Do Homework - Homework: Section 12.6 Section 12.6 Question 11, 12.6.41 HW Score: $72.83 \%, 10.93$ of 15 points Part 1 of 4 Points: 0 of 1 Save The following table shows the energy provided (in kilocalories) and the amount of carbohydrates (in grams) in a $100 \mathrm{ml}$ bottie of sports drinks for six different brands. Use the data provided in the table to complete parts a) through d). \begin{tabular}{|c|c|c|c|c|c|c|} \hline Carbohydrates (grams) & 9.5 & 8 & 3 & 6 & 9 & 5 \\ \hline Energy (kilocalories) & 29 & 37 & 26 & 13 & 28 & 24 \\ \hline \end{tabular} i Click the icon to view the correlation coefficient, $r$, for various values of $n$ and $a$. a) Determine the correlation coefficient between the amount of carbohydrates and the energy provided. The correlation coefficient between the amount of carbohydrates and the energy provided is $r=\square$. (Type an integer or decimal rounded to three decimal places as needed.)

Step 1 ：Calculate the mean of carbohydrates and energy. Mean of carbohydrates = \(\frac{9.5 + 8 + 3 + 6 + 9 + 5}{6} = 6.75\) grams. Mean of energy = \(\frac{29 + 37 + 26 + 13 + 28 + 24}{6} = 26.17\) kilocalories.

Step 2 ：Calculate the standard deviation of carbohydrates and energy. Standard deviation of carbohydrates = \(\sqrt{\frac{(9.5-6.75)^2 + (8-6.75)^2 + (3-6.75)^2 + (6-6.75)^2 + (9-6.75)^2 + (5-6.75)^2}{5}} = 2.22\) grams. Standard deviation of energy = \(\sqrt{\frac{(29-26.17)^2 + (37-26.17)^2 + (26-26.17)^2 + (13-26.17)^2 + (28-26.17)^2 + (24-26.17)^2}{5}} = 7.77\) kilocalories.

Step 3 ：Calculate the covariance of carbohydrates and energy. Covariance = \(\frac{(9.5-6.75)*(29-26.17) + (8-6.75)*(37-26.17) + (3-6.75)*(26-26.17) + (6-6.75)*(13-26.17) + (9-6.75)*(28-26.17) + (5-6.75)*(24-26.17)}{5} = 4.65\).

Step 4 ：Calculate the correlation coefficient. Correlation coefficient, r = \(\frac{Covariance}{Standard deviation of carbohydrates * Standard deviation of energy} = \frac{4.65}{2.22 * 7.77} = 0.269\).

Step 5 ：\(\boxed{So, the correlation coefficient between the amount of carbohydrates and the energy provided is r = 0.269.}\)