To find a regression model that best models the data, we can use the least squares method. The function that best models the data is y = 1.18x - 812.9. The amount of methane emissions in 2000 was about -152.9 million metric tons, and the amount of methane emissions in the years immediately following 2008 would most likely increase.
Step-by-step explanation:
To find a regression model that best models the data, we can use the least squares method. This involves finding a linear equation of the form y = mx + b, where y represents the methane emissions and x represents the years. We can use the given data points to calculate the values of m and b.
Let's calculate the mean of the years and methane emissions. The mean of the years is (2002 + 2003 + 2004 + 2005 + 2006 + 2007 + 2008) / 7 = 2005, and the mean of the methane emissions is (673.3 + 660.6 + 661.6 + 669.2 + 678.5 + 690.9 + 724.2) / 7 = 681.5.
Next, let's calculate the differences between each year and the mean year, and each methane emission and the mean methane emission. The differences for the years are: 2002 - 2005 = -3, 2003 - 2005 = -2, 2004 - 2005 = -1, 2005 - 2005 = 0, 2006 - 2005 = 1, 2007 - 2005 = 2, 2008 - 2005 = 3. The differences for the methane emissions are: 673.3 - 681.5 = -8.2, 660.6 - 681.5 = -20.9, 661.6 - 681.5 = -19.9, 669.2 - 681.5 = -12.3, 678.5 - 681.5 = -3, 690.9 - 681.5 = 9.4, 724.2 - 681.5 = 42.7.
Then, let's calculate the product of the differences for each year and methane emission. For example, for 2002, the product is -3 * -8.2 = 24.6. We do this for all the years and methane emissions.
Finally, let's calculate the sum of the products of the differences for the years and methane emissions, and the sum of the squares of the differences for the years. The sum of the products is 24.6 + (-8.2) + (-19.9) + (-12.3) + (-3) + 9.4 + 42.7 = 33.3, and the sum of the squares of the differences for the years is (-3)^2 + (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 + 3^2 = 28.
Using these values, we can calculate the slope of the regression line as m = sum of the products / sum of the squares of the differences for the years = 33.3 / 28 = 1.18. To find the y-intercept, we can use the equation b = mean of the methane emissions - (m * mean of the years) = 681.5 - (1.18 * 2005) = -812.9.
The function that best models the data is y = 1.18x - 812.9.
Now, let's use this model to complete the statements:
The amount of methane emissions in 2000 was about -152.9 million metric tons. (Using the equation: y = 1.18 * 2000 - 812.9)
The amount of methane emissions in the years immediately following 2008 would most likely increase. (Since the slope of the regression line is positive, the methane emissions would likely continue to increase after 2008)