a) Year is on the horizontal axis and population is on the vertical axis. b) Let us first determine the necessary sums: $\sum \text{}{x}_{i}=3990$

$\sum \text{}{x}_{i}^{2}=79960670$

$\sum \text{}{y}_{i}=5592$

$\sum \text{}{x}_{i}\text{}{y}_{i}=11183199$ Next, we can determine $S}_{xx$ and $S}_{xy$

${S}_{xx}=\text{}\sum \text{}{x}_{i}^{2}\text{}-\text{}\frac{{(\sum \text{}{x}_{i})}^{2}}{n}=79960670\text{}-\text{}\frac{{3990}^{2}}{20}=665$

${S}_{xy}=\text{}\sum \text{}{x}_{i}{y}_{i}\text{}-\text{}\frac{(\sum \text{}{x}_{i})(\sum \text{}{y}_{i})}{n}=11183199\text{}-\text{}\frac{3990\text{}\cdot \text{}5592}{20}=1995$ The estimate b of the slope $\beta$ is the ratio of $S}_{xy$ and $S}_{xx$: $b=\text{}\frac{{S}_{xy}}{{S}_{xx}}=\text{}\frac{1995}{665}=3$ The mean is the sum of all values divided by the number of values: $\stackrel{\u2015}{x}=\text{}\frac{\sum \text{}{x}_{i}}{n}=\text{}\frac{6990}{20}=1999.5$

$\stackrel{\u2015}{y}=\text{}\frac{\sum \text{}{y}_{i}}{n}=\text{}\frac{5592}{20}=279.6$ The estimate a of the intercept $\alpha$ is the average of u decreased by the product of the estimate of the slope and the average of x. $a=\text{}\stackrel{\u2015}{y}\text{}-\text{}b\stackrel{\u2015}{x}=279.6\text{}-\text{}3\text{}\cdot \text{}1999.5=\text{}-5718.9$ General least-squares equation: $\hat{y}=\text{}\alpha \text{}+\text{}\beta \text{}x$. Replace $\alpha$ by $a=\text{}-5505.3432$ and $\beta$ by $b=2.8930$ in the general least-squares equation: $y=a\text{}+\text{}bx=\text{}-5718.9\text{}+\text{}3x$

c) Let us evaluate the regression line of part (b) at $x=2010$ and $x=2011$. $y=\text{}-5718.9\text{}+\text{}3\left(2010\right)=311.1$

$y=\text{}-5718.9\text{}+\text{}3\left(2011\right)=314.1$ Thus the predicted U.S. population in 2010 is 311.1 million and predicted U.S. population in 2011 is 314.1 million.