# VB Simple Linear Regression Example

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```Imports System
Imports System.IO
Imports System.Collections

Imports CenterSpace.NMath.Core

Namespace CenterSpace.NMath.Core.Examples.VisualBasic

' A .NET example in Visual Basic showing how to use the linear regression class to perform a simple
' linear regression.

' A simple linear regression fits a straight line through a series of data
' points. Specifically, the points <c>(x,y)</c> are assumed to conform to
' the equation <c>y = mx + b</c>. A simple linear regression uses a least
' squares fit to compute the slope, <c>m</c>, and the y-intercept, <c>b</c>.

Module SimpleLinearRegressionExample

Sub Main()

' Read in data from the file. The data comes from The Data and Story
' Library (http:'lib.stat.cmu.edu/DASL) and is described below:
'
' Results of a study of gas chromatography, a technique which is used
' to detect very small amounts of a substance. Five measurements were
' taken for each of four specimens containing different amounts of the
' substance. The amounts of the substance in each specimen was determined
' before the experiment. The responses variable is the output reading
' from the gas chromatograph. The purpose of the study is to calibrate
' the chromatograph by relating the actual amounts of the substance to

' First read in the independent (or predictor) values. This is a matrix
' with one column and a row for each amounts measurement.
Dim Amounts As DoubleMatrix = New DoubleMatrix(DataStream)

' Next, read in the responses. These are the readings of the gas
' chromatograph
Dim Responses As DoubleVector = New DoubleVector(DataStream)

' Print out the amounts and responses values.
Console.WriteLine()
Console.WriteLine("amounts = " & Amounts.ToString())
Console.WriteLine("responses = " & Responses.ToString())

' Construct a linear regression. If we want our regression to calculate a
' y-intercept we must send in true for the "addIntercept" parameter (the
' third parameter in the constructor).
Dim Regression As LinearRegression = New LinearRegression(Amounts, Responses, True)

' The y-intercept is the first element of the parameter array returned by
' the regression, and the slope is the second.
Console.WriteLine("y-intercept = " & Regression.Parameters(0))
Console.WriteLine("Slope = " & Regression.Parameters(1))

' What would the model predict for the chromatograph reading for an
' input amounts of 0.75 and 1.25?
Dim Amount As DoubleVector = New DoubleVector(0.75)
Dim PredictedReading As Double = Regression.PredictedObservation(Amount)
Console.WriteLine("Predicted reading for amount " & Amount(0) & " is " & PredictedReading)
Amount(0) = 1.25
Console.WriteLine("Predicted reading for amount " & Amount(0) & " is " & PredictedReading)

' Let's look at the coefficient of determination for the model - Rsquared.
' To do this we will need to construct an Analysis Of VAriance object for
' the regression.
Dim RegressionAnova As New LinearRegressionAnova(Regression)
Console.WriteLine("Rsquare = " & RegressionAnova.RSquared)

Console.WriteLine()
Console.WriteLine("Press Enter Key")