Chapter 1.7, Problem 50E

Solution

a.

To plot: The scatter plot of ${\text{CO}}_2$ concentration as a function of the years since 1950.

Given information:

The atmospheric concentration of carbon dioxide (in parts per million) recorded by the observatory in years between 1960 and 2015.

x	y
1960	317
1970	326
1980	339
1990	354
2000	369
2005	380
2010	390
2015	401

Plot:

Take the year as x coordinate and concentration as y coordinate, and then draw the scatter plot as shown below:

  

Interpretation:

From the scatter plot it seems that the linear regression will best fit the scenario.

b.

To find: If a linear or a quadratic model fit more appropriate.

A linear model seems more appropriate for the scatter plot.

Given information:

The atmospheric concentration of carbon dioxide (in parts per million) recorded by the observatory in years between 1960 and 2015.

x	y
1960	317
1970	326
1980	339
1990	354
2000	369
2005	380
2010	390
2015	401

Calculation:

Observe that the points do not fall nearly along a well-known curve. But, they do seem to fall near a line with upward slope.

Since the curve drawn seems to be a straight line, a linear model seems more appropriate for the scatter plot.

c.

To find: Find the regression equation for the appropriate model, and superimpose the graph of the regression equation on the scatter plot.

The regression can line by written as $Y=-2695.5685+1.5342X$ . The model seems to work appropriately.

Given information:

The atmospheric concentration of carbon dioxide (in parts per million) recorded by the observatory in years between 1960 and 2015.

x	y
1960	317
1970	326
1980	339
1990	354
2000	369
2005	380
2010	390
2015	401

Calculation:

Prepare the table to find the regression coefficients as follows;

  

Use the table to perform the calculations as shown below.

  $\begin{array}{c}\overline{X}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i}\\ =\frac{15930}{8}\\ =1991.25\\ \overline{Y}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i}\\ =\frac{2876}{8}\\ =359.5\\ S{S}_{XX}={\displaystyle \sum _{i=1}^n{X}_i^2}-\frac{1}{n}{\left({\displaystyle \sum _{i=1}^n{X}_i}\right)}^2\\ =31723350-{15930}^2/8\\ =2737.5\\ S{S}_{YY}={\displaystyle \sum _{i=1}^n{Y}_i^2}-\frac{1}{n}{\left({\displaystyle \sum _{i=1}^n{Y}_i}\right)}^2\\ =1040464-{2876}^2/8\\ =6542\\ S{S}_{XY}={\displaystyle \sum _{i=1}^n{X}_i}{Y}_i-\frac{1}{n}\left({\displaystyle \sum _{i=1}^n{X}_i}\right)\left({\displaystyle \sum _{i=1}^n{Y}_i}\right)\\ =5731035-15930\times 2876/8\\ =4200\end{array}$

Then,

  $\begin{array}{c}m=\frac{S{S}_{XY}}{S{S}_{XX}}\\ =\frac{4200}{2737.5}\\ =1.5342\end{array}$

Therefore,

  $\begin{array}{c}n=\overline{Y}-\overline{X}\cdot m\\ =359.5-1991.25\times 1.5342\\ =-2695.5685\end{array}$

So, the regression can line by written as $Y=-2695.5685+1.5342X$ .

Now, plot the regression line as shown below.

  

The graph of the regression line best fit the data. That means, the graph shows that the model appears to work well.

d.

To find: The approximate concentration level of carbon dioxide in the year 2025.

The concentration of carbon dioxide in the year 2025 will be $\text{411}\text{.1865}$ ppm.

Given information:

The atmospheric concentration of carbon dioxide (in parts per million) recorded by the observatory in years between 1960 and 2015.

x	y
1960	317
1970	326
1980	339
1990	354
2000	369
2005	380
2010	390
2015	401

Calculation:

The model for the given situation is $Y=-2695.5685+1.5342X$ .

Substitute $X=2025$ in the equation to find the concentration in the year 2025.

  $\begin{array}{c}Y=-2695.5685+1.5342X\\ =-2695.5685+1.5342\times 2025\\ =\text{411}\text{.1865}\end{array}$

Conclusion:

The concentration of carbon dioxide in the year 2025 will be $\text{411}\text{.1865}$ ppm.

e.

To explain: The circumstances that might caused the predication in part (d) of the mark and tell whether the true value will be higher or lower than the predicted value.

Given information:

The atmospheric concentration of carbon dioxide (in parts per million) recorded by the observatory in years between 1960 and 2015.

x	y
1960	317
1970	326
1980	339
1990	354
2000	369
2005	380
2010	390
2015	401

Explanation:

If the model is predicted to be an upward quadratic, then the prediction in part (d) would have been off the mark.

Since an upward quadratic function gives a higher value than a linear function with upward slope, the true value would have been lower than the predicated value.