A Complete Example: Answers | Saylor Academy

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.

Answers

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: $(168.2974,176.0938)$ .

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