To find the correlation coefficient for the linear regression of the given data, we can follow these steps:
1. Organize the Data: Note the pairs of hours and save percentages provided.
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
\text{Hours (x)} & 5 & 8 & 10 & 6 & 4 & 10 & 13 & 8 \\
\hline
\text{Save Percentage (y)} & 0.875 & 0.892 & 0.931 & 0.883 & 0.846 & 0.918 & 0.92 & 0.927 \\
\hline
\end{array}
\][/tex]
2. Calculate the Means of x and y:
[tex]\[
\bar{x} = \frac{5 + 8 + 10 + 6 + 4 + 10 + 13 + 8}{8} = \frac{64}{8} = 8
\][/tex]
[tex]\[
\bar{y} = \frac{0.875 + 0.892 + 0.931 + 0.883 + 0.846 + 0.918 + 0.92 + 0.927}{8} = \frac{7.192}{8} \approx 0.899
\][/tex]
3. Compute the Differences from the Mean and their Products:
Calculate each [tex]\((x_i - \bar{x})\)[/tex], [tex]\((y_i - \bar{y})\)[/tex], and their product:
[tex]\[
\begin{array}{|c|c|c|c|}
\hline
x_i & (x_i - \bar{x}) & y_i & (y_i - \bar{y}) & (x_i - \bar{x})(y_i - \bar{y}) \\
\hline
5 & -3 & 0.875 & -0.024 & 0.072 \\
8 & 0 & 0.892 & -0.007 & 0 \\
10 & 2 & 0.931 & 0.032 & 0.064 \\
6 & -2 & 0.883 & -0.016 & 0.032 \\
4 & -4 & 0.846 & -0.053 & 0.212 \\
10 & 2 & 0.918 & 0.019 & 0.038 \\
13 & 5 & 0.92 & 0.021 & 0.105 \\
8 & 0 & 0.927 & 0.028 & 0 \\
\hline
\end{array}
\][/tex]
4. Calculate Sum of Products and Variances:
[tex]\[
\sum (x_i - \bar{x})(y_i - \bar{y}) = 0.072 + 0 + 0.064 + 0.032 + 0.212 + 0.038 + 0.105 + 0 = 0.523
\][/tex]
[tex]\[
S_x = \sqrt{\sum (x_i - \bar{x})^2 / (n - 1)}
\][/tex]
[tex]\[
S_y = \sqrt{\sum (y_i - \bar{y})^2 / (n - 1)}
\][/tex]
5. Calculate Correlation Coefficient r:
[tex]\[
r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{(n-1)S_x S_y} = \frac{0.523}{\sqrt{\sum (x_i - \bar{x})^2}\sqrt{\sum (y_i - \bar{y})^2}}
\][/tex]
6. Simplify to find r:
For the given data, after simplifying the correlation coefficient, the rounded result to the nearest thousandth is obtained.
Therefore, the correlation coefficient [tex]\( r \approx 0.837 \)[/tex].
