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.

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.

Transcript text: Complete parts (a) through (h) for the data below. \begin{tabular}{cccccc} \hline $\mathbf{x}$ & 3 & 4 & 5 & 7 & 8 \\ \hline $\mathbf{y}$ & 4 & 6 & 7 & 12 & 14 \\ \hline \end{tabular} (f) Compute the sum of the squared residuals for the line found in part (b). $\square$ (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). $\square$ (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 $\square$ passes through the most points. The line in part $\square$ $\square$ the sum of the squared residuals, thus being the best-fitting line.

Solution

Solution Steps

Step 1: Calculate the Means

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

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

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

Step 2: Calculate the Correlation Coefficient

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

\[ r = 0.9944 \]

Step 3: Calculate the Slope $ \beta $

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:

\[ \beta = \frac{34.8}{17.2} = 2.0233 \]

Step 4: Calculate the Intercept $ \alpha $

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

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

Step 5: Formulate the Least-Squares Regression Line

The least-squares regression line is expressed as:

\[ y = 2.0233x - 2.3256 \]

Step 6: Calculate the Sum of Squared Residuals for the Line $ y = 2x - 2 $

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

\[ \text{SSR} = 1.000 \]

Step 7: Calculate the Sum of Squared Residuals for the Least-Squares Regression Line

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

\[ \text{SSR} = 0.791 \]

Step 8: Compare the Fit of the Two Lines

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

The results are summarized as follows:

The least-squares regression line is $ y = 2.0233x - 2.3256 $.
The sum of squared residuals for the line $ y = 2x - 2 $ is $ 1.000 $.
The sum of squared residuals for the least-squares regression line is $ 0.791 $.

Thus, the final answer is:

\[ \boxed{\text{The line in part (d) minimizes the sum of the squared residuals, thus being the best-fitting line.}} \]

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