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"An interpretation based on coefficient magnitude also suggests that an increase of x (independent variable) by one standard deviation is associated with the increase of y (dependent variable by 2.63% of its standard deviation."

Could somebody explain to me the formula for calculating 2.63%? The coefficient of the x is 0.05 and I have other information like mean, std, and standard error

Thanks

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You have some variable $x$ in the regression. Calculate the standard deviation of $x$, which we will call $s_x$.

You have the coefficient on $x$ in the regression, which we will call $\hat\beta_x$.

To determine how much changes in $y$ is associated with a change in $x$ of $s_x$, multiply $s_x\times\hat\beta_x$.

Now we know by how much $y$ changes when $x$ changes by one standard deviation. To determine how many standard deviations of $y$ that is, divide by the standard deviation of $y$. If you then want this in terms of percent, multiply by $100\%$.

Overall, this corresponds to $\dfrac{s_x\hat\beta_x}{s_y}\times 100\%$.

Note that all of this assumes a linear model. If you have a nonlinear model or a linear model that acts on nonlinear functions of the original variables (e.g., $x$ and $x^2$ are in the model), then this does not work, as one characteristic of nonlinearity is that the rate of change (related to $\hat\beta_x$) is not constant.

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    $\begingroup$ Thank you so much. Yes, I have a linear regression. But the dependent variable (y) is in Ln or log format. Is this change the formula? @Dave $\endgroup$
    – Roxanna
    Feb 19 at 19:24
  • $\begingroup$ You would take the standard deviation of whatever you feed into the models, so if that’s a logged value, take the standard deviation of the logged $y$. However, taking a logarithm usually implies an interest in percent changes. How to consider percent standard deviation changes when the model is of percent changes in $y$ probably warrants its own posted question. $\endgroup$
    – Dave
    Feb 19 at 19:32
  • $\begingroup$ Thanks. I added a question here. I would appreciate it if you could explain it to me stats.stackexchange.com/questions/605960/… Thanks @Dave $\endgroup$
    – Roxanna
    Feb 20 at 16:51

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