using an online linear regression graphing calculator, i get:
Regression line equation: y = 2.0240963855422 * x + 0.53413654618474
the results from the use of this calculator look like this:
you can probably round this to y = 2.024 * x + .534 without any great loss of accuracy.
the rounded equation would look like this:
my inputs to the regression calculator look like this:
2,4
3,7
4,9
7,15
8,16
9,19
the equation is in the form of y = mx + b.
m is the slope
b is the y-intercept.
the slope is the average change in the value of y for each corresponding change in the value of x.
since y represents the number of pineapple plants in a garden in a given year, then the slope represents the average change in the number of pineapples in a given year divided by the average change in the number of years.
since the average change in the number of years is 1, then the slope represents the average change in the number of pineapple plants in the garden for each additional year.
the y-intercept is the value of y when the value of x is equal to 0.
since you did not have an actual point pair for when x = 0, this is just an approximation based on the regression line.
the actual value could have been 0 or it could have been 1.
a value for when x = 0 would have changed the equation of the regression line somewhat.
since the value was .534, you would probably round to 1, making the point (0,1).
the results of doing that are shown below:
the rounded formula becomes y = 1.99 * x + .754
you can see that the y-intercept became closer to 1 and the slope was changed slightly.
the line is the line of best fit.
it's not right on, but the average distance between the points on the line and the actual points is minimized.
an example is when the value of x = 3.
the rounded formula in the original graph is y = 2.024 * x + .534 y =
when x = 3, the formula becomes y = 2.024 * 3 + .534.
the resulting value of y is 6.606.
you can see that on the graph using the rounded values.
the actual point is (3,7), so the linear regression line is close but not right on.
the closer the actual points are to the line, the better the approximation.
r is the correlation coefficient.
r^2 tells you how good of a fit you have.
the r for this data is 0.996 rounded to 3 decimal places.
the r^2 is the square of this.
r^2 = .996^2 = .992 rounded to 3 decimal places.
that's a pretty good fit.
the following reference explains the difference between r and r^2.
r is called the coefficient of correlation.
r^2 is called the coefficient of determination.
http://blog.uwgb.edu/bansalg/statistics-data-analytics/linear-regression/what-is-the-difference-between-coefficient-of-determination-and-coefficient-of-correlation/