


Visually, a trendline looks somewhat similar to a line chart, but it doesn't connect the actual data points as a line chart does. This analytical tool is most often used to show data movements over a period of time or correlation between two variables. Here are some intermediate results to check against your code.A trendline, also referred to as a line of best fit, is a straight or curved line in a chart that shows the general pattern or overall direction of the data. To get "a" we calculate e ln(a) = a.įinally, follow the equation above, to calculate Rsq. Reason: we took the log of a which results in ln(a).

Reason: we took the log of x B which results in B*ln(X). Recall that our power model is in the form y=ax Bī =. With this form, it is now possible to fit the regression line with linear regression and use the equation for R Squared.įit a linear model to the new dataset. You might think that the ln(x) part would make linear regression inappropriate, however, linear regression means that it is linear in parameters. This model can now be fit with linear regression.

The Power regression model is in the form. However, the power equation can be converted to linear. 6782.Īdditionally, Rsquared is only valid for linear regression Models. The correct equation for Rsq is Also, in the most current version of Excel, the Rsquared for your dataset is. SsReg += math.pow(((intercept * math.pow(xValue, slope)) -yMean), 2).toDouble() SumOfYSquare += math.pow(yValue, 2).toInt() SumOfXSquare += math.pow(xValue, 2).toInt() Int sumOfX = 0,sumOfY = 0,sumOfXY = 0, sumOfXSquare = 0, sumOfYSquare = 0 ĭouble slope = 0.5204, intercept = 2.3133 I have implemented the following code snippet logic to achieve the R squared value List xValue = Here ȳ is the mean value of y and ŷ is the regression equation, here ŷ forĮxponential is y = ae^xb and for power is y = ax^b Tried to calculate the R-squared value for the exponential and power trendline type using the following equation.
