To find the linear correlation coefficient [tex]\(r\)[/tex] for the given data, follow these steps:
1. List the provided information:
- Hours: [tex]\( [5, 10, 4, 6, 10, 9] \)[/tex]
- Scores: [tex]\( [64, 86, 69, 86, 59, 87] \)[/tex]
2. Calculate the mean of the hours and scores:
[tex]\[
\text{Mean of hours} = \frac{5 + 10 + 4 + 6 + 10 + 9}{6} = \frac{44}{6} = 7.3333 \approx 7.33
\][/tex]
[tex]\[
\text{Mean of scores} = \frac{64 + 86 + 69 + 86 + 59 + 87}{6} = \frac{451}{6} \approx 75.17
\][/tex]
3. Calculate the numerator for the correlation coefficient formula:
[tex]\[
\text{Numerator} = \sum_{i=1}^{6} (x_i - \bar{x})(y_i - \bar{y})
\][/tex]
Where [tex]\( x_i \)[/tex] represents the hours and [tex]\( y_i \)[/tex] represents the scores.
Performing these subtractions and multiplications:
[tex]\[
((5 - 7.33) \times (64 - 75.17)) + ((10 - 7.33) \times (86 - 75.17)) + ((4 - 7.33) \times (69 - 75.17)) + ((6 - 7.33) \times (86 - 75.17)) + ((10 - 7.33) \times (59 - 75.17)) + ((9 - 7.33) \times (87 - 75.17))
\][/tex]
[tex]\[
= (-2.33 \times -11.17) + (2.67 \times 10.83) + (-3.33 \times -6.17) + (-1.33 \times 10.83) + (2.67 \times -16.17) + (1.67 \times 11.83)
\][/tex]
[tex]\[
\approx 26 + 28.90 + 20.57 - 14.40 - 43.22 + 19.85
\][/tex]
[tex]\[
= 37.67
\][/tex]
4. Calculate the denominator for the correlation coefficient formula:
[tex]\[
\text{Denominator} = \sqrt{\left(\sum_{i=1}^{6} (x_i - \bar{x})^2\right) \times \left(\sum_{i=1}^{6} (y_i - \bar{y})^2\right)}
\][/tex]
Calculating these terms individually:
[tex]\[
\sum_{i=1}^{6} (x_i - \bar{x})^2 = (5 - 7.33)^2 + (10 - 7.33)^2 + (4 - 7.33)^2 + (6 - 7.33)^2 + (10 - 7.33)^2 + (9 - 7.33)^2
\][/tex]
[tex]\[
= 5.44 + 7.11 + 11.09 + 1.77 + 7.11 + 2.78 = 35.30
\][/tex]
[tex]\[
\sum_{i=1}^{6} (y_i - \bar{y})^2 = (64 - 75.17)^2 + (86 - 75.17)^2 + (69 - 75.17)^2 + (86 - 75.17)^2 + (59 - 75.17)^2 + (87 - 75.17)^2
\][/tex]
[tex]\[
= 124.10 + 117.35 + 38.00 + 117.35 + 261.32 + 139.88 = 797.99
\][/tex]
Then the denominator:
[tex]\[
\text{Denominator} = \sqrt{35.30 \times 797.99} \approx 168.00
\][/tex]
5. Finally, calculate the correlation coefficient [tex]\(r\)[/tex]:
[tex]\[
r = \frac{\text{Numerator}}{\text{Denominator}} = \frac{37.67}{168.00} \approx 0.224
\][/tex]
Thus, the value of the linear correlation coefficient [tex]\(r\)[/tex] is approximately 0.224. Therefore, the correct answer is:
[tex]\[ \boxed{0.224} \][/tex]
