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Biostatistics Case Studies 2008 Peter D. Christenson Biostatistician http://gcrc.labiomed.org/ biostat Session 3: Replicates
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### Transcript of Biostatistics Case Studies 2008 Peter D. Christenson Biostatistician Session 3: Replicates.

• Biostatistics Case Studies 2008Peter D. ChristensonBiostatisticianhttp://gcrc.labiomed.org/biostatSession 3: Replicates

• Question #1

• Question #1Fig 6: Fat in Males19.0=3.515.5Strength of Treatment Effect:Signal:Noise Ratio t= 3.5SD(1/NCtrl + 1/NAb)1/2Are the Ns the # of dams or # of offspring?What is correct SD?Ctrl Ab

• Question #1Group N Mean SD

Ctrl 33 18.784 1.5148 Ab 13 15.643 1.3693 Diff 3.1411 1.4766

t-test: t Value Pr > |t| 6.50

• Question #2

• Question #3

• Basic Issue for Using Offspring as Replicates Dams vary. Overall, offspring vary. Do offspring from a dam vary less than offspring from different dams (positive correlation)? Do offspring from a dam vary more than offspring from different dams (negative correlation)? What could cause this?

• Intra-Dam Correlation Among OffspringExample: Four dams - A,B,C,D - with 2 offspring each:AABBCCDDAAABBBCCCDDDOffspring FatDam MeansStrong Negative CorrelationStrong Positive CorrelationOverall MeanNo CorrelationAABBCCDD

• Intra-Dam and Inter-Dam VariationExample: Four dams - A,B,C,D - with 2 offspring each:AABBCCDDAAABBBCCCDDDOffspring FatOverall MeanCorrelation = Scaled VInter - VIntraCan be calculated from the data.Denote correlation by r.AABBCCDDVInterVIntra

• Correct SD Uses Both VariationsTable 6: Fat in Males19.0=3.515.5Strength of Treatment Effect:Signal/Noise Ratio t= 3.5SD(1/NCtrl + 1/NAb)1/2Are the Ns the # of dams or # of offspring?What is correct SD?Ctrl AbSD2 = V(1 + (n-1)r), where n=# offspring/dam

• Correct AnalysisNs are #s of offspring.Incorporate offspring correlation by using:SD2 = V(1 + (n-1)r), where n=# offspring/damSignal/Noise Ratio t= SD(1/NCtrl + 1/NAb)1/2If r=0, then SD2=V and same as t-test.If r>0, then SD2>V, so t-test overstates effect.If rV, so t-test understates effect.

• Correct AnalysisThus, the reasoning is that the dams are clusters of correlated outcomes (offspring).If offspring were completely correlated (r=1), i.e., identical in a dam, then the correct analysis is the same as using dam means. [SD2 = nV] If there is no correlation (r=0), the analysis is the same as ignoring dams and using offspring results. [SD2 = V] If there is some correlation, then SD incorporates that correlation, i.e., relative intra- and inter-.

• Correct Analysis in SoftwareIf we have the same # of offspring for every dam, we can use repeated measures ANOVA. Specify the dam as a subject and the offspring as the repeated values.Otherwise, use Mixed Model for Repeated Measures.

Both of these methods consider the dams as clusters of correlated outcomes (offspring).

• Numerical Illustrations1. All Offspring for a Dam Identical2. All Offspring for a Dam are Unique3. Offspring for a Dam are Negatively Correlated

• 1. All Offspring for a Dam Identical

• Recall Paper Uses OffspringGroup N Mean SD

Ctrl 33 18.784 1.5148 Ab 13 15.643 1.3693 Diff 3.1411 1.4766

t-test: t Value Pr > |t| 6.50

• Analysis on Dam MeansGroup N Mean SD

Ctrl 9 19.000 1.500Ab 9 15.494 1.500 Diff 3.5061 1.500

t-test: t Value Pr > |t| 4.96

• Analysis using Calculated CorrelationSame data using mixed model gives: CovParm Subject Estimate CS id 2.2485 Residual 1.365E-6

Num DenEffect DF DF F Value Pr > Fgroup 1 16 24.61 0.0001 group Estimate Std Err Lower UpperCtrl 19.0006 0.4998 17.9410 20.0602Ab 15.4940 0.4998 14.4344 16.5536Square root of 24.61 is t = 4.96, same as analysis on means.R = 1 =2.2485 (2.2485 + 0)

• 2. All Offspring for a Dam are Unique

• Second Set of Simulated DataGroup N Mean SD

Ctrl 33 19.000 1.5000 Ab 13 15.500 1.5000 Diff 3.5000 1.5000

t-test: t Value Pr > |t| 7.13

• Analysis using Calculated CorrelationSame data using mixed model gives: CovParm Subject Estimate CS id -0.1860 Residual 2.4278

Num DenEffect DF DF F Value Pr > Fgroup 1 16 56.20 0.0001 group Estimate Std Err Lower UpperCtrl 18.9690 0.2225 18.4972 19.4407Ab 15.5102 0.4042 14.6534 16.3670R = -0.083 =-0.186 (-0.186+2.428)Square root of 56.20 is t = 7.50, close to t-test ignoring dams.

• 3. Offspring for a Dam are Negatively Correlated

• Third Set of Simulated DataGroup N Mean SD

Ctrl 32 19.000 1.4756 Ab 12 15.500 1.4302 Diff 3.500 1.4639

t-test: t Value Pr > |t| 7.06

• Analysis using Calculated CorrelationSame data using mixed model gives: CovParm Subject Estimate CS id -1.5780 Residual 3.6458

Num DenEffect DF DF F Value Pr > Fgroup 1 20 218.33 0.0001 group Estimate Std Err Lower UpperCtrl 19.0000 0.1237 18.7419 19.2580Ab 15.5000 0.2020 15.0786 15.9214R = -0.76 =-1.578 (-1.578+3.646)Square root of 218.33 is t = 14.8, twice the t-test.But, with neg corr, probably would not have a 3.5 difference.