# ps10 - Statistics 403 Problem Set 10 Due in lab on Friday...

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Unformatted text preview: Statistics 403 Problem Set 10 Due in lab on Friday, December 3rd 1. Suppose we are using stratification to estimate a treatment effect (i.e. we want to estimate EY T- EY U while controlling for the effect of a known confounder X ). We use five strata, and for simplicity assume that the standard deviation of Y T is 1 in every stratum, and the standard deviation of Y U is 2 in every stratum. The proportions of treated and untreated subjects in the five strata are T 0.1 0.3 0.2 0.2 0.2 U 0.3 0.2 0.3 0.1 0.1 You can also assume that the numbers of treated and untreated subjects in the sample are the same. What sample size is needed so that a confidence interval for the treatment effect is 0.5 units wide? Solution: If there are n treated subjects and n untreated subjects, the sample sizes are T . 1 n . 3 n . 2 n . 2 n . 2 n U . 3 n . 2 n . 3 n . 1 n . 1 n Within each stratum, we will be estimating the mean difference between treated and untreated subjects using a statistic of the form D i = ¯ Y ( i ) T- ¯ Y ( i ) U where ¯ Y ( i ) T and ¯ Y ( i ) U are the sample means for treated and for untreated subjects in the i th stratum. The variance of this statistic is V i = σ 2( i ) T /n ( i ) T + σ 2( i ) U /n ( i ) U , where σ 2( i ) T and σ 2( i ) U are the corresponding variances, and n ( i ) T and n ( i ) U are the corresponding sample sizes. We get the following V i values V 1 = 1 / (0 . 1 n ) + 4 / (0 . 3 n ) = 23 . 3 /n V 2 = 1 / (0 . 3 n ) + 4 / (0 . 2 n ) = 23 . 3 /n V 3 = 1 / (0 . 2 n ) + 4 / (0 . 3 n ) = 18 . 3 /n V 4 = 1 / (0 . 2 n ) + 4 / (0 . 1 n ) =...
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## This note was uploaded on 02/06/2012 for the course STAT 403 taught by Professor Kerbyshedden during the Winter '12 term at University of Michigan-Dearborn.

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ps10 - Statistics 403 Problem Set 10 Due in lab on Friday...

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