Расчет омеги для факторного анализа: результат NA

Я пытаюсь вычислить оценки омега после исследовательского факторного анализа, чтобы оценить надежность компонентов, которые я нашел. Используя функцию omega() из пакета psych, я получаю следующий результат:

Вывод для функции омега

       Alpha:                 0.8 
        G.6:                   0.86 
        Omega Hierarchical:    0.37 
        Omega H asymptotic:    0.43 
        Omega Total            0.86 

    Schmid Leiman Factor loadings greater than 

 0.2 
              g   F1*   F2*   F3*   h2   u2   p2
    EMS1   0.30        0.71       0.59 0.41 0.15
    EMS3        -0.21  0.64       0.53 0.47 0.05
    EMS4         0.62             0.41 0.59 0.04
    EMS7   0.34        0.62       0.50 0.50 0.23
    EMS8   0.36        0.42       0.32 0.68 0.40
    EMS9         0.57             0.33 0.67 0.00
    EMS10              0.39       0.20 0.80 0.11
    EMS11        0.72             0.51 0.49 0.02
    EMS12        0.68             0.46 0.54 0.02
    EMS15        0.54 -0.24       0.41 0.59 0.02
    EMS16  0.22        0.77       0.63 0.37 0.08
    EMS19        0.65             0.52 0.48 0.01
    EMS20  0.27        0.53       0.36 0.64 0.21
    EMS21        0.62             0.40 0.60 0.04
    EMS23        0.63             0.42 0.58 0.07
    EMS24  0.68                   0.45 0.55 1.02
    EMS25  0.73                   0.56 0.44 0.95
    EMS27  0.45        0.20       0.25 0.75 0.83
    EMS28  0.78                   0.59 0.41 1.02
    EMS34  0.26  0.31  0.48       0.34 0.66 0.20

    With eigenvalues of:
      g F1* F2* F3* 
    2.5 3.4 2.9 0.0 

    general/max  0.73   max/min =   Inf
    mean percent general =  0.27    with sd =  0.36 and cv of  1.33 
    Explained Common Variance of the general factor =  0.28 

    The degrees of freedom are 133  and the fit is  0.8 
    The number of observations was  601  with Chi Square =  471.81  with prob <  1.9e-39
    The root mean square of the residuals is  0.04 
    The df corrected root mean square of the residuals is  0.05
    RMSEA index =  0.066  and the 10 % confidence intervals are  0.059 0.072
    BIC =  -379.21

    Compare this with the adequacy of just a general factor and no group factors
    The degrees of freedom for just the general factor are 170  and the fit is  5.4 
    The number of observations was  601  with Chi Square =  3195.63  with prob <  0
    The root mean square of the residuals is  0.22 
    The df corrected root mean square of the residuals is  0.24 

    RMSEA index =  0.173  and the 10 % confidence intervals are  0.167 0.177
    BIC =  2107.87

    Measures of factor score adequacy             
                                                    g  F1*  F2* F3*
    Correlation of scores with factors            0.9 0.94 0.93   0
    Multiple R square of scores with factors      0.8 0.89 0.86   0
    Minimum correlation of factor score estimates 0.6 0.78 0.73  -1

     Total, General and Subset omega for each subset
                                                     g  F1*  F2* F3*
    Omega total for total scores and subscales    0.86 0.82 0.85  NA
    Omega general for total scores and subscales  0.37 0.08 0.34  NA
    Omega group for total scores and subscales    0.58 0.75 0.51  NA
    Warning messages:
    1: In fac(r = r, nfactors = nfactors, n.obs = n.obs, rotate = rotate,  :
       A loading greater than abs(1) was detected.  Examine the loadings carefully.
    2: In fac(r = r, nfactors = nfactors, n.obs = n.obs, rotate = rotate,  :
      An ultra-Heywood case was detected.  Examine the results carefully
    3: In cov2cor(t(w) %*% r %*% w) :
      diag(.) had 0 or NA entries; non-finite result is doubtful

Вот как я вызываю функцию: omega(df[,items],nfactors=3)

После поиска рекомендаций я не смог найти, почему омега не была рассчитана для 3-го фактора. Я не уверен, что это проблема, связанная с одним из предупреждающих сообщений:

Warning messages:
1: In fac(r = r, nfactors = nfactors, n.obs = n.obs, rotate = rotate,  :
   A loading greater than abs(1) was detected.  Examine the loadings carefully.
2: In fac(r = r, nfactors = nfactors, n.obs = n.obs, rotate = rotate,  :
  An ultra-Heywood case was detected.  Examine the results carefully
3: In cov2cor(t(w) %*% r %*% w) :
  diag(.) had 0 or NA entries; non-finite result is doubtful

person Renan Saraiva    schedule 31.10.2018    source источник


Ответы (1)


Это может быть связано с тем, что Омега рассчитывается путем подбора модели CFA, и в вашем случае с 3 факторами фактор номер 3 используется для целей идентификации. Так что вы не ожидаете, что Омега будет рассчитана на это

person PsychometStats    schedule 08.05.2019