To perform a logarithmic transformation on the dependent variable $y$, we need to take the natural logarithm (ln) of each $y$ value, then develop a linear regression equation using $X$ as the independent variable and $\ln(y)$ as the dependent variable.

Step 1: Logarithmic Transformation of $y$

We compute $\ln(y)$ for each value of $y$:

X & y & \ln(y) \\ \hline 2 & 4 & \ln(4) \approx 1.386 \\ 3 & 5 & \ln(5) \approx 1.609 \\ 4 & 4 & \ln(4) \approx 1.386 \\ 5 & 6 & \ln(6) \approx 1.792 \\ 7 & 4 & \ln(4) \approx 1.386 \\ 7 & 6 & \ln(6) \approx 1.792 \\ 7 & 9 & \ln(9) \approx 2.197 \\ 8 & 5 & \ln(5) \approx 1.609 \\ 9 & 11 & \ln(11) \approx 2.398 \\ \end{array}$$ ### Step 2: Perform Linear Regression Now, we need to estimate a linear regression model of the form: $$\ln(y) = a + bX$$ Using the data points $$X$$ and $$\ln(y)$$, we can compute the coefficients $$a$$ (intercept) and $$b$$ (slope) using the least-squares method. Let me calculate the regression coefficients for you.The estimated regression equation after performing the logarithmic transformation on $$y$$ is: $$\ln(y) = 1.182 + 0.095X$$ This equation represents the relationship between $$X$$ and the natural logarithm of $$y$$, where the intercept is approximately 1.182 and the slope is 0.095. Would you like more details on this process or have any questions? Here are 5 related questions to deepen understanding: 1. How does a logarithmic transformation help in regression analysis? 2. What are the assumptions of linear regression? 3. How can we interpret the slope and intercept in this transformed model? 4. Can you revert the logarithmic transformation to predict actual $$y$$ values? 5. What are the advantages of transforming the dependent variable in non-linear data? **Tip:** When interpreting coefficients in a log-transformed model, a one-unit increase in $$X$$ corresponds to approximately a percentage change in $$y$$, not a simple linear increase.