0
$\begingroup$

I'm investigating adherence to a special diet (that is scored from 0-18) in relation to C-reactive protein level and am in the process of building multiple linear regression models: To achieve a normal distribution, I log transformed CRP. Now I'm having issues interpreting my results:

Log(C-reactive protein) = intercept + diet score + age + gender (This is just my basic model and I will be adding variables later ...)

My R output:

Call:
lm(formula = log_CRP ~ total_score + age + gender_T0, 
    data = final_regression_table)

Residuals:
    Min      1Q  Median      3Q     Max 
-8.6813 -0.2796  0.6229  1.4016  3.3035 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)
(Intercept)   -0.40926    1.33728  -0.306    0.760
total_score   -0.01755    0.03166  -0.554    0.579
age           -0.01537    0.01816  -0.847    0.397
gender_T0male -0.08077    0.14462  -0.559    0.577

Residual standard error: 2.415 on 1133 degrees of freedom
Multiple R-squared:  0.001141,  Adjusted R-squared:  -0.001504 
F-statistic: 0.4314 on 3 and 1133 DF,  p-value: 0.7305

                      5 %       95 %
(Intercept)   -2.61068556 1.79216737
total_score   -0.06967919 0.03457092
age           -0.04526635 0.01451645
gender_T0male -0.31885146 0.15730319

Do I exponentiate the betas individually? The geometric mean is exp (-2.61) = 0.073? So can I say that a one unit increase in diet score is expected to decrease CRP by exp(-0.40926) = diet score * 0.9326 (keeping age and gender constant)?

Would the standard error also be exponentiated?

would the confidence intervals be exponentiated? so they would range from 0.93 to 1.035?? & this would mean that I cannot reject null hypothesis as CI cross 0?

$\endgroup$

1 Answer 1

0
$\begingroup$

Short answer - yes.

The logic here is that if we say ln(y)=a+b*x+u (this is simple linear regression but applies just the same to multiple linear regression) and defining $y_1$ and $y_2$ as follows:

$ln(\hat{y}_1)=a+b*x$ and $ln(\hat{y}_2)=a+b*(x+1)$ for some fixed x,

then the relative change in y is

$\hat{y}_2/\hat{y}_1=exp(a+b*x)/exp(a+b*(x+1)=exp(b)$

In your case, as you said, a unit increase in total diet score means that the level of protein is 0.98 times as high, or each unit increase in diet score reduces the level of protein by 2%.

The interesting thing you'll notice about the transformed confindance intervals is that usually, the coefficient is the midpoint of the confidence interval. For total diet score, the untransformed 95% confidence interval is:

$CI=[-0.01755-1.645*0.03166, -0.01755+1.645*0.03166]=[-0.070,0.035]$

and obviously, the coefficient -0.01755 is its midpoint.

But the transformed interval is:

$CI=[exp(-0.070),exp(0.035)]=[0.933,1.035]$

and the transformed coefficient 0.983 is not the midpoint. This also means you can't just transform the standard errors because they aren't symmetric. This is really important because the reason why the log transformation achieves normal distribution is because the original distribution is essentially:

$y=exp(a+b*x+u)=e^a*e^{b*x}*e^u$

And it's u that is normally distributed, so $e^u$ clearly isn't.

You indeed can't reject the null. You can see this from the p-value of 0.579 for the untransformed coefficient. You can also see it in the transformed CI which contains the value 1.

Hope this helps at all!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.