Question

predict what the amount spent will be in 2008.
amount $ type your answer... billion

Explanation:

Step1: Find the trend

Looking at the data from 1997 - 2003, we can calculate the differences between consecutive years. For example, from 1997 ($6.3$) to 1998 ($6.9$), the increase is $6.9 - 6.3=0.6$. From 1998 ($6.9$) to 1999 ($7.9$), the increase is $7.9 - 6.9 = 1$. From 1999 ($7.9$) to 2000 ($8.6$), the increase is $8.6-7.9 = 0.7$. From 2000 ($8.6$) to 2001 ($9.0$), the increase is $9.0 - 8.6=0.4$. From 2001 ($9.0$) to 2002 ($9.6$), the increase is $9.6 - 9.0=0.6$. From 2002 ($9.6$) to 2003 ($9.9$), the increase is $9.9 - 9.6 = 0.3$.

Step2: Assume a linear trend (a simple approach for prediction)

Let \(x\) be the number of years after 1997. In 1997, \(x = 0\), in 2003, \(x=6\). We can use the least - squares regression formula for a line \(y=ax + b\). The formula for \(a=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}\), \(b=\frac{\sum_{i = 1}^{n}y_i - a\sum_{i = 1}^{n}x_i}{n}\) where \(n = 7\) (number of data points from 1997 - 2003), \(x_i\) are \(0,1,2,3,4,5,6\) and \(y_i\) are \(6.3,6.9,7.9,8.6,9.0,9.6,9.9\)

\(\sum_{i=1}^{7}x_i=0 + 1+2 + 3+4+5+6=\frac{6\times(6 + 1)}{2}=21\)

\(\sum_{i=1}^{7}y_i=6.3+6.9+7.9+8.6+9.0+9.6+9.9=58.2\)

\(\sum_{i=1}^{7}x_iy_i=0\times6.3+1\times6.9+2\times7.9+3\times8.6+4\times9.0+5\times9.6+6\times9.9\)
\(=6.9+15.8+25.8+36+48+59.4=191.9\)

\(\sum_{i=1}^{7}x_i^{2}=0^{2}+1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}=\frac{6\times(6 + 1)\times(2\times6 + 1)}{6}=91\)

\(a=\frac{7\times191.9-21\times58.2}{7\times91-21^{2}}=\frac{1343.3-1222.2}{637 - 441}=\frac{121.1}{196}\approx0.62\)

\(b=\frac{58.2-0.62\times21}{7}=\frac{58.2 - 13.02}{7}=\frac{45.18}{7}\approx6.45\)

The regression line is \(y = 0.62x+6.45\). For 2008 (\(x = 11\))

\(y=0.62\times11 + 6.45=6.82+6.45 = 13.27\)

Another simple approach: The average increase from 2003 to 2008 is 5 years. From 1997 - 2003 (6 years), the total increase is \(9.9 - 6.3=3.6\), average annual increase is \(\frac{3.6}{6}=0.6\). If we assume the average annual increase of \(0.6\) continues:

In 2003, the amount is \(9.9\). For 2008 (5 years after 2003), the amount is \(9.9+0.6\times5=9.9 + 3=12.9\)

Answer:

12.9