我在一个构造上有五个项目,当我在它上面运行alpha时,我得到以下结果,没有任何错误
psych::alpha(construct,
na.rm = TRUE,
title = 'myscale',
n.iter = 1000)
Reliability analysis myscale
Call: psych::alpha(x = construct, title = "myscale", na.rm = TRUE,
n.iter = 1000)
raw_alpha std.alpha G6(smc) average_r S/N ase mean sd
0.81 0.81 0.78 0.46 4.3 0.013 2.6 0.89
lower alpha upper 95% confidence boundaries
0.78 0.81 0.84
lower median upper bootstrapped confidence intervals
0.77 0.81 0.84
我一直在读论文从阿尔法到欧米茄：内部一致性评估的普遍问题的实用解决方案link
它建议使用以下代码
MBESS::ci.reliability(construct, interval.type="bca", B=1000, type = "omega")
$est
[1] 0.8107376
$se
[1] 0.01651936
$ci.lower
[1] 0.7764029
$ci.upper
[1] 0.839944
$conf.level
[1] 0.95
$type
[1] "omega"
$interval.type
[1] "bca bootstrap"
我一直试图使用心理包在我的样本集上运行omega,以便在我的分析中保持一致
psych::omega(m = construct,
nfactors = 1, fm = "pa", n.iter = 1000, p = 0.05,
title = "Omega", plot = FALSE, n.obs = 506)
我收到两条错误消息
In factor.scores, the correlation matrix is singular, an approximation is used
Omega_h for 1 factor is not meaningful, just omega_t
发生此警告是因为Omega_h的列数小了两个.关于SO的上一个问题有点回答
McDonalds omega: warnings in R
我所遇到的错误如下
Error in fac(r = r, nfactors = nfactors, n.obs = n.obs, rotate = rotate, :
I am sorry: missing values (NAs) in the correlation matrix do not allow me to continue.
Please drop those variables and try again.
In addition: There were 50 or more warnings (use warnings() to see the first 50)
没有缺失值,所以我不确定错误二
我的构造的细节是
Q1 Q2 Q3
Min. :0.000 Min. :0.000 Min. :0.000
1st Qu.:2.000 1st Qu.:2.000 1st Qu.:2.000
Median :3.000 Median :2.000 Median :3.000
Mean :2.597 Mean :2.393 Mean :3.227
3rd Qu.:3.000 3rd Qu.:3.000 3rd Qu.:4.000
Max. :6.000 Max. :6.000 Max. :6.000
Q4 Q5
Min. :0.00 Min. :0.000
1st Qu.:1.00 1st Qu.:2.000
Median :2.00 Median :2.000
Mean :2.17 Mean :2.445
3rd Qu.:3.00 3rd Qu.:3.000
Max. :6.00 Max. :6.000
编辑
创建具有相同属性的数据–100个条目(Alpha大约0.56),但它在omega上生成相同的错误
structure(list(Q1 = c(4, 5, 3, 5, 4, 5, 3, 5, 5, 5, 6,
3, 5, 4, 6, 5, 5, 6, 7, 4, 5, 5, 3, 4, 4, 5, 4, 3, 5, 4, 5, 5,
6, 6, 3, 6, 3, 4, 4, 4, 6, 5, 3, 2, 6, 6, 4, 5, 4, 3, 6, 4, 4,
5, 6, 2, 4, 3, 4, 6, 4, 6, 4, 5, 5, 6, 4, 6, 5, 5, 4, 5, 6, 6,
2, 5, 4, 3, 4, 4, 4, 6, 3, 3, 5, 4, 4, 4, 5, 5, 5, 3, 6, 6, 6,
6, 5, 4, 3, 5), Q2 = c(7, 4, 4, 4, 4, 6, 6, 6, 7, 6, 5,
6, 5, 4, 5, 6, 6, 6, 7, 5, 4, 4, 6, 6, 4, 4, 6, 2, 6, 5, 4, 6,
4, 6, 6, 6, 5, 4, 4, 4, 4, 3, 3, 4, 4, 4, 4, 6, 2, 6, 6, 5, 4,
6, 6, 4, 4, 7, 6, 5, 5, 5, 5, 6, 5, 5, 4, 5, 5, 5, 4, 6, 7, 5,
5, 5, 6, 5, 6, 5, 6, 7, 2, 6, 5, 7, 3, 5, 5, 3, 3, 3, 7, 4, 5,
6, 6, 6, 5, 7), Q3 = c(5, 4, 5, 6, 4, 4, 5, 4, 2, 6, 5,
5, 5, 5, 7, 5, 5, 6, 7, 6, 3, 6, 6, 6, 5, 6, 6, 5, 5, 4, 5, 5,
6, 6, 5, 6, 5, 5, 4, 4, 6, 4, 4, 4, 4, 4, 4, 5, 5, 4, 5, 5, 4,
3, 5, 4, 5, 6, 6, 6, 4, 5, 5, 5, 6, 4, 5, 5, 7, 4, 5, 6, 6, 5,
5, 3, 3, 5, 4, 6, 5, 5, 1, 3, 5, 3, 2, 5, 4, 6, 6, 6, 6, 4, 6,
3, 6, 6, 6, 5), Q4 = c(6, 6, 4, 7, 4, 6, 7, 6, 7, 6, 6,
6, 5, 7, 7, 6, 6, 5, 7, 7, 6, 6, 7, 7, 6, 6, 6, 5, 6, 7, 5, 6,
7, 5, 4, 6, 4, 3, 6, 4, 6, 6, 6, 3, 5, 7, 5, 6, 4, 6, 7, 6, 7,
4, 6, 3, 5, 7, 5, 4, 6, 6, 4, 6, 5, 5, 5, 5, 7, 7, 7, 6, 6, 6,
5, 6, 6, 4, 5, 7, 6, 7, 3, 5, 6, 5, 6, 5, 5, 7, 7, 6, 6, 2, 7,
6, 6, 7, 7, 5)), .Names = c("Q1", "Q2", "Q3",
"Q4"), row.names = c(NA, 100L), class = "data.frame")
谁能看到我跌倒的地方？
感谢您的时间
最佳答案 所以我试过这个：
psych::omega(m = construct)
它适用于这个结果：
Omega
Call: psych::omega(m = construct)
Alpha: 0.56
G.6: 0.49
Omega Hierarchical: 0.53
Omega H asymptotic: 0.89
Omega Total 0.6
Schmid Leiman Factor loadings greater than 0.2
g F1* F2* F3* h2 u2 p2
Q1 0.41 0.30 0.26 0.74 0.65
Q2 0.37 0.25 0.20 0.80 0.67
Q3 0.50 0.25 0.31 0.69 0.80
Q4 0.64 0.23 0.46 0.54 0.89
With eigenvalues of:
g F1* F2* F3*
0.95 0.15 0.06 0.05
general/max 6.35 max/min = 2.83
mean percent general = 0.75 with sd = 0.11 and cv of 0.15
Explained Common Variance of the general factor = 0.78
The degrees of freedom are -3 and the fit is 0
The number of observations was 100 with Chi Square = 0 with prob < NA
The root mean square of the residuals is 0
The df corrected root mean square of the residuals is NA
Compare this with the adequacy of just a general factor and no group factors
The degrees of freedom for just the general factor are 2 and the fit is 0.01
The number of observations was 100 with Chi Square = 0.62 with prob < 0.73
The root mean square of the residuals is 0.03
The df corrected root mean square of the residuals is 0.05
RMSEA index = 0 and the 90 % confidence intervals are NA 0.14
BIC = -8.59
Measures of factor score adequacy
g F1* F2* F3*
Correlation of scores with factors 0.75 0.37 0.27 0.24
Multiple R square of scores with factors 0.57 0.14 0.07 0.06
Minimum correlation of factor score estimates 0.14 -0.72 -0.86 -0.88
Total, General and Subset omega for each subset
g F1* F2* F3*
Omega total for total scores and subscales 0.60 0.37 0.31 0.46
Omega general for total scores and subscales 0.53 0.25 0.25 0.41
Omega group for total scores and subscales 0.06 0.12 0.06 0.05
我检查了默认值和nfactors = 3和n.iter = 1.然后我慢慢增加了n.iter,并减少了n.factor,它一直工作直到n.iter = 7,并保持nfactors为3
psych::omega(m = construct, n.iter = 7, p = 0.05, nfactors = 3)
使用完整的数据集,您应该能够获得更高的数据
