frac { 6 ^ { 2 } } { 65 } } \right)$
D) 0 ± 1.96 $( \sqrt { ( 55 + 65 } )$

Question

Quizplus · Accepted Answer

The correct confidence interval for the ATE (Average Treatment Effect) is calculated by taking the difference between the average outcomes of the treated and control groups, which is $12 - 10 = 2$, and then adding and subtracting the margin of error. The margin of error is calculated using the formula for the standard error of the difference between two means, which involves the sample standard deviations and sizes of both groups. Choice C correctly applies the 1.96 multiplier for a 95% confidence interval and the formula for the standard error of the difference between two means.

Suppose You Want to Build a 95% Confidence Interval for the ATE