if the number of bankruptcies (inthoussands) fot the years 1991 through 2008 wer
ID: 2868409 • Letter: I
Question
if the number of bankruptcies (inthoussands) fot the years 1991 through 2008 were tp ne as follows:
1. A use the regression capabilities of a graphing utility to find a model of form b(t)=at^4+bt^3+bt^2+bt+e forthe data
(Let T= 1represemt 1991)
2. use a graphing utility to plot the data and graph the model.
3.Analytically find the minimum of the model and compare the result with the actual data.
1991 345.8
1992 347.9
1993 354.7
1994 328.9
1995 344.1
1996 435.5
1997 500.8
1998 588.3
1999 633.2
2001 868.9
2002 950.2
2003 901.1
2004 837.5
2005 860.1
2006 1050.2
2007 1322.8
2008 1411.6
Explanation / Answer
1) Regression :
I used this link to find the 4th degree polynomial regression equation
I entered in coordinates like this.....
1 345.8
2 347.9
3 354.7
4 328.9
5 344.1
6 435.5
7 500.8
8 588.3
9 633.2
11 868.9
12 950.2
13 901.1
14 837.5
15 860.1
16 1050.2
17 1322.8
18 1411.6
And obtained the equation :
y = 0.1225t^4 - 4.6265t^3 + 59.4404t^2 - 237.9551t + 581.5525 ---> FIRST ANSWER
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2) This needs a graphing utility. You have to simply plot this graph and then plot the individual points given in the table to measure the accuracy of the regression equation
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3)
y = 0.1225t^4 - 4.6265t^3 + 59.4404t^2 - 237.9551t + 581.5525
Deriving :
dy/dt = 0.49t^3 - 13.8795t^2 + 118.881t - 237.955 = 0
When solved, this gives us t = 2.86042 as the only real solution
Now, to confirm that it indeed is a minimum, we find the second derivative
y'' = 1.47t^2 - 27.759t + 118.881
Plug in 2.86 --> y'' = 1.47(2.86)^2 - 27.759(2.86) + 118.881 --> 51.514272 --> positive
Since the second derivative is positive, we know that t = 2.86 constitutes a minimum
Now, check form the table and confirm that the minimum occurs around t = 2.86
t = 2.86 indicates the year 1993 approx
But from the table, the minimum occurs at the year 1994.
So, the regression was not very accurate
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