Questions: Complete parts (a) through (h) for the data below. x 3 4 5 7 8 y 4 6 7 12 14 (f) Compute the sum of the squared residuals for the line found in part (b). (Round to three decimal places as needed.) (g) Compute the sum of the squared residuals for the least-squares regression line found in part (d). (Round to three decimal places as needed.) (h) Comment on the fit of the line found in part (b) versus the least-squares regression line found in part (d). The line in part passes through the most points. The line in part has the sum of the squared residuals, thus being the best-fitting line.

Solution Steps

The means of the variables $x$ and $y$ are calculated as follows:

$$\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = \frac{3 + 4 + 5 + 7 + 8}{5} = 5.4$$

$$\bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = \frac{4 + 6 + 7 + 12 + 14}{5} = 8.6$$

The correlation coefficient $r$ is computed to assess the strength of the linear relationship between $x$ and $y$:

$$r = 0.9944$$

The slope $\beta$ is determined using the following formulas:

Numerator for $\beta$:

$$\sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 267 - 5 \cdot 5.4 \cdot 8.6 = 34.8$$

Denominator for $\beta$:

$$\sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 163 - 5 \cdot (5.4)^2 = 17.2$$

Thus, the slope $\beta$ is calculated as:

$$\beta = \frac{34.8}{17.2} = 2.0233$$

The intercept $\alpha$ is calculated using the formula:

$$\alpha = \bar{y} - \beta \bar{x} = 8.6 - 2.0233 \cdot 5.4 = -2.3256$$

The least-squares regression line is expressed as:

$$y = 2.0233x - 2.3256$$

The sum of squared residuals (SSR) for the line $y = 2x - 2$ is computed as:

$$\text{SSR} = 1.000$$

The sum of squared residuals (SSR) for the least-squares regression line is:

$$\text{SSR} = 0.791$$

The line in part (d) passes through the most points. The line in part (d) minimizes the sum of the squared residuals, thus being the best-fitting line.

Final Answer