How to interpret transformed independent and dependent variables in summary(lm)?

Question

Call:                   
lm(formula = GROWTH ~ log(X1) + log(X2) + log(X3) + log(X4) +                   
    log(X5) + log(1 +X6) + log(1 + X7) +                    
    log(X8) + log(X9) + log(X10) + log(X11) +                   
    log(X12) + log(X13) + X14 + X14:X9 +                    
    X14:X10                 
    data = Data)                    

Residuals:                  
  Min          1Q        Median     3Q       Max    
-3.04237    -0.31965    0.05351   0.36639   2.52087 

Coefficients:                   
                    Estimate    Std. Error  t value Pr(>|t|)    
(Intercept)         2.837487    9.543146    0.297   0.766217    
log(X1)             0.377957    0.008647    43.71   < 2e-16 ***
log(X2)             0.363631    0.008906    40.829  < 2e-16 ***
log(X3)             0.337246    0.024202    13.934  < 2e-16 ***
log(X4)            -0.19371     0.029786   -6.503   8.11E-11    ***
log(X5)             0.01227     0.00437     2.808   0.004995    **
log(1 + X6)         0.006533    0.036977    0.177   0.859759    
log(1 + X7)         0.426738    0.191617    2.227   0.02596 *
log(X8)            -0.020741    0.009424    -2.201  0.027759    *
log(X9)             11.303514   2.745818    -4.117  3.87E-05    ***
log(X10)           -7.466939    0.814056    -9.173  < 2e-16 ***
log(X11)           -0.004444    0.00885    -0.502   0.615567    
log(X13)            0.067205    0.010626    6.325   2.61E-10    ***
log(X12)            1.711401    0.580518    2.948   0.003203    **
X14 [LEVEL 1]       18.422627   9.391444    -1.962  0.049823    *
X14 [LEVEL 2]       20.160172   9.386903    -2.148  0.031755    *
X14 [LEVEL 3]       12.78601    15.33008    0.834   0.404268    
X14 [LEVEL 4]       19.937816   9.679742    -2.06   0.03944 *
X14 [LEVEL 5]       13.83603    10.916449   -1.267  0.205015    
X14 [LEVEL 6]       23.939136   9.47908     -2.525  0.011565    *
X14 [LEVEL 7]       20.220041   11.217758  -1.803   0.071487    .
X14 [LEVEL 8]:X9    6.652888    4.17066     1.595   0.110697    
X14 [LEVEL 1]:X9    7.560706    1.981892    3.815   0.000137    ***
X14 [LEVEL 2]:X9    8.124572    1.857204    4.375   1.22E-05    ***
X14 [LEVEL 3]:X9    0.765371    5.173577    0.148   0.882393    
X14 [LEVEL 4]:X9    8.415016    2.337441    3.6 0.000319    ***
X14 [LEVEL 5]:X9    8.760546    3.293728    2.66    0.007828    **
X14 [LEVEL 6]:X9    10.727086   1.950529    5.5 3.87E-08    ***
X14 [LEVEL 7]:X9    8.913338    3.62592 2.458   0.013974    *
X14 [LEVEL 8]:X10   -9.409351   6.665734    -1.412  0.158089    
X14 [LEVEL 1]:X10   5.600412    0.628323    8.913   < 2e-16 ***
X14 [LEVEL 2]:X10   6.308849    0.669047    9.43    < 2e-16 ***
X14 [LEVEL 3]:X10   12.890973   5.191096    -2.483  0.013029    *
X14 [LEVEL 4]:X10   6.008453    0.835861    7.188   6.88E-13    ***
X14 [LEVEL 5]:X10   -0.174229   2.401866    -0.073  0.942174    
X14 [LEVEL 6]:X10   6.335575    0.774041    8.185   2.95E-16    ***
X14 [LEVEL 7]:X10   5.391272    2.226843    2.421   0.015488    *
---                 
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1                  

Residual standard error: 0.563 on 14573 degrees of freedom                  
(31913 observations deleted due to missingness)                 
Multiple R-squared:  0.5652                 
"   Adjusted R-squared:  0.5642 "                   
F-statistic: 526.3 on 36 and 14573 DF                   
p-value: < 2.2e-16**    

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Above is a linear GROWTH model. I have substituted in independent variable 'labels' for privacy purposes. In the example all numeric variables have been logarithmically transformed, and the dependent growth variable has had a box cox transformation applied to it. In the case of the independents this was done to normalize input variables, and the box cox transformation was applied to the dependent to correct increasing variance in the output. While i am certainly new to R, i believe this to be a better fit than the data with no transformations. However, please, let me know if I'm off base here. NOW, my question is, how do i interpret these values? Is there a way to 'un'transform outputs, so that the coefficient estimates and standard errors are valuable to me? They mean little in their current state.

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This question was originally posted on stack overflow, but i was advised I might get better feedback from exchange. I am still new to these forums, and hope that holds true.

Answer 1

Considering just the simplest case:

$$y = \beta_0 + \beta_1 \ln(x)$$

Notice that log() in R is natural log, so I just use ln() in the formula. Now consider two scenario, between which x was multiplied by a factor, k:

$$\hat{y} = \beta_0 + \beta_1 \ln(x)$$

$$\hat{y}' = \beta_0 + \beta_1 \ln(kx)$$

Subtracting the two:

$$\hat{y}' - \hat{y} = \beta_1 (\ln(kx) - \ln(x))$$

Because log(a) - log(b) = log(a/b):

$$\hat{y}' - \hat{y} = \beta_1 \ln(kx/x)$$

$$\hat{y}' - \hat{y} = \beta_1 \ln(k)$$

So, the interpretation is then a k-time change in x is associated with $\beta_1 * ln(k)$ unit difference in mean of y.

Let's say if the regression is:

$$Growth = 2.831 + 0.0120 \ln(EducationInYear)$$

A 10% increase in years of education is associated with 0.0114 ($0.12 * \ln(1.10)$) unit increase in mean growth.

A 10% decrease in years of education is associated with 0.0126 ($0.12 * \ln(0.90)$) unit decrease in mean growth.

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Onto another point, ask yourself what is the motive of logarithmic transformation, with more than 14,000 cases, mild to even moderate skewness in predictors seldom matters. Do run them with and without transformation and see if the outcomes differ at all.

Other techniques such as robust regression would allow you to keep the original functional form while alleviate the outlier-like effects of the extreme value in the skewed predictors.

Adding one and then transform is, to me, a bad idea because time to time, it's that jumping from 0 to 1 that matters. Assuming them all to be non-zero does not sit well with me conceptually. Another usual phenomenon is that when we need to add a small constant to the variable, there is usually a big column of zeros at the start of the histogram; even with the small constant added, transformation is unlikely to do any better.