Question

linear regression - christmas trees
the following table gives the total number of live christmas trees sold, in millions, in the united states from 2004 to 2011. (source: statista.com).

t = number of years since 2004	0	2	4	6	7
c = total number of christmas trees sold in the u.s. (in millions)	27.1	28.6	28.2	27	30.8

1. use the table to determine the number of live christmas trees sold in the year 2008.

in 2008, there were approximately

1. determine the linear regression equation that models the set of data above. use the indicated variables and round to the nearest hundredth as needed.

c=

1. interpret the meaning of the slope of your (rounded) regression model.

from 2004 to 2011, the number of christmas trees sold in the u.s. increased by approximately

1. use the regression equation to determine the number of live christmas trees sold in the year 2008.

in 2008, there were approximately

1. your answers to parts 1 and 4 should be different. why is this the case?

the data are not perfectly linear. the regression equation gives a more precise answer than the value from the table.
the data are linear. the regression equation gives only an approximation of the actual value, not the exact value.
the data are not perfectly linear. the regression equation gives only an approximation of the actual value, not the exact value.
the data are linear. the regression equation gives a more precise answer than the value from the table.

Explanation:

Step1: Find years since 2004 for 2008

2008 - 2004 = 4. From the table, when \(t = 4\), \(C=28.2\).

Step2: Calculate linear - regression coefficients

Let \(t\) be the number of years since 2004 and \(C\) be the total number of Christmas trees sold. The formula for the linear regression equation is \(C=at + b\), where \(a=\frac{n\sum_{i = 1}^{n}t_{i}C_{i}-\sum_{i = 1}^{n}t_{i}\sum_{i = 1}^{n}C_{i}}{n\sum_{i = 1}^{n}t_{i}^{2}-(\sum_{i = 1}^{n}t_{i})^{2}}\) and \(b=\overline{C}-a\overline{t}\).
We have \(n = 5\), \(\sum_{i = 1}^{5}t_{i}=0 + 2+4 + 6+7=19\), \(\sum_{i = 1}^{5}C_{i}=27.1+28.6+28.2+27+30.8 = 141.7\), \(\sum_{i = 1}^{5}t_{i}C_{i}=0\times27.1+2\times28.6+4\times28.2+6\times27+7\times30.8=0 + 57.2+112.8+162+215.6 = 547.6\), \(\sum_{i = 1}^{5}t_{i}^{2}=0^{2}+2^{2}+4^{2}+6^{2}+7^{2}=0 + 4+16+36+49 = 105\).
\(\overline{t}=\frac{\sum_{i = 1}^{n}t_{i}}{n}=\frac{19}{5}=3.8\), \(\overline{C}=\frac{\sum_{i = 1}^{n}C_{i}}{n}=\frac{141.7}{5}=28.34\).
\(a=\frac{5\times547.6-19\times141.7}{5\times105 - 19^{2}}=\frac{2738-2692.3}{525 - 361}=\frac{45.7}{164}\approx0.28\).
\(b=28.34-0.28\times3.8=28.34 - 1.064 = 27.28\). So the linear - regression equation is \(C = 0.28t+27.28\).

Step3: Interpret the slope

The slope \(a = 0.28\) means that from 2004 to 2011, the number of Christmas trees sold in the U.S. increased by approximately \(0.28\) million trees each year.

Step4: Use the regression equation for 2008

For 2008, \(t = 4\). Substitute \(t = 4\) into \(C = 0.28t+27.28\), we get \(C=0.28\times4+27.28=1.12 + 27.28=28.4\).

Step5: Explain the difference

The data are not perfectly linear. The regression equation gives only an approximation of the actual value, not the exact value.

Answer:

1. 28.2
2. \(C = 0.28t+27.28\)
3. 0.28
4. 28.4
5. The data are not perfectly linear. The regression equation gives only an approximation of the actual value, not the exact value.