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2_sample_power

# 2_sample_power - The power of the two-sample mean...

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The power of the two-sample mean comparison is maximized when we maximize ET , where ET = μ x - μ y q σ 2 x /n x + σ 2 y /n y = q n x + n y μ x - μ y q σ 2 x /p x + σ 2 y /p y = q n x + n y μ x - μ y q σ 2 x /p x + σ 2 y / (1 - p x ) Suppose the raw effect size μ x - μ y , the total sample size n x + n y , and the variances σ 2 x and σ 2 y are all fixed, and we want to maximize the power by setting the value of p x to its ideal point. This becomes equivalent to minimizing σ 2 x /p x + σ 2 y / (1 - p x ) . Note that the minimum of this expression cannot occur on the boundary ( p x = 0 or p x = 1), since it becomes equal to + in either case. If we differentiate with respect to p x , we get - σ 2 x /p 2 x + σ 2 y / (1 - p x ) 2 , so the minimum occurs when - σ 2 x /p 2 x + σ 2 y / (1 - p x ) 2 = 0 . Thus we have - σ 2 x (1 - p x ) 2 + σ 2 y p 2 x = 0 , which simplifies to ( σ 2 y - σ 2 x ) p 2 x + 2 σ 2 x p x - σ 2 x = 0 . This is quadratic in p x , so the minimum occurs when 1

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p x = - 2 σ 2 x ± q 4 σ 4 x + 4 σ 2 x ( σ 2 y - σ 2 x ) 2( σ 2 y - σ 2 x ) = - σ 2 x ± q σ 4 x + σ 2 x ( σ 2 y - σ 2 x ) σ 2 y - σ 2 x = - σ 2 x ± σ x σ y σ 2 y - σ 2 x = - σ 2 x ± σ x σ y ( σ y - σ x )( σ y + σ x ) The two solutions are - σ 2 x + σ x σ y ( σ y - σ x )( σ y + σ x ) = σ x σ y + σ x and - σ 2 x - σ x σ y ( σ y - σ x )( σ y + σ
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2_sample_power - The power of the two-sample mean...

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