Look first at the differences between the data. Call these D1, D2, etc. No need for $ sign. All figures in cents.
D1: 31, 23, 45, 109, 43, 75, 27, -11, -38, -108
D2: -8, 22, -36, 34, 32, -48, -38, -27, -70
D3: 30, -58, 70, -2, -80, 10, 11, -43
D4: -88, 128, -72, -78, 90, 1, -54
No pattern, so perhaps statistical analysis will help. Linear regression, perhaps.
Mean is $2.892 for money data (Y). Mean is 2010 for other data (X).
A table will help:
X |
Y |
XY |
X^2 |
2005 |
2.27 |
4551.35 |
4020025 |
2006 |
2.58 |
5175.48 |
4024036 |
2007 |
2.81 |
5639.67 |
4028049 |
2008 |
3.26 |
6546.08 |
4032064 |
2009 |
2.35 |
4721.15 |
4036081 |
2010 |
2.78 |
5587.80 |
4040100 |
2011 |
3.53 |
7098.83 |
4044121 |
2012 |
3.60 |
7243.20 |
4048144 |
2013 |
3.49 |
7025.37 |
4052169 |
2014 |
3.11 |
6263.54 |
4056196 |
2015 |
2.03 |
4090.45 |
4060225 |
TOTAL: 22110 |
31.81 |
63942.92 |
44441210 |
From these totals the slope=(11*(3rd column)-(1st col)*(2nd col))/(11*(4th col)-(1st col)^2)=
(11*63942.92-22110*31.81)/(11*44441210-488852100)=0.04382.
Intercept=((2nd col)-slope*(1st col))/11=(31.81-0.04382*22110)/11=-85.1827.
The equation is Y=0.04382X-85.1827, which represents the best model to fit the data, assuming a linear relationship.
Using this equation we get (2005,2.67), (2006,2.72), (2007,2.76), (2008,2.80), (2009,2.85), (2010,2.89), (2011,2.94), (2012,2.98), (2013,3.02), (2014,3.07), (2015,3.11) as the pairs of (X,Y) values.
If the relation is non-linear, plotting the values may give us a clue. If we omit figures for 2009 and 2010, the graph resembles a parabola.
We can find the average slope between consecutive pairs of points, ignoring 2009 and 2010:
2005-2006: 31; 2006-2007: 23; 2007-2008: 45; 2008-2011: 27/3=9; 2011-2012: 7; 2012-2013: -11; 2013-2014: -38; 2014-2015: -108.
If this were a parabola, we would expect the negative and positive slopes to be opposites after the maximum at 2012, but this doesn't appear to be the case. However, this points to a maximum for the range at $3.60. The domain goes from 2005 to 2015 and we don't have a satisfactory model for the figures.