The means of the independent variable \( x \) and the dependent variable \( y \) are calculated as follows:
\[
\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = \frac{1 + 2 + 3 + 4 + 5}{5} = 3.0
\]
\[
\bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = \frac{2 + 4 + 5 + 4 + 5}{5} = 4.0
\]
The correlation coefficient \( r \) is calculated to measure the strength of the linear relationship between \( x \) and \( y \):
\[
r = 0.7746
\]
The slope \( \beta \) of the regression line is determined using the following formulas:
Numerator for \( \beta \):
\[
\sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 66 - 5 \cdot 3.0 \cdot 4.0 = 6.0
\]
Denominator for \( \beta \):
\[
\sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 55 - 5 \cdot 3.0^2 = 10.0
\]
Thus, the slope \( \beta \) is calculated as:
\[
\beta = \frac{6.0}{10.0} = 0.6
\]
The intercept \( \alpha \) is calculated using the formula:
\[
\alpha = \bar{y} - \beta \bar{x} = 4.0 - 0.6 \cdot 3.0 = 2.2
\]
The equation of the line of best fit is given by:
\[
y = 2.2 + 0.6x
\]
The \( R^2 \) value, which indicates the proportion of variance explained by the regression line, is calculated as:
\[
R^2 = r^2 = (0.7746)^2 = 0.6000
\]
Since \( R^2 \) is indeed used to determine the strength of a linear fit, the answer to the question is:
\[
\boxed{\text{True}}
\]
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