R中的响应行为建模

我试图对极端反应行为进行建模，即对来自20个不同国家的受访者的10个问卷项目进行1个强烈同意或5个强烈不同意的倾向检查，并提供有关其教育背景和性别的额外信息。我想确认一下 1有极端的反应行为， 2不同国家的反应行为不同 3.受教育程度不同，反应行为也不同 4反应行为存在性别差异 5 2x3、2x4、2x5、3x4、3x5和4x5之间存在交互作用

我不知道如何从R开始。我一直在使用潜在黄金，但不能在我的模型中包含国家、教育背景或性别变量

我是否可以将反应行为建模为潜在变量，然后使用常规ols回归检查2到5

我甚至不知道从哪里开始，如果你们中的一些人能把我推向正确的方向，我会非常高兴

以下是一些示例数据：

+--------+-------+-------+-------+-------+-------+-------+-------+-------+-------+--------+---------+------+-----+
|   id   | item1 | item2 | item3 | item4 | item5 | item6 | item7 | item8 | item9 | item10 | country | educ | gen |
+--------+-------+-------+-------+-------+-------+-------+-------+-------+-------+--------+---------+------+-----+
| 123512 |     3 |     2 |     3 |     1 |     1 |     4 |     1 |     4 |     4 |      1 | DE      |    1 |   0 |
| 123513 |     4 |     4 |     2 |     5 |     3 |     3 |     3 |     5 |     3 |      5 | DE      |    2 |   0 |
| 123514 |     5 |     1 |     4 |     5 |     4 |     4 |     4 |     1 |     1 |      4 | DE      |    3 |   0 |
| 123515 |     2 |     3 |     1 |     2 |     5 |     2 |     1 |     5 |     3 |      2 | E       |    1 |   0 |
| 123516 |     2 |     5 |     5 |     3 |     3 |     5 |     3 |     5 |     4 |      3 | E       |    2 |   1 |
| 123517 |     2 |     4 |     3 |     2 |     2 |     5 |     2 |     1 |     1 |      3 | E       |    3 |   1 |
| 123518 |     1 |     4 |     2 |     2 |     3 |     3 |     1 |     5 |     2 |      2 | E       |    1 |   0 |
| 123519 |     5 |     1 |     5 |     2 |     5 |     3 |     2 |     5 |     4 |      3 | E       |    1 |   1 |
| 123520 |     4 |     5 |     1 |     2 |     3 |     2 |     4 |     3 |     1 |      4 | E       |    1 |   1 |
| 123521 |     5 |     5 |     3 |     5 |     3 |     5 |     3 |     4 |     5 |      1 | F       |    1 |   0 |
| 123522 |     2 |     2 |     5 |     3 |     1 |     2 |     3 |     1 |     2 |      5 | F       |    1 |   1 |
| 123523 |     3 |     3 |     5 |     5 |     1 |     2 |     2 |     1 |     4 |      3 | F       |    2 |   1 |
| 123524 |     3 |     2 |     5 |     2 |     1 |     3 |     3 |     4 |     4 |      3 | F       |    3 |   1 |
| 123525 |     3 |     3 |     3 |     3 |     5 |     2 |     2 |     2 |     2 |      2 | F       |    1 |   1 |
| 123526 |     4 |     3 |     1 |     2 |     1 |     3 |     3 |     4 |     4 |      1 | F       |    2 |   0 |
| 123527 |     5 |     3 |     4 |     5 |     4 |     3 |     4 |     2 |     5 |      2 | F       |    4 |   0 |
| 123528 |     3 |     5 |     3 |     4 |     2 |     3 |     1 |     5 |     3 |      4 | F       |    1 |   1 |
| 123529 |     1 |     1 |     2 |     4 |     4 |     3 |     3 |     1 |     4 |      1 | F       |    1 |   0 |
| 123530 |     5 |     1 |     4 |     4 |     5 |     4 |     4 |     5 |     3 |      1 | RUS     |    2 |   1 |
| 123531 |     2 |     2 |     3 |     1 |     2 |     4 |     1 |     4 |     1 |      1 | RUS     |    2 |   0 |
| 123532 |     5 |     5 |     2 |     4 |     2 |     3 |     1 |     1 |     5 |      3 | RUS     |    1 |   1 |
| 123533 |     4 |     5 |     2 |     1 |     3 |     2 |     4 |     2 |     1 |      1 | RUS     |    1 |   0 |
| 123534 |     1 |     1 |     3 |     2 |     3 |     3 |     1 |     2 |     4 |      5 | RUS     |    2 |   0 |
| 123535 |     2 |     1 |     1 |     1 |     1 |     1 |     3 |     1 |     2 |      4 | RUS     |    3 |   1 |
| 123536 |     5 |     1 |     4 |     2 |     1 |     3 |     2 |     2 |     5 |      4 | RUS     |    3 |   1 |
| 123537 |     5 |     5 |     5 |     1 |     5 |     5 |     4 |     2 |     2 |      4 | RUS     |    3 |   1 |
| 123538 |     2 |     1 |     3 |     1 |     4 |     5 |     2 |     1 |     3 |      2 | RUS     |    1 |   0 |
| 123539 |     2 |     4 |     2 |     4 |     5 |     5 |     5 |     3 |     1 |      4 | RUS     |    2 |   0 |
+--------+-------+-------+-------+-------+-------+-------+-------+-------+-------+--------+---------+------+-----+

非常感谢你的帮助，我希望能找到一些建议

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致以最诚挚的问候

如果您正在寻找关于反应行为分析的现代方法，我想您会从我们这里得到关于这一问题的更专业的答案。然而，我的两分钱是：

要优化对问题的回答，请查看和

让我们将您的数据结构用作

resp <- structure(list(id = 123512:123539, 
                   item1 = c(3L, 4L, 5L, 2L, 2L, 2L, 1L, 5L, 4L, 5L, 2L, 3L, 3L, 3L, 
                             4L, 5L, 3L, 1L, 5L, 2L, 5L, 4L, 1L, 2L, 5L, 5L, 2L, 2L), 
                   item2 = c(2L, 4L, 1L, 3L, 5L, 4L, 4L, 1L, 5L, 5L, 2L, 3L, 2L, 3L, 
                             3L, 3L, 5L, 1L, 1L, 2L, 5L, 5L, 1L, 1L, 1L, 5L, 1L, 4L), 
                   item3 = c(3L, 2L, 4L, 1L, 5L, 3L, 2L, 5L, 1L, 3L, 5L, 5L, 5L, 3L, 
                             1L, 4L, 3L, 2L, 4L, 3L, 2L, 2L, 3L, 1L, 4L, 5L, 3L, 2L), 
                   item4 = c(1L, 5L, 5L, 2L, 3L, 2L, 2L, 2L, 2L, 5L, 3L, 5L, 2L, 3L, 
                             2L, 5L, 4L, 4L, 4L, 1L, 4L, 1L, 2L, 1L, 2L, 1L, 1L, 4L), 
                   item5 = c(1L, 3L, 4L, 5L, 3L, 2L, 3L, 5L, 3L, 3L, 1L, 1L, 1L, 5L, 
                             1L, 4L, 2L, 4L, 5L, 2L, 2L, 3L, 3L, 1L, 1L, 5L, 4L, 5L), 
                   item6 = c(4L, 3L, 4L, 2L, 5L, 5L, 3L, 3L, 2L, 5L, 2L, 2L, 3L, 2L, 
                             3L, 3L, 3L, 3L, 4L, 4L, 3L, 2L, 3L, 1L, 3L, 5L, 5L, 5L), 
                   item7 = c(1L, 3L, 4L, 1L, 3L, 2L, 1L, 2L, 4L, 3L, 3L, 2L, 3L, 2L, 
                             3L, 4L, 1L, 3L, 4L, 1L, 1L, 4L, 1L, 3L, 2L, 4L, 2L, 5L), 
                   item8 = c(4L, 5L, 1L, 5L, 5L, 1L, 5L, 5L, 3L, 4L, 1L, 1L, 4L, 2L, 
                             4L, 2L, 5L, 1L, 5L, 4L, 1L, 2L, 2L, 1L, 2L, 2L, 1L, 3L), 
                   item9 = c(4L, 3L, 1L, 3L, 4L, 1L, 2L, 4L, 1L, 5L, 2L, 4L, 4L, 2L, 
                             4L, 5L, 3L, 4L, 3L, 1L, 5L, 1L, 4L, 2L, 5L, 2L, 3L, 1L), 
                   item10 = c(1L, 5L, 4L, 2L, 3L, 3L, 2L, 3L, 4L, 1L, 5L, 3L, 3L, 2L, 
                              1L, 2L, 4L, 1L, 1L, 1L, 3L, 1L, 5L, 4L, 4L, 4L, 2L, 4L), 
                   country = c("DE", "DE", "DE", "E", "E", "E", "E", "E", "E", "F", 
                               "F", "F", "F", "F", "F", "F", "F", "F", "RUS", "RUS", 
                               "RUS", "RUS", "RUS", "RUS", "RUS", "RUS", "RUS", "RUS"), 
                   educ = c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 1L, 1L, 1L, 1L, 2L, 3L, 1L, 
                            2L, 4L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 3L, 3L, 3L, 1L, 2L), 
                   gen = c(0L, 0L, 0L, 0L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 
                           0L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 0L, 1L, 1L, 1L, 0L, 0L)), 
              row.names = c(NA, -28L), class = "data.frame")

例如，对于第一个陈述，通过标准化潜在回归系数，我们得到了一个提示，表明F国的受访者的极端反应倾向较低

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大家好，有人能告诉我我的问题是否措辞错误，或者是否不够具体吗？我需要提供更多信息吗？请让我知道，如果我需要重申我的问题或我需要改变什么才能得到答案。非常感谢你

items <- paste0("item", 1:10)
resp[, items] <- 1 * (resp[, items] == 1 | resp[, items] == 5)

summary(TAM::tam.mml(resp[, items], formulaY = ~country, dataY = resp[, c("country", "educ", "gen")]))
summary(TAM::tam.mml(resp[, items], formulaY = ~country*educ, dataY = resp[, c("country", "educ", "gen")]))
summary(TAM::tam.mml(resp[, items], formulaY = ~country*gen, dataY = resp[, c("country", "educ", "gen")]))
summary(TAM::tam.mml(resp[, items], formulaY = ~educ*gen, dataY = resp[, c("country", "educ", "gen")]))

------------------------------------------------------------
Standardized Coefficients
        parm dim  est   StdYX    StdX    StdY
1  Intercept   1  0.0      NA      NA      NA
2   countryE   1  0.0  0.0000  0.0000  0.0000
3   countryF   1 -0.6 -0.9939 -0.2854 -2.0898
4 countryRUS   1  0.0  0.0000  0.0000  0.0000

** Explained Variance R^2
[1] 0.9879
** SD Theta
[1] 0.2871
** SD Predictors
 Intercept   countryE   countryF countryRUS 
    0.0000     0.4179     0.4756     0.4880 
------------------------------------------------------------