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Unformatted text preview: Statistics 403 Problem Set 7 Due in lab on Friday, November 5th 1. Suppose we are planning to studying a treatment for depression. We will obtain base line depression scores X i , i = 1 ,..., 30 on a set of research subjects. Three months later, after the treatment is complete, we will obtain a new set of depression scores Y i , i = 1 ,..., 30 on the same subjects. We anticipate that the treatment will reduce average depression scores by 1 . 6 raw units. For power analysis, we assume that the pretreatment measures will have a standard deviation of 3 units, the posttreatment measures will have a standard deviation of 2 units, and the pretreatment and post treatment measures will be correlated 0 . 4 with each other. (i) What is the power for detecting the anticipated effect of 1 . 6 units when using a paired ttest; (ii) What would the power be if we used an unpaired ttest? Solution: The standard deviation of the D i = X i Y i values is q var( X ) + var( Y ) 2cov( X,Y ) = √ 3 2 + 2 2 2 · . 4 · 3 · 2 = 2 . 86 . Thus the expected test statistic under the paired analysis is ET = √ 30 · 1 . 6 / 2 . 86 = 3 . 1 . This gives us a power of P ( T > 2) = P ( Z > 2 ET ) = P ( Z > 1 . 1) = 0 . 86 . Under the unpaired analysis, the expected value of the test statistic is ET = E ¯ X ¯ Y q σ 2 X / 30 + σ 2 Y / 30 = 1 . 6 q 3 2 / 30 + 2 2 / 30 = 2 . 43 , and the variance of the test statistic is var( T ) = var ¯ X ¯ Y q σ 2 X /n + σ 2 Y /n = n σ 2 X + σ 2 Y · var( ¯ D ) 1 = n σ 2 X + σ 2 Y · σ 2 X + σ 2 Y 2cov( X,Y ) n = σ 2 X + σ 2 Y 2cov( X,Y ) σ 2 X + σ 2 Y = 13 2 · . 4 · 3 · 2 13 ≈ . 63 ....
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 Winter '12
 KerbyShedden
 Statistics, Normal Distribution, Variance, Conditional expectation, coverage probability

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