Tema17 - 17.19 Heres VBA code to implement linear...

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17.19 Here’s VBA code to implement linear regression: Option Explicit Sub Regres() Dim n As Integer Dim x(20) As Single, y(20) As Single, a1 As Single, a0 As Single Dim syx As Single, r2 As Single n = 7 x(1) = 1: x(2) = 2: x(3) = 3: x(4) = 4: x(5) = 5 x(6) = 6: x(7) = 7 y(1) = 0.5: y(2) = 2.5: y(3) = 2: y(4) = 4: y(5) = 3.5 y(6) = 6: y(7) = 5.5 Call Linreg(x(), y(), n, a1, a0, syx, r2) MsgBox "slope= " & a1 MsgBox "intercept= " & a0 MsgBox "standard error= " & syx MsgBox "coefficient of determination= " & r2 MsgBox "correlation coefficient= " & Sqr(r2) End Sub Sub Linreg(x, y, n, a1, a0, syx, r2) Dim i As Integer Dim sumx As Single, sumy As Single, sumxy As Single Dim sumx2 As Single, st As Single, sr As Single Dim xm As Single, ym As Single sumx = 0 sumy = 0 sumxy = 0 sumx2 = 0 st = 0 sr = 0 For i = 1 To n sumx = sumx + x(i) sumy = sumy + y(i) sumxy = sumxy + x(i) * y(i) sumx2 = sumx2 + x(i) ^ 2 Next i xm = sumx / n ym = sumy / n a1 = (n * sumxy - sumx * sumy) / (n * sumx2 - sumx * sumx) a0 = ym - a1 * xm For i = 1 To n st = st + (y(i) - ym) ^ 2 sr = sr + (y(i) - a1 * x(i) - a0) ^ 2 Next i syx = (sr / (n - 2)) ^ 0.5 r2 = (st - sr) / st End Sub
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17.20 log N log Stress 0 3.053463 1 3.024486 2 2.996949 3 2.903633 4 2.813581 5 2.749736 6 2.630428 n =7 = 514 . 58 i i y x = 91 2 i x = 21 i x
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Tema17 - 17.19 Heres VBA code to implement linear...

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