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Using the 8 non-shaded values above, find a_0 and a_1 for the least squares line

ID: 3677690 • Letter: U

Question

Using the 8 non-shaded values above, find a_0 and a_1 for the least squares linear regression. We will save the shaded values for our test data, that is, data points that are known but we will not include in the information used to make a representative curve. We will use these points to sec how close our curve fit is to predicting actual values that were not used to derive the curve. Compute the overall squared-error. Write the completed polynomial. Using the 8 non-shaded values from part A, find a_0, a_1, and a_2 for a parabolic least squares regression (polynomial of degree 2). Use MS Excel to solve for these coefficients. Compute the overall squared-error. Write the completed polynomial. Include a printout of your Excel spreadsheet. On two separate graphs, plot the non-shaded data points and show the resulting curves from Part A and Part B; a separate graph for each curve. Use graph paper. Fill in the following test table: which method(s) performed the best? Would you have expected the outcomes? How do these perform for these data points vs. the linear and parabolic curve's squared errors? Discuss your answer.

Explanation / Answer

PART-A

n = 8                                       ( x) 2 = 46225

x = 215                                mean of x = 26.875                            

y = 277                                mean of y = 34.625

x y = 59555          x y = 8505

x2 = 7625

a1 = n xy - x y

           n x2 - ( x) 2

a1 = 8(8505) – (215)(277)

               8(7625) – 46225

a1 = 0.5743

a0 = mean of y – (a1)(mean of x)

a0 = 34.625 – (0.5743)(26.875)

a0 = 19.1912

Least squares linear regression:

Y = 19.1912 + 0.5743X

Overall Squared Error = e 2 = (y – ao – a1x)2

= (yactual – y model)2

= (17 – 19.1912 – 0.5743 (5))2 + ...+ (46 – 19.1912 – 0.5743(50))2

Overall Squared Error = 60.7783

PART-B:-

MS Excel worksheet

8

215

7625

215

7625

306125

M

7625

306125

13113125

1.41605314

-0.1094736

0.00173225

-0.1094736

0.01055302

-0.0001827

M-1

0.00173225

-0.0001827

3.3342E-06

277

8505

N

318825

13.4580689

1.17896251

M-1 x N

-0.0110349

Least squares parabolic regression:

Y = 13.4581 + 1.1790X – 0.0110X2

Overall Squared Error = e 2 = (y – ao – a1x – a2x2)2

= (yactual – y model)2

                                    = (17 – 13.4581 – 1.1790(5) + 0.0110(5)2)2 + . . . + (46 –13.4581 – 1.1790(50) + 0.0110(50)2)2

Overall Squared Error = 24.2566

13.4580689

1.17896251

M-1 x N

-0.0110349


Least squares parabolic regression:

Y = 13.4581 + 1.1790X – 0.0110X2

Overall Squared Error = e 2 = (y – ao – a1x – a2x2)2

                                                                                                = (yactual – y model)2

                                    = (17 – 13.4581 – 1.1790(5) + 0.0110(5)2)2 + . . . + (46 –13.4581 – 1.1790(50) + 0.0110(50)2)2

Overall Squared Error = 24.2566

MS Excel worksheet

8

215

7625

215

7625

306125

M

7625

306125

13113125

1.41605314

-0.1094736

0.00173225

-0.1094736

0.01055302

-0.0001827

M-1

0.00173225

-0.0001827

3.3342E-06

277

8505

N

318825

13.4580689

1.17896251

M-1 x N

-0.0110349


Least squares parabolic regression:

Y = 13.4581 + 1.1790X – 0.0110X2

Overall Squared Error = e 2 = (y – ao – a1x – a2x2)2

                                                                                                = (yactual – y model)2

                                    = (17 – 13.4581 – 1.1790(5) + 0.0110(5)2)2 + . . . + (46 –13.4581 – 1.1790(50) + 0.0110(50)2)2

Overall Squared Error = 24.2566

MS Excel worksheet

8

215

7625

215

7625

306125

M

7625

306125

13113125

1.41605314

-0.1094736

0.00173225

-0.1094736

0.01055302

-0.0001827

M-1

0.00173225

-0.0001827

3.3342E-06

277

8505

N

318825

13.4580689

1.17896251

M-1 x N

-0.0110349


Least squares parabolic regression:

Y = 13.4581 + 1.1790X – 0.0110X2

Overall Squared Error = e 2 = (y – ao – a1x – a2x2)2

= (yactual – y model)2

= (17 – 13.4581 – 1.1790(5) + 0.0110(5)2)2 + . . . + (46 –13.4581 – 1.1790(50) + 0.0110(50)2)2

Overall Squared Error = 24.2566

MS Excel worksheet

8

215

7625

215

7625

306125

M

7625

306125

13113125

1.41605314

-0.1094736

0.00173225

-0.1094736

0.01055302

-0.0001827

M-1

0.00173225

-0.0001827

3.3342E-06

277

8505

N

318825

13.4580689

1.17896251

M-1 x N

-0.0110349


Least squares parabolic regression:

Y = 13.4581 + 1.1790X – 0.0110X2

Overall Squared Error = e 2 = (y – ao – a1x – a2x2)2

                                                                                                = (yactual – y model)2

                                    = (17 – 13.4581 – 1.1790(5) + 0.0110(5)2)2 + . . . + (46 –13.4581 – 1.1790(50) + 0.0110(50)2)2

Overall Squared Error = 24.2566

MS Excel worksheet

8

215

7625

215

7625

306125

M

7625

306125

13113125

1.41605314

-0.1094736

0.00173225

-0.1094736

0.01055302

-0.0001827

M-1

0.00173225

-0.0001827

3.3342E-06

277

8505

N

318825

13.4580689

1.17896251

M-1 x N

-0.0110349


Least squares parabolic regression:

Y = 13.4581 + 1.1790X – 0.0110X2

Overall Squared Error = e 2 = (y – ao – a1x – a2x2)2

                                                                                                = (yactual – y model)2

                                    = (17 – 13.4581 – 1.1790(5) + 0.0110(5)2)2 + . . . + (46 –13.4581 – 1.1790(50) + 0.0110(50)2)2

Overall Squared Error = 24.2566

MS Excel worksheet

8

215

7625

215

7625

306125

M

7625

306125

13113125

1.41605314

-0.1094736

0.00173225

-0.1094736

0.01055302

-0.0001827

M-1

0.00173225

-0.0001827

3.3342E-06

277

8505

N

318825

13.4580689

1.17896251

M-1 x N

-0.0110349


Least squares parabolic regression:

Y = 13.4581 + 1.1790X – 0.0110X2

Overall Squared Error = e 2 = (y – ao – a1x – a2x2)2

                                                                                                = (yactual – y model)2

                                    = (17 – 13.4581 – 1.1790(5) + 0.0110(5)2)2 + . . . + (46 –13.4581 – 1.1790(50) + 0.0110(50)2)2

Overall Squared Error = 24.2566

MS Excel worksheet

8

215

7625

215

7625

306125

M

7625

306125

13113125

1.41605314

-0.1094736

0.00173225

-0.1094736

0.01055302

-0.0001827

M-1

0.00173225

-0.0001827

3.3342E-06

277

8505

N

318825

13.4580689

1.17896251

M-1 x N

-0.0110349


Least squares parabolic regression:

Y = 13.4581 + 1.1790X – 0.0110X2

Overall Squared Error = e 2 = (y – ao – a1x – a2x2)2

                                                                                                = (yactual – y model)2

                                    = (17 – 13.4581 – 1.1790(5) + 0.0110(5)2)2 + . . . + (46 –13.4581 – 1.1790(50) + 0.0110(50)2)2

Overall Squared Error = 24.2566

MS Excel worksheet

8

215

7625

215

7625

306125

M

7625

306125

13113125

1.41605314

-0.1094736

0.00173225

-0.1094736

0.01055302

-0.0001827

M-1

0.00173225

-0.0001827

3.3342E-06

277

8505

N

318825

13.4580689

1.17896251

M-1 x N

-0.0110349


Least squares parabolic regression:

Y = 13.4581 + 1.1790X – 0.0110X2

Overall Squared Error = e 2 = (y – ao – a1x – a2x2)2

                                                                                                = (yactual – y model)2

                                    = (17 – 13.4581 – 1.1790(5) + 0.0110(5)2)2 + . . . + (46 –13.4581 – 1.1790(50) + 0.0110(50)2)2

Overall Squared Error = 24.2566

PART-D:-

Y values

Absolute Error:

|Actual - Predicted|

Results

X

linear interpolation

linear fit

parabolic fit

linear interpolation

linear fit

Parabolic fit

actual

best value

best method

24

38.5

34.68

36.64

3.5

4.32

2.36

37

36.64

Parabolic

44

45.5

46.16

47.02

2.5

2.16

2.02

44

46.16

Linear

  

The parabolic fit (regression) was the best method for (X = 24). The linear fit was the best method for (X = 44). Based on the graph which appeared to provide a better ‘fit’ to the data and overall squared error, which was less for the parabolic regression, the expectation would be that the parabolic regression would have the best fit to the actual data. But that was only true for the one set of data (X = 24). Perhaps other ways to obtain or access the error, besides absolute error (which just took the difference), are needed to determine which regression provides the best fit for all the data.

Part-E:-

Private Sub GoExcel_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles GoExcel.Click

        '------------------------------------------------------

        'The Setup - always the same

        Dim oXL As Excel.Application

        Dim oWB As Excel.Workbook

        Dim oSheet As Excel.Worksheet

        oXL = CreateObject("Excel.Application")

        oXL.Visible = True 'this makes it so they can see what's happening

        oWB = oXL.Workbooks.Add

        oSheet = oWB.ActiveSheet

        '-------------------------------------------------------

        'SOLVING MX=N

        'now you have a sheet open

        'ENTER MATRIX M

        oSheet.Cells(2, "A") = "=8"         'row, column -- refers to cell A2

        oSheet.Cells(3, "A") = "=215"       'row, column -- refers to cell A3

        oSheet.Cells(4, "A") = "=7625"      'row, column -- refers to cell A4

        oSheet.Cells(2, "B") = "=215"       'row, column -- refers to cell B2

        oSheet.Cells(3, "B") = "=7625"      'row, column -- refers to cell B3

        oSheet.Cells(4, "B") = "=306125"    'row, column -- refers to cell B4

        oSheet.Cells(2, "C") = "=7625"      'row, column -- refers to cell C2

        oSheet.Cells(3, "C") = "=306125"    'row, column -- refers to cell C3

        oSheet.Cells(4, "C") = "=13113125" 'row, column -- refers to cell C4

        'matrix has been entered in A2..C4

        'ENTER MATRIX N

        oSheet.Cells(2, "E") = "=277"

        oSheet.Cells(3, "E") = "=8505"

        oSheet.Cells(4, "E") = "=318825"

        'matrix has been entered in E2..E4

        'now put the inverse command in a cell

        'Since this has to be entered as an array formula

        'follow the directions in MS Excel

        '1. Put the formula for the inverse =MINVERSE(A2:C4) in the upper       

            ‘left cell A6

        '2. Select a range (A6:C8) in this example

        '3. Press F2

        '4. Press CTRL+SHIFT+ENTER

  

'1

        oSheet.Cells(6, "A") = "=MINVERSE(A2:C4)"

        '2

        oSheet.Range("A6", "C8").Select()

        '3 - the true tells VB .NET to wait until the keys have been

             ‘processed by MS Excel before moving on

        oXL.SendKeys("{F2}", True)

        '4: CTRL is ^ (carrot)

        '    SHIFT is + (plus)

        '    ENTER is {ENTER}

        oXL.SendKeys("^+{ENTER}", True)

        'the inverse is there in A6:C8

        'To solve MX = N, take X=(Minverse)N

        'so Matrix is A6:C8 times E2:E4 and put answer in A10:A12

        'This is also an array operation just like above

        'so we must follow steps 1 - 4 for MMULT

        '1

        oSheet.Cells("10", "A") = "=MMULT(A6:C8, E2:E4)"

        '2

        oSheet.Range("A10", "A12").Select()

        '3 - the true tells VB .NET to wait until the keys have been

              ‘processed by MS Excel before moving on

        oXL.SendKeys("{F2}", True)

        '4: CTRL is ^ (carrot)

        '    SHIFT is + (plus)

        '    ENTER is {ENTER}

        oXL.SendKeys("^+{ENTER}", True)

        'the solution matrix X is now in A10:A12

        'now save the Excel File

        'NOTE THAT .VISIBLE must be true since it uses SENDKEYS

        'othewise they could normally use false to do it in the background

        oWB.SaveAs("F:CSES141.xls")

        oXL.Quit()

    End Sub

Part-C:-Take the graph paper and plot the values it is very easy.

NOTE:-The above values are approximate values so please calculate once again the values according to the given formulas but the procedure is same.

8

215

7625

215

7625

306125

M

7625

306125

13113125

1.41605314

-0.1094736

0.00173225

-0.1094736

0.01055302

-0.0001827

M-1

0.00173225

-0.0001827

3.3342E-06

277

8505

N

318825

13.4580689

1.17896251

M-1 x N

-0.0110349

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