Linear Least-Squares Analysis: Determining the Best Fit Line through Minimizing Deviations, Study notes of Chemistry

The linear least-squares analysis method used to determine the best linear fit to a set of data points. The method minimizes the squares of the deviations of individual data points from the line. The mathematical equations to calculate the slope (m), y-intercept (b), standard deviation of residuals (sr), and standard deviation of slope (sm) of the line. An example is given using data from a chemistry laboratory.

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Appendix B
Linear Least-Squares Analysis
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Appendix B

Linear Least-Squares Analysis

Linear Least-Squares Analysis^1 In many of the Chem 145 laboratories, we will be confronted with an analysis of our data where a linear relationship will be theoretically predicted to exist between two variables (y and x) such that:

y = b + mx

In the above expression, b is known as the y-intercept and m is known as the slope of the line. The method of linear lest-squares analysis allows us to determine from an experimental measure of y and x what the best linear fit to the data is. This method is based on the idea that the best linear fit to the data will be one in which the squares of the deviations for the individual points relative to the line are minimized (sounds reasonable!). We define the square of the deviation (Qi ) as:

Qi = { y i − ( a + bx i )}^2

where xi , yi represent an individual data point. To determine the least-squares fit, we define the following quantities:

A^2 = ∑ xi^2 − (∑^ xi )

2 n

B^2 = ∑ yi^2 − (∑^ yi )

2 n

AB = ∑ xi yi − ∑^ x^ i^ ∑ yin

Where xi , yi represent the individual pairs of xy points. The quantity n is the number of pairs of data used in determination of the line. In actuality, A 2 and B^2 are simply the sum of the squares of the deviations from the means of the individual values for x and y; however, the more mathematically convenient (if less obvious) expressions given above are generally used.

Given the above definitions, we define four quantities of interest as follows:

m = 2.39669 /1.14537 = 2. b = 12.51 / 5 − 2.0925(5.365 / 5) = 0.

sr = 5.07748^ −^ (2.0925)

s (^) m = 0.144 / 1.14537 = 0.

A plot of this data (using Kalidagraph, Excel, or another spreadsheet program) yields:

Note that the slope and y-intercept parameters we determined by hand are in exact agreement with those given by the spreadsheet (Kalidagraph in this case).


  1. “Fundamental of Analytical Chemistry,” D. A. Skoog and D. M. West. Philadelphia: Saunders College Publishing 1982. Pages 70-72. “Introduction to Statistical Analysis,” W. J. dixon and F. J. Massey, Jr,New York: McGraw-Hill 1969.

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0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.

Peak Area

Mole Percent

Y = M0 + M1*X M0 0. M1 2. R 0.