What if $SS_{xx} = 0$?

This occurs if all $x$ values in your dataset are identical. A linear regression line cannot be calculated because the slope formula divides by zero.

Can I use standard deviations and the correlation coefficient to find the line?

Yes. If those summary statistics are provided instead of raw data, calculate the slope as $b_1 = r \frac{s_y}{s_x}$ and the intercept as $b_0 = \bar{y} - b_1\bar{x}$.

Does swapping $x$ and $y$ simply invert the slope?

No. The regression of $x$ on $y$ yields a different line because it minimizes horizontal residuals instead of vertical residuals. The slopes are only strict inverses if the correlation is exactly 1 or -1.

How to calculate a linear regression line by hand

The linear regression line, $\hat{y} = b_0 + b_1x$ , is calculated by minimizing the sum of squared vertical residuals between observed data and the line. This method applies when you have a set of bivariate numerical data $(x, y)$ and need to model the linear relationship without statistical software.

The setup

Define your data pairs as $(x_i, y_i)$ and count the number of pairs $n$ . Compute the five fundamental sums: $\Sigma x$ , $\Sigma y$ , $\Sigma x^2$ , $\Sigma y^2$ , and $\Sigma xy$ . Calculate the sample means $\bar{x} = \frac{\Sigma x}{n}$ and $\bar{y} = \frac{\Sigma y}{n}$ .

The steps

Calculate the sums of squares for $x$ and $xy$ : $SS_{xx} = \Sigma x^2 - \frac{(\Sigma x)^2}{n}$ and $SS_{xy} = \Sigma xy - \frac{(\Sigma x)(\Sigma y)}{n}$ . 2. Calculate the slope $b_1 = \frac{SS_{xy}}{SS_{xx}}$ . 3. Calculate the y-intercept $b_0 = \bar{y} - b_1\bar{x}$ . 4. Write the final equation as $\hat{y} = b_0 + b_1x$ .

Checking the result

Verify that the point $(\bar{x}, \bar{y})$ lies exactly on your calculated line by testing if $\bar{y} = b_0 + b_1\bar{x}$ . Additionally, verify that the sign of your slope $b_1$ matches the sign of $SS_{xy}$ .

Common errors

Rounding intermediate calculations, especially the means or sums of squares, causes severe compounding errors in the final slope and intercept. Another common error is swapping the independent variable $x$ and dependent variable $y$ , which produces an entirely different regression line.

Worked example

Calculate the linear regression line for the following five data points: $(1, 2), (2, 4), (3, 5), (4, 4), (5, 5)$ .

First, find $n = 5$ . Compute the sums: $\Sigma x = 1 + 2 + 3 + 4 + 5 = 15$ . $\Sigma y = 2 + 4 + 5 + 4 + 5 = 20$ . $\Sigma x^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$ . $\Sigma xy = (1)(2) + (2)(4) + (3)(5) + (4)(4) + (5)(5) = 2 + 8 + 15 + 16 + 25 = 66$ . Compute the means: $\bar{x} = 15 / 5 = 3$ . $\bar{y} = 20 / 5 = 4$ . Next, calculate the sums of squares: $SS_{xx} = 55 - \frac{15^2}{5} = 55 - 45 = 10$ . $SS_{xy} = 66 - \frac{(15)(20)}{5} = 66 - 60 = 6$ . Calculate the slope: $b_1 = \frac{6}{10} = 0.6$ . Calculate the intercept: $b_0 = 4 - (0.6)(3) = 4 - 1.8 = 2.2$ . The final regression line is $\hat{y} = 2.2 + 0.6x$ .

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References: OpenStax Introductory Statistics, Chapter 12: Linear Regression and Correlation · Moore, McCabe, Craig: Introduction to the Practice of Statistics