Question

i neeed help for my qz2

QZ2: Linear Regression (100 points) Given: A data table (see next page) Find: 1. Use one-dimensional array to find the equation of least-squares-fit line (slope and y-intercept; 10 points each), correlation coefficient and coefficient of the determination (10 points each) 2. Use two-dimensional array to find the equation of least-squares-fit line (slope and y-intercept; 10 points each), correlation coefficient and coefficient of the determination (10 points each) 3. Is using linear model to present these data appropriately? Why? (10 points) Requirements: a) As always, you need to submit your QZ2-VB code using a flash drive. b) Your VB code needs to be able to handle ANY nine-data table! c) You need to output the slope, y-intercept, correlation coefficient and coefficient of the determination for both using 1-D array and 2-D array and the answer of part 3 (see sample output on page 4). Note: The needed formulas are shown on page 3.

this one is visual basic , i dont understand how to write code

Answer 1

Module Module1

Sub Main()

Console.WriteLine("============= Quiz2: Linear Regression =============")
Console.WriteLine("++++++++++++++++++++++++++++++++++++++++++++++++++++")
Console.WriteLine("")

' write your name here ,
Console.WriteLine(" written by , March, 2020")
Console.WriteLine("")
Console.WriteLine("++++++++++++++++++++++++++++++++++++++++++++++++++++")
Console.WriteLine("====================================================")
Console.WriteLine("")
Console.WriteLine("")
Console.WriteLine("")
Console.WriteLine("")

' using 1-D array

' X array
Dim X() As Integer = {0, 5, 10, 15, 20, 25, 30, 35, 40}

' Y array
Dim Y() As Integer = {24, 33, 62, 77, 105, 123, 151, 170, 188}

' local variables
Dim m As Double = 0

Dim n As Integer = X.Length
Dim XY_s As Integer = 0
Dim X_s As Integer = 0
Dim Y_s As Integer = 0
Dim X2_s As Integer = 0
Dim Y2_s As Integer = 0
' calculate
For index As Integer = 0 To n - 1
XY_s += X(index) * Y(index)
X_s += X(index)
Y_s += Y(index)
X2_s += X(index) * X(index)
Y2_s += Y(index) * Y(index)
Next

m = (n * XY_s - X_s * Y_s) / (n * X2_s - X_s * X_s)

Dim b As Double = 0
b = (Y_s - m * X_s) / n

Dim r As Double = 0

r = (n * XY_s - X_s * Y_s) / (Math.Sqrt(n * X2_s - X_s * X_s) * Math.Sqrt(n * Y2_s - Y_s * Y_s))

' display

Console.WriteLine(" Output the best fit line: using 1-D array")
Console.WriteLine("====================================================")
Console.Write(" Slope: ")
Console.WriteLine(m)
Console.Write(" Y-Intercept: ")
Console.WriteLine(b)
Console.Write(" Correlation Coefficient: ")
Console.WriteLine(r)
Console.Write(" Correlation of Determination: ")
Console.WriteLine((r * r))
Console.WriteLine("====================================================")

' using 2-D array

Dim Ary2D(,) As Integer = {{0, 24}, {5, 33}, {10, 62}, {15, 77}, {20, 105}, {25, 123}, {30, 151}, {35, 170}, {40, 188}}

' local variables

Dim m2D As Double = 0

' upper dimension od 2D array
Dim n2D As Integer = Ary2D.GetUpperBound(0)

Dim XY2D_s As Integer = 0
Dim X2D_s As Integer = 0
Dim Y2D_s As Integer = 0
Dim X2_2D_s As Integer = 0
Dim Y2_2D_s As Integer = 0

' calculation

For index As Integer = 0 To n2D - 1
XY2D_s += Ary2D(index, 0) * Ary2D(index, 1)
X2D_s += Ary2D(index, 0)
Y2D_s += Ary2D(index, 1)
X2_2D_s += Ary2D(index, 0) * Ary2D(index, 0)
Y2_2D_s += Ary2D(index, 1) * Ary2D(index, 1)
Next

m2D = (n2D * XY2D_s - X2D_s * Y2D_s) / (n2D * X2_2D_s - X2D_s * X2D_s)

Dim b2D As Double = 0
b2D = (Y2D_s - m2D * X2D_s) / n2D

Dim r2D As Double = 0

r2D = (n2D * XY2D_s - X2D_s * Y2D_s) / (Math.Sqrt(n2D * X2_2D_s - X2D_s * X2D_s) * Math.Sqrt(n2D * Y2_2D_s - Y2D_s * Y2D_s))

' display

Console.WriteLine("")
Console.WriteLine("")
Console.WriteLine(" Output the best fit line: using 2-D array")
Console.WriteLine("====================================================")
Console.Write(" Slope: ")
Console.WriteLine(m2D)
Console.Write(" Y-Intercept: ")
Console.WriteLine(b2D)
Console.Write(" Correlation Coefficient: ")
Console.WriteLine(r2D)
Console.Write(" Correlation of Determination: ")
Console.WriteLine((r2D * r2D))
Console.WriteLine("====================================================")

' r and r2D both greater than .5 , that mean near to 1

If r >= 0.5 And r2D >= 0.5 Then

Console.WriteLine("")
Console.WriteLine("")
Console.WriteLine("====================================================")
Console.WriteLine(" Since the correlation coefficient is nearly")
Console.WriteLine(" one, using linear model to present these")
Console.WriteLine(" data is appropriate.")
Console.WriteLine("====================================================")

End If

Console.WriteLine("")
Console.WriteLine("")
Console.WriteLine(" Application ends.")

Console.ReadKey()

End Sub

End Module

============= Quiz2: Linear Regression ============= ++++++++++++++++++++++++++++++++++++++++++++++++++++ written by, March, 2020 ++++++++++++++++++++++++++++++++++++++++++++++++++++ ==================================================== Output the best fit line: using 1-D array ======= =========================================== Slope: 4.30333333333333 Y-Intercept: 17.6 Correlation coefficient: 0.997525536212337 Correlation of Determination: 0.995057195395711 ==================================================== Output the best fit line: using 2-D array ======== =========================================== Slope: 4.34047619047619 Y-Intercept: 17.1666666666667 Correlation coefficient: 0.996651294978394, Correlation of Determination: 0.99331380378211 ==================================================== ==================================================== Since the correlation coefficient is nearly one, using linear model to present these data is appropriate. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Application ends.

============= Quiz2: Linear Regression ============= ++++++++++++++++++++++++++++++++++++++++++++++++++++ written by, March, 2020 ++++++++++++++++++++++++++++++++++++++++++++++++++++ ==================================================== Output the best fit line: using 1-D array ======= =========================================== Slope: 4.30333333333333 Y-Intercept: 17.6 Correlation coefficient: 0.997525536212337 Correlation of Determination: 0.995057195395711 ==================================================== Output the best fit line: using 2-D array ======== =========================================== Slope: 4.34047619047619 Y-Intercept: 17.1666666666667 Correlation coefficient: 0.996651294978394, Correlation of Determination: 0.99331380378211 ==================================================== ==================================================== Since the correlation coefficient is nearly one, using linear model to present these data is appropriate. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Application ends.