Linear regression with independent variable plus 1 Standard Deviation

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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：

  1. Depv      Indv1 Indv2   Indv3    Indv3_plusSD
  2. 1   1.1555864       48    1  77.07593       0
  3. 2   1.0596864       61    2  69.51333       0
  4. 3   0.8380413       51    1  87.38040       0
  5. 4   1.5305489       53    2  67.43750       0
  6. 5   1.0619884       55    1 165.99977       1
  7. 6   0.8474507       56    2 229.14570       1
  8. 7   0.9579580       64    2 121.89550       0
  9. 8   0.7432210       58    1 211.17690       1
  10. 9   0.8374197       60    1 139.69577       0
  11. 10  0.7378349       65    1 277.03920       1
  12. 11  0.6971632       61    1 195.72100       1
  13. 12  0.5227076       64    2 194.63220       1
  14. 13  0.9900380       52    1 138.25417       0
  15. 14  0.8954233       52    2 237.39020       1
  16. 15  0.9058147       56    1 123.42930       0
  17. 16  0.9436135       55    2 152.75953       1
  18. 17  0.7123374       55    1 190.34547       1
  19. 18  1.1928167       58    1 166.50990       1
  20. 19  1.3342048       47    2  76.35120       0
  21. 20  1.0881865       49    1 135.71740       0
  22. 21  2.9028876       48    2  61.83147       0
  23. 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 Linear regression with independent variable plus 1 Standard Deviation

  1.        Depv      Indv1 Indv2   Indv3    Indv3_plusSD
  2. 1   1.1555864       48    1  77.07593       0
  3. 2   1.0596864       61    2  69.51333       0
  4. 3   0.8380413       51    1  87.38040       0
  5. 4   1.5305489       53    2  67.43750       0
  6. 5   1.0619884       55    1 165.99977       1
  7. 6   0.8474507       56    2 229.14570       1
  8. 7   0.9579580       64    2 121.89550       0
  9. 8   0.7432210       58    1 211.17690       1
  10. 9   0.8374197       60    1 139.69577       0
  11. 10  0.7378349       65    1 277.03920       1
  12. 11  0.6971632       61    1 195.72100       1
  13. 12  0.5227076       64    2 194.63220       1
  14. 13  0.9900380       52    1 138.25417       0
  15. 14  0.8954233       52    2 237.39020       1
  16. 15  0.9058147       56    1 123.42930       0
  17. 16  0.9436135       55    2 152.75953       1
  18. 17  0.7123374       55    1 190.34547       1
  19. 18  1.1928167       58    1 166.50990       1
  20. 19  1.3342048       47    2  76.35120       0
  21. 20  1.0881865       49    1 135.71740       0
  22. 21  2.9028876       48    2  61.83147       0
  23. 22  0.6661121       61    1 139.68627       0
  1. 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）。

  1. library(tidyverse)
  2. df = read_table2('Depv      Indv1 Indv2   Indv3
  3. 1.1555864       48    1  77.07593
  4. 1.0596864       61    2  69.51333
  5. 0.8380413       51    1  87.38040
  6. 1.5305489       53    2  67.43750
  7. 1.0619884       55    1 165.99977
  8. 0.8474507       56    2 229.14570
  9. 0.9579580       64    2 121.89550
  10. 0.7432210       58    1 211.17690
  11. 0.8374197       60    1 139.69577
  12. 0.7378349       65    1 277.03920
  13. 0.6971632       61    1 195.72100
  14. 0.5227076       64    2 194.63220
  15. 0.9900380       52    1 138.25417
  16. 0.8954233       52    2 237.39020
  17. 0.9058147       56    1 123.42930
  18. 0.9436135       55    2 152.75953
  19. 0.7123374       55    1 190.34547
  20. 1.1928167       58    1 166.50990
  21. 1.3342048       47    2  76.35120
  22. 1.0881865       49    1 135.71740
  23. 2.9028876       48    2  61.83147
  24. 0.6661121       61    1 139.68627') %>%
  25.   mutate(Indv3_scale = scale(Indv3))
  27. (sd3 = sd(df$Indv3))
  28. #> [1] 60.84117
  30. model1 =  lm(Depv ~ Indv1 + Indv2 + Indv3, data = df)   
  31. model2 =  lm(Depv ~ Indv1 + Indv2 + Indv3_scale, data = df)   
  33. coef(model1)['Indv3'] * sd3
  34. #>      Indv3 
  35. #> -0.1609104
  36. coef(model2)['Indv3_scale']
  37. #> Indv3_scale 
  38. #>  -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 &lt;- 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 -->

  1. library(tidyverse)
  2. df = read_table2(&#39;Depv      Indv1 Indv2   Indv3
  3. 1.1555864       48    1  77.07593
  4. 1.0596864       61    2  69.51333
  5. 0.8380413       51    1  87.38040
  6. 1.5305489       53    2  67.43750
  7. 1.0619884       55    1 165.99977
  8. 0.8474507       56    2 229.14570
  9. 0.9579580       64    2 121.89550
  10. 0.7432210       58    1 211.17690
  11. 0.8374197       60    1 139.69577
  12. 0.7378349       65    1 277.03920
  13. 0.6971632       61    1 195.72100
  14. 0.5227076       64    2 194.63220
  15. 0.9900380       52    1 138.25417
  16. 0.8954233       52    2 237.39020
  17. 0.9058147       56    1 123.42930
  18. 0.9436135       55    2 152.75953
  19. 0.7123374       55    1 190.34547
  20. 1.1928167       58    1 166.50990
  21. 1.3342048       47    2  76.35120
  22. 1.0881865       49    1 135.71740
  23. 2.9028876       48    2  61.83147
  24. 0.6661121       61    1 139.68627&#39;) %&gt;% 
  25.   mutate(Indv3_scale = scale(Indv3))
  27. (sd3 = sd(df$Indv3))
  28. #&gt; [1] 60.84117
  30. model1 =  lm(Depv ~ Indv1 + Indv2 + Indv3, data = df)   
  31. model2 =  lm(Depv ~ Indv1 + Indv2 + Indv3_scale, data = df)   
  33. coef(model1)[&#39;Indv3&#39;] * sd3
  34. #&gt;      Indv3 
  35. #&gt; -0.1609104
  36. coef(model2)[&#39;Indv3_scale&#39;]
  37. #&gt; Indv3_scale 
  38. #&gt;  -0.1609104

<sup>Created on 2020-01-14 by the reprex package (v0.3.0)</sup>

huangapple
  • 本文由 发表于 2020年1月3日 21:27:38
  • 转载请务必保留本文链接：https://go.coder-hub.com/59579434.html
  • dummy-variable
  • r
  • regression
  • standard-deviation
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