Расчет омега для факторного анализа: результат NA
Я пытаюсь вычислить оценки омега после разведочного факторного анализа, чтобы оценить надежность компонентов, которые я нашел. С использованием omega() функция от psych Пакет я получаю этот вывод:
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 руководства я не мог найти, почему омега не была вычислена 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
0 ответов
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