What would you like to understand?

The y-deviations are calculated by subtracting the sample mean of the y-values, $\bar{y}=4$, from each individual y observation in the table. This produces a column of signed deviations measuring how far each point lies above or below the horizontal mean line.

Conditions: The y-values in the table are 2, 4, 5, 4, and 5.; The sample mean $\bar{y}$ has been computed as 4.

The y-deviations are calculated by subtracting the sample mean of the y-values, $\bar{y}=4$, from each individual y observation in the table. This produces a column of signed deviations measuring how far each point lies above or below the horizontal mean line.

Conditions: The y-values in the table are 2, 4, 5, 4, and 5.; The sample mean $\bar{y}$ has been computed as 4.

The intercept $b_0$ is calculated by substituting the mean point $(\bar{x}, \bar{y})$ and the known slope $b_1$ into the regression equation $\hat{y} = b_0 + b_1x$, and then solving algebraically for $b_0$. In the example, substituting $x=3$, $\hat{y}=4$, and $b_1=0.6$ yields $4 = b_0 + 0.6(3)$, which simplifies to $b_0 = 2.2$.

Conditions: The slope $b_1$ has already been computed.; The sample means $\bar{x}$ and $\bar{y}$ are known.; The regression line is assumed to pass through $(\bar{x}, \bar{y})$.

The intercept $b_0$ is calculated by substituting the mean point $(\bar{x}, \bar{y})$ and the known slope $b_1$ into the regression equation $\hat{y} = b_0 + b_1x$, and then solving algebraically for $b_0$. In the example, substituting $x=3$, $\hat{y}=4$, and $b_1=0.6$ yields $4 = b_0 + 0.6(3)$, which simplifies to $b_0 = 2.2$.

Conditions: The slope $b_1$ has already been computed.; The sample means $\bar{x}$ and $\bar{y}$ are known.; The regression line is assumed to pass through $(\bar{x}, \bar{y})$.