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The slope $b_1$ is computed by dividing the sum of the cross-products of deviations by the sum of the squared x-deviations. In the worked example, the sum of the cross-products is 6 and the sum of the squared x-deviations is 10, yielding $b_1 = 6/10 = 0.6$.

Conditions: The data set consists of paired observations.; The sample means $\bar{x}$ and $\bar{y}$ have been computed.; The denominator $\sum (x_i-\bar{x})^2$ is nonzero.

The slope $b_1$ is computed by dividing the sum of the cross-products of deviations by the sum of the squared x-deviations. In the worked example, the sum of the cross-products is 6 and the sum of the squared x-deviations is 10, yielding $b_1 = 6/10 = 0.6$.

Conditions: The data set consists of paired observations.; The sample means $\bar{x}$ and $\bar{y}$ have been computed.; The denominator $\sum (x_i-\bar{x})^2$ is nonzero.

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.