Not quite.

Your interpretation for the first model is correct, but your explanation isn't quite right. Note that
$$ \begin{equation*} \beta_1 = \frac{\partial \log y}{\partial x}, \end{equation*}$$

but that isn't very easy to interpret. So, we recall the calculus result that
$$ \begin{equation*} \frac{\partial \log y}{\partial y} = \frac{1}{y} \end{equation*}$$
or
$$ \begin{equation*} \partial \log y = \frac{\partial y}{y}. \end{equation*}$$

Plugging this into the equation for $\beta_1$, we have

$$ \begin{equation*} \beta_1 = \frac{\partial y / y}{\partial x}. \end{equation*}$$

If we multiply both sides by 100, we have

$$ \begin{equation*} 100\beta_1 = \frac{100 \times \partial y / y}{\partial x}. \end{equation*}$$

We realize that $100 \times \partial y/y$ is just the percentage change in $y$, giving the interpretation that $100 \beta_1$ is the percentage change in the outcome for a one unit increase in $x$.

The correct interpretation for your second model would be that a 1 unit increase in GDP leads to a 10 **percentage point** increase in sales. It's easiest to understand this by thinking of your outcome as being measured in percentage points, rather than percent. Then, a 1 unit change in $x$ leads to a $\beta_1$ unit change in $y$, just as we normally get.

This is an important distinction. An increase in sales from 1% to 5% is a $5 - 1 = 4$ percentage point increase, but a $(5 - 1)/1 \times 100 =400$ percent change.

Consider the simple regression model $y=\beta_0 + \beta_1x + u$ where $u \sim N(0,1)$ is iid.

# Rescaling $x$ only

- $\beta_1$
*will* change if you change the units of $x$ only. For
example, if $x$ is the dose of a drug in mg. You might want to
express the $\beta_1$ as the unit change in $y$ for a 50mg change in
dose. Then you multiply $x*50$ but that's just the same thing as
dividing $\beta_1$ by 50 (vice versa). So scaling the $x$ is the same as scaling the estimated $\beta_1$ correspondingly. The interpretation of $\beta_1$ changes only because you're dealing with different units. The degree (and statistical significance) of the association with $y$ will not change.
- You'll notice if you do a substitution that rescaling $x$ in no way affects $\beta_0$.
- The $R^2$ metric will not be affected either. It's a measure of fit. It doesn't matter what units your variables are in.

# Rescaling $x$ and $y$ equivalently

- This shouldn't change the $\beta_1$ or $R^2$ but it will rescale the $\beta_0$ correspondingly.

# Examples in R

Lets just go through an example in R.

```
# generate x and y with beta_1=4 and beta_0=10 and i.i.d standard normal errors
set.seed(1)
x<-rnorm(100,30,10)
y<-10+4*x+rnorm(100,0,1)
# estimate basic regression
summary(lm(y~x))
```

Call:
lm(formula = y ~ x)

```
Residuals:
Min 1Q Median 3Q Max
-1.8768 -0.6138 -0.1395 0.5394 2.3462
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 9.96549 0.34847 28.6 <2e-16 ***
x 3.99989 0.01077 371.3 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.9628 on 98 degrees of freedom
Multiple R-squared: 0.9993, Adjusted R-squared: 0.9993
F-statistic: 1.379e+05 on 1 and 98 DF, p-value: < 2.2e-16
```

Now let's rescale only x.

```
# Now rescale only x
# notice the coefficient and standard errors scale correspondingly
# but the R^2 and t-statistics/p-values do not.
x_rescale<-x/100
summary(lm(y~x_rescale))
Call:
lm(formula = y ~ x_rescale)
Residuals:
Min 1Q Median 3Q Max
-1.8768 -0.6138 -0.1395 0.5394 2.3462
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 9.9655 0.3485 28.6 <2e-16 ***
x_rescale 399.9894 1.0773 371.3 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.9628 on 98 degrees of freedom
Multiple R-squared: 0.9993, Adjusted R-squared: 0.9993
F-statistic: 1.379e+05 on 1 and 98 DF, p-value: < 2.2e-16
```

Now we'll rescale both x and y

```
# now rescale y and run a regression where both x and y are rescaled.
y_rescale<-y/100
summary(lm(y_rescale~x_rescale))
Call:
lm(formula = y_rescale ~ x_rescale)
Residuals:
Min 1Q Median 3Q Max
-0.018768 -0.006138 -0.001395 0.005394 0.023462
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.099655 0.003485 28.6 <2e-16 ***
x_rescale 3.999894 0.010773 371.3 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.009628 on 98 degrees of freedom
Multiple R-squared: 0.9993, Adjusted R-squared: 0.9993
F-statistic: 1.379e+05 on 1 and 98 DF, p-value: < 2.2e-16
```

## Best Answer

You are almost correct to interpret the regression coefficients this way. Instead of saying "for every 1-unit change" there is a decrease, it should be: "For every 1-unit increase in currency" there is a decrease. Specifying the direction of the relationship is important for readers.

A more pedantically correct way to interpret these coefficient relationships would be:

"For every one unit increase in currency, the expected y decreases by 420 kgs. For every one unit decrease in currency, the expected y increases by 420 kgs."