英文:
Linear regression with independent variable plus 1 Standard Deviation
问题
这是您提供的代码部分,无需翻译:
Depv Indv1 Indv2 Indv3 Indv3_plusSD
1 1.1555864 48 1 77.07593 0
2 1.0596864 61 2 69.51333 0
3 0.8380413 51 1 87.38040 0
4 1.5305489 53 2 67.43750 0
5 1.0619884 55 1 165.99977 1
6 0.8474507 56 2 229.14570 1
7 0.9579580 64 2 121.89550 0
8 0.7432210 58 1 211.17690 1
9 0.8374197 60 1 139.69577 0
10 0.7378349 65 1 277.03920 1
11 0.6971632 61 1 195.72100 1
12 0.5227076 64 2 194.63220 1
13 0.9900380 52 1 138.25417 0
14 0.8954233 52 2 237.39020 1
15 0.9058147 56 1 123.42930 0
16 0.9436135 55 2 152.75953 1
17 0.7123374 55 1 190.34547 1
18 1.1928167 58 1 166.50990 1
19 1.3342048 47 2 76.35120 0
20 1.0881865 49 1 135.71740 0
21 2.9028876 48 2 61.83147 0
22 0.6661121 61 1 139.68627 0
请继续您的工作。
英文:
This must be a really simple question though Im not sure if I do it correctly:
I want to perform a multiple lineair regression where I want to include the effect of an independent variable (Indv3) change in 1 standard deviation (SD)
In other words: if 'Indv3' changes 1SD, how is the dependent (Depv) variable associated to it?
What I did was: calculate the SD-value of 'Indv3' and make a dummy variable (Indv3_plusSD) with 'Indv3' + 1SD-value = 1 and the rest gets value 0.
Then to do the lineair regression I add the 'Indv3_plusSD' dummy and execute the regression. However when I do this I get another beta-coefficient for the 'Depv' compared to an analysis with the same data already published in a paper...(so prob Im doing it wrong with the SD analysis
Depv Indv1 Indv2 Indv3 Indv3_plusSD
1 1.1555864 48 1 77.07593 0
2 1.0596864 61 2 69.51333 0
3 0.8380413 51 1 87.38040 0
4 1.5305489 53 2 67.43750 0
5 1.0619884 55 1 165.99977 1
6 0.8474507 56 2 229.14570 1
7 0.9579580 64 2 121.89550 0
8 0.7432210 58 1 211.17690 1
9 0.8374197 60 1 139.69577 0
10 0.7378349 65 1 277.03920 1
11 0.6971632 61 1 195.72100 1
12 0.5227076 64 2 194.63220 1
13 0.9900380 52 1 138.25417 0
14 0.8954233 52 2 237.39020 1
15 0.9058147 56 1 123.42930 0
16 0.9436135 55 2 152.75953 1
17 0.7123374 55 1 190.34547 1
18 1.1928167 58 1 166.50990 1
19 1.3342048 47 2 76.35120 0
20 1.0881865 49 1 135.71740 0
21 2.9028876 48 2 61.83147 0
22 0.6661121 61 1 139.68627 0
linregr <- lm(Depv ~ Indv1 + Indv2 + Indv3_plusSD, data = df)
答案1
得分: 0
以下是您提供的内容的中文翻译:
在不使用您的标准差项的情况下,对Indv1
、Indv2
和Indv3
进行回归:
linregr <- lm(Depv ~ Indv1 + Indv2 + Indv3, data = df)
Indv3
的回归系数是预测在Indv3
单位变化时Depv
将会变化的数量,因此在Indv3
变化1个标准差时,Depv
将变化的数量为标准差乘以(Indv3的系数)。
library(tidyverse)
df = read_table2('Depv Indv1 Indv2 Indv3
1.1555864 48 1 77.07593
1.0596864 61 2 69.51333
0.8380413 51 1 87.38040
1.5305489 53 2 67.43750
1.0619884 55 1 165.99977
0.8474507 56 2 229.14570
0.9579580 64 2 121.89550
0.7432210 58 1 211.17690
0.8374197 60 1 139.69577
0.7378349 65 1 277.03920
0.6971632 61 1 195.72100
0.5227076 64 2 194.63220
0.9900380 52 1 138.25417
0.8954233 52 2 237.39020
0.9058147 56 1 123.42930
0.9436135 55 2 152.75953
0.7123374 55 1 190.34547
1.1928167 58 1 166.50990
1.3342048 47 2 76.35120
1.0881865 49 1 135.71740
2.9028876 48 2 61.83147
0.6661121 61 1 139.68627') %>%
mutate(Indv3_scale = scale(Indv3))
(sd3 = sd(df$Indv3))
#> [1] 60.84117
model1 = lm(Depv ~ Indv1 + Indv2 + Indv3, data = df)
model2 = lm(Depv ~ Indv1 + Indv2 + Indv3_scale, data = df)
coef(model1)['Indv3'] * sd3
#> Indv3
#> -0.1609104
coef(model2)['Indv3_scale']
#> Indv3_scale
#> -0.1609104
创建于2020年01月14日,使用reprex包 (v0.3.0)
英文:
Regress against Indv1
, Indv2
and Indv3
without your SD term:
linregr <- lm(Depv ~ Indv1 + Indv2 + Indv3, data = df)
The regression coefficient for Indv3
is the amount Depv
is predicted to change for a unit change in Indv3
, so the amount Depv
will change for a change of 1 SD in Indv3
is SD * (coefficient of Indv3).
<!-- language-all: lang-r -->
library(tidyverse)
df = read_table2('Depv Indv1 Indv2 Indv3
1.1555864 48 1 77.07593
1.0596864 61 2 69.51333
0.8380413 51 1 87.38040
1.5305489 53 2 67.43750
1.0619884 55 1 165.99977
0.8474507 56 2 229.14570
0.9579580 64 2 121.89550
0.7432210 58 1 211.17690
0.8374197 60 1 139.69577
0.7378349 65 1 277.03920
0.6971632 61 1 195.72100
0.5227076 64 2 194.63220
0.9900380 52 1 138.25417
0.8954233 52 2 237.39020
0.9058147 56 1 123.42930
0.9436135 55 2 152.75953
0.7123374 55 1 190.34547
1.1928167 58 1 166.50990
1.3342048 47 2 76.35120
1.0881865 49 1 135.71740
2.9028876 48 2 61.83147
0.6661121 61 1 139.68627') %>%
mutate(Indv3_scale = scale(Indv3))
(sd3 = sd(df$Indv3))
#> [1] 60.84117
model1 = lm(Depv ~ Indv1 + Indv2 + Indv3, data = df)
model2 = lm(Depv ~ Indv1 + Indv2 + Indv3_scale, data = df)
coef(model1)['Indv3'] * sd3
#> Indv3
#> -0.1609104
coef(model2)['Indv3_scale']
#> Indv3_scale
#> -0.1609104
<sup>Created on 2020-01-14 by the reprex package (v0.3.0)</sup>
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