Statistics Meeting Book (June 20, 2018)

where ‘z 0.975 = 1.96. The limits of eq.(4) should be truncated to fall within the values -1 to +1. (The need for such truncation should indicate caution in the use of the resulting interval, which may not retain accurate coverage.) ’ is the 97.5% quantile of the standard normal distribution, or z 0.975

A similar issue arises with the estimation of the across-collaborator mean difference in methods, or dLPOD(A,B). The point estimate is the mean value of the individual dPOD values, or

Σ dPOD k / L

dLPOD

=

(5)

Under the large-sample approximation the standard error of dLPOD may be estimated by ( note change in notation from TR296 ):

= √ { [ Σ ( dPOD k – dLPOD ) 2 / (L – 1) ] / L }

SE(dLPOD)

(6)

In TR310, is was recommended that a 95% confidence interval (LCL, UCL) for dLPOD be estimated by

LCL =

max{ -2, dLPOD - t 0.975

SE(dLPOD) }

(7a)

UCL =

min{ +2, dLPOD + t 0.975

SE(dLPOD) }

(7b)

where t 0.975

is the 97.5% quantile of the student t distribution with L - 1 degrees of freedom.

(Alternatively, it should be noted that a less conservative alternative to t 0.975

is z 0.975

= 1.96.)

The confidence interval of eq.(7) for dLPOD suffers from the same sort of degeneracy as that for dPOD. I.e., whenever all dPOD values across collaborators have the same value, the estimate of SE(dLPOD) has the value zero in eq.(6), which leads to an interval of zero length in eq.(7).

Some type of continuity-correction for the dLPOD formulae is needed to fix this problem.

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