## A Complete Example

This section explains linear regression, from presenting the data to using scatter plots to identify the linear pattern. It then fits a linear model using least squares estimation and addresses statistical inferences on correlation coefficient and slope parameter.

1.

$Σx=14.2$, $Σy=49.6$, $Σxy=91.73$, $Σx^2=26.3$, $Σy^2=333.86$.

$SS_{xx}=6.136$, $SS_{xy}=21.298$, $SS_{yy}=87.844$.

$\overline x =1.42$, $\overline y =4.96$.

$\hat β_1=3.47$, $\hat β_0=0.03$.

$SSE=13.92$.

$sε=1.32$.

$r = 0.9174$, $r^2 = 0.8416$.

$df=8$, $T = 6.518$.

The 95% confidence interval for $β_1$ is: $(2.24,4.70)$.

At $x_p=2$, the 95% confidence interval for $E(y)$ is $(5.77,8.17)$.

At $x_p=2$, the 95% prediction interval for $y$ is $(3.73,10.21)$.

3. The positively correlated trend seems less profound than that in each of the previous plots.

5. The regression line: $\hat y=3.3426x+138.7692$. Coefficient of Correlation: $r = 0.9431$. Coefficient of Determination: $r^2 = 0.8894$. $SSE=283.2473$. $s_e=1.9305$. A 95% confidence interval for $β_1: (3-.0733,3.6120)$. Test Statistic for $H_0:β_1=0: T = 24.7209$. At $x_p=10$, $\hat y=172.1956$; a 95% confidence interval for the mean value of $y$ is: $(171.5577,172.8335)$; and a 95% prediction interval for an individual value of $y$ is: $(168.2974,176.0938)$.