Remember that the equation of a straight line is y = mx + b
where m is the slope (gradient) of the line and b is the intercept on the y axis (that is, when x =0).
Microsoft excel will calculate both the slope of the line, m, and the y-intercept, b, for you.
First you will need to open up a new excel spreadsheet and set up your data.
Here we have used Row 1 of columns A and B for our headings (x values and y values).
Various values for x and y have then been entered into the cells:
cell | A | B |
1 | x values | y values |
---|---|---|
2 | 1.02 | 11.0 |
3 | 2.05 | 21.3 |
4 | 2.96 | 31.2 |
5 | 4.12 | 39.7 |
6 | 4.86 | 53.4 |
Next, we highlight any two adjacent cells (empty cells) somewhere else in the same spreadsheet, for example A8 and B8
cell | A | B |
1 | x values | y values |
---|---|---|
2 | 1.02 | 11.0 |
3 | 2.05 | 21.3 |
4 | 2.96 | 31.2 |
5 | 4.12 | 39.7 |
6 | 4.86 | 53.4 |
7 | ||
8 |
Next to the fx symbol above the cell headings A, B, etc, type in =LINEST(B2:B6, A2:A6)
then press Ctrl+Shift+Enter simultaneously.
(B2 is the first y value to be used in the calculation and B6 is the last y value to be used in the calculation. Similarly, A2 is the first of the x values to be used in the calculation and A6 is the last of the x values to be used in the calculation.)
Your spreadsheet should now look this:
cell | A | B |
1 | x values | y values |
---|---|---|
2 | 1.02 | 11.0 |
3 | 2.05 | 21.3 |
4 | 2.96 | 31.2 |
5 | 4.12 | 39.7 |
6 | 4.86 | 53.4 |
7 | ||
8 | 10.50606 | -0.21918 |
The number in cell A8 is the slope (m) of the line of best fit.
The number in cell B8 is the y-intercept (b) of the line of best fit, that is, the point with coordinates (0,b)
The equation for the line of best fit for this data is y = 10.50606x -0.21918
In order to draw the line of best fit for the experimental data above, we only need to calculate new values for y using the equation for the line of best fit (that is, the first and last points on the line of best fit).:
given x=1.02 ,then, y = (10.50606 x 1.02) -0.21918 = 10.5
and x=4.86 , then, y =(10.50606 x 4.86) -0.21918 = 50.8
cell | A | B | C |
1 | x values | y values | new y values |
---|---|---|---|
2 | 1.02 | 11.0 | 10.5 |
3 | 2.05 | 21.3 | |
4 | 2.96 | 31.2 | |
5 | 4.12 | 39.7 | |
6 | 4.86 | 53.4 | 50.8 |
7 | |||
8 | 10.50606 | -0.21918 |
Plot the points (1.02, 10.5) and (4.86, 50.8) on the graph of experimental data and draw a straight line between these two points. This is the line of best fit for the experimental data in the table.
Further Activities:
Go to http://www.ausetute.com.au/members/bestfitline.html
1/ Enter the data above into the tool for calculating the two points needed to draw a line of best fit. Check the new y values above, and see what the graph looks like.
Graph the data points in questions 2 to 5 below and draw the line of best fit:
2/ (0.23, 9.75), (0.94, 6.39), (1.77, 5.73), (4.59, 4.91), (11.52, 0.33)
3/
time / min | mass / g |
---|---|
0 | 102.8 |
2 | 73.4 |
4 | 48.7 |
6 | 27.3 |
8 | 10.9 |
4/
volume / mL | 5.0 | 10.0 | 15.0 | 20.0 | 25.0 |
mass / g | 3.2 | 7.1 | 9.4 | 13.5 | 14.8 |
5/ At time 0, the temperature of a reaction mixture was 25oC. The temperature of the reaction mixture was then recorded every minute for 4 minutes. The results of the experiment were temperatures of 29oC, 32oC, 35oC and 40 oC.
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