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.