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. Author manuscript; available in PMC: 2015 Nov 10.
Published in final edited form as: Stat Med. 2013 Aug 23;33(2):330–360. doi: 10.1002/sim.5926

Table 6.

Mm-ANOVA approach for typical test×reader×case factorial study design

  1. Derive the mm-ANOVA model
    1. Conventional ANOVA model: Yijk = μ + τi + Rj + Ck + (τR)ij + (τC)ik + (RC)jk + (τRC)ijk + εijk, i = 1, …, t; j = 1, …, r; k = 1, …, c, with variance components σR2,σC2,σTR2,σTC2,σRC2,στRC2, and σε2 and constraint i=1tτi=0. Define σ2=σTRC2+σε2.
    2. Mm-ANOVA model (note: ij = Yij):
      ij=μ+τi+Rj+(τR)ij+ε˜ij where ε˜ij=C+(τC)i+(RC)j+(τRC)ij+εij and i=1tτi=0
    3. Mm-ANOVA error variance and covariances expressed in terms of conventional ANOVA variance components: σ˜ε2=1c(σC2+σTC2+σRC2+σ2), Cov1cov(ε˜ij,ε˜ij)=1c(σC2+σRC2), Cov2cov(ε˜ij,ε˜ij)=1c(σC2+στC2), Cov3cov(ε˜ij,ε˜ij)=1cσC2, where ii′, jj
    4. Covariance constraints: Cov1 ≥ Cov3; Cov2 ≥ Cov3; Cov3 ≥ 0
  2. Derive the mm-ANOVA test statistic and its null distribution
    1. Mm-ANOVA model hypothesis of equal test accuracies: H0 : θ1 = ⋯ = θt where θi = E (i)
    2. Conventional ANOVA model hypothesis: θi = E (Yi••) = μ + τi ⇒ H0 : τ1 = ⋯ = τt
    3. Conventional ANOVA expected mean squares
      Mean square Expected mean square
      MS(T)
      rc(t1)i=1tτi2+cσTR2+rσTC2+σ2
      MS(R)
      tcσR2+cσTR2+tσRC2+σ2
      MS(C)
      trσC2+rσTC2+tσRC2+σ2
      MS(T * R)
      cσTR2+σ2
      MS(T * C)
      rσTC2+σ2
      MS(R * C)
      tσRC2+σ2
      MS(T * R * C)
      σ2σTRC2+σε2
    4. Conventional ANOVA test statistic: F=MS(T)MS(T*R)+MS(T*C)MS(T*R*C)
    5. MS˜(T)=1cMS(T),MS˜(T*R)=1cMS(T*R),MS˜(R)=1cMS(R)
    6. F=MS˜(T)MS˜(T*R)+U where U=1c{MS(T*C)MS(T*R*C)}
    7. E{MS(T*C)}=rσTC2+σ2,E{MS[T*R*C]}=σ2E(U)=1c(rσTC2)=r(Cov2Cov3).
    8. FOR*=MS˜(T)MS˜(T*R)+r(Cov2Cov3)
    9. FOR=MS˜(T)MS˜(T*R)+rmax(Cov^2Cov^3,0)
    10. Under H0, FORFt−1,df2 where df2=[MS˜[T*R]+rmax(Cov^2Cov^3,0)]2[MS˜[T*R]]2/[(t1)(r1)]
  3. Derive confidence intervals
    • (a)
      Mm-ANOVA test accuracy parameters: θ = (θ1, …, θt)′, with θi = E (i), i = 1, …, t
    • (b)
      Corresponding conventional ANOVA parameters: θi = E (Yi••) = μ + τi
    • (c)
      Conventional ANOVA estimate: θ̂i = Yi••
      • CI for l′ (θ) with l = (l1, …, lt)′, i=1tli=0:
    • (d)
      l(θ^)=i=1tliθ^i=i=1tliYi=i=1tliτi+i=1tli[(τR)i+(τC)i+(τRC)i+εi]V=i=1tli2[σTR2r+σTC2c+σ2rc]=1rci=1tli2[cσTR2+rσTC2+σ2]
    • (e)
      V=1rci=1tli2E[MS(T*R)+MS(T*C)MS(T*R*C)]
    • (f)
      V=1ri=1tli2E[MS˜(T*R)+U] where U=1c{MS(T*C)MS(T*R*C)}
    • (g)
      E(U)=rσTC2c=r(Cov2Cov3)V=1ri=1tli2{E[MS˜(T*R)]+r(Cov2Cov3)}
    • (h)
      V^=1ri=1tli2{MS˜(T*R)+max[r(Cov^2Cov^3),0]}
    • (i)
      df2=[MS˜(T*R)+rmax(Cov^2Cov^3,0)]2[MS˜(T*R)]2/[(t1)(r1)] (same as df2 in step 2j)
    • (j)
      θ̂i = i
    • (k)
      CI:i=1tlii±tα/2;df21ri=1tli2{MS˜(T*R)+max[r(Cov^2Cov^3),0]}
      • CI for θi
    • (d)
      θ^i=Yi=τi+R+C+(τR)i+(τC)i+(RC)+(τRC)i+εiV=σR2r+σC2c+σTR2r+σTC2c+σRC2rc+σ2rc=1rc(cσR2+rσC2+cσTR2+rσTC2+σRC2+σ2)
    • (e)
      V=1trcE[MS(R)+(t1)MS(T*R)+MS(C)MS(R*C)+(t1)MS(T*C)(t1)MS(T*R*C)]
    • (f)
      V=1trE[MS˜(R)+(t1)MS˜(T*R)+U]
      where
      U=1c{MS(C)MS(R*C)+(t1)MS(T*C)(t1)MS(T*R*C)}
    • (g)
      E(U)=trc(σC2+σTC2)=trCov2V=1tr{E[MS˜(R)+(t1)MS˜(T*R)]+trCov2}
    • (h)
      V^=1tr[MS˜(R)+(t1)MS˜(T*R)+trmax(Cov^2,0)]
    • (i)
      df2=[MS˜(R)+(t1)MS˜(T*R)+trmax(Cov^2,0)]2[MS˜(R)]2r1+[(t1)MS˜(T*R)]2(t1)(r1)
    • (j)
      θ̂i = i
    • (k)
      CI:i±tα/2;df21tr[MS˜(R)+(t1)MS˜(T*R)+trmax(Cov^2,0)]
  4. Derive the non-null distribution Fdf1,df2 of the step-2 F statistic
    1. Step 2d F numerator: MSnum = MS(T), E[MS(T)]=rc(t1)i=1tτi2+cσTR2+rσTC2+σ2, df (MS (T)) = t − 1, E(Yijk)=μ+τiλ=df(MSnum)MSnum|Y=E(Y)E(MSnum|H0)=rci=1tτi2cσTR2+rσTC2+σ2
    2. rσTC2+σ2=c[σε˜2Cov1+(r1)(Cov2Cov3)]λ=ri=1tτi2σTR2+σε˜2Cov1+(r1)(Cov2Cov3)
    3. Step 2h FOR*denominator=MS˜(T*R)+r(Cov2Cov3), E(MS˜(T*R))=1cE(MS˜(T*R))=1c(cσTR2+σ2)=(σTR2+σε˜2Cov1Cov2+Cov3)df2=[σTR2+σε˜2Cov1+(r1)(Cov2Cov3)]2[σTR2+σε˜2Cov1Cov2+Cov3]2(t1)(r1)
    4. FORFt−1,df2