Questions: Research was conducted on the amount of oxygen runners could utilize during training, known as their VO2 max, and their finishing times for a 5 K race to see if there was a significant relationship between these variables. The results for 8 runners were compiled and placed in the following table. VO2 Max (ml / kg / min)(x) 5K Finishing Time (min)(y) 38.45 24.75 28.22 33.35 42.89 20.79 44.27 23.43 39.25 26.03 17.35 34.98 15.04 37.28 21.32 33.38 Calculate the correlation coefficient. Round to three decimals r= Find the equation of the regression line. Round each value to two decimals Predict the time in the 5 K for a runner whose VO2 max is 25.86 ml / kg / min. Round your answer to two decimals

Research was conducted on the amount of oxygen runners could utilize during training, known as their VO2 max, and their finishing times for a 5 K race to see if there was a significant relationship between these variables. The results for 8 runners were compiled and placed in the following table. VO2 Max (ml / kg / min)(x) 5K Finishing Time (min)(y) 38.45 24.75 28.22 33.35 42.89 20.79 44.27 23.43 39.25 26.03 17.35 34.98 15.04 37.28 21.32 33.38 Calculate the correlation coefficient. Round to three decimals r= Find the equation of the regression line. Round each value to two decimals Predict the time in the 5 K for a runner whose VO2 max is 25.86 ml / kg / min. Round your answer to two decimals
Transcript text: Research was conducted on the amount of oxygen runners could utilize during training, known as their $\mathrm{VO}_{2}$ max, and their finishing times for a 5 K race to see if there was a significant relationship between these variables. The results for 8 runners were compiled and placed in the following table. \begin{tabular}{|c|c|} \hline $\mathrm{VO}_{2}$ Max $(\mathrm{ml} / \mathrm{kg} / \mathrm{min})(x)$ & 5K Finishing Time $(\mathrm{min})(\boldsymbol{y})$ \\ \hline 38.45 & 24.75 \\ \hline 28.22 & 33.35 \\ \hline 42.89 & 20.79 \\ \hline 44.27 & 23.43 \\ \hline 39.25 & 26.03 \\ \hline 17.35 & 34.98 \\ \hline 15.04 & 37.28 \\ \hline 21.32 & 33.38 \\ \hline \end{tabular} Calculate the correlation coefficient. Round to three decimals $r=$ $\square$ Find the equation of the regression line. Round each value to two decimals $\square$ Predict the time in the 5 K for a runner whose $\mathrm{VO}_{2} \mathrm{max}$ is $25.86 \mathrm{ml} / \mathrm{kg} / \mathrm{min}$. Round your answer to two decimals $\square$
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Solution

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Solution Steps

Step 1: Calculate the Correlation Coefficient

The correlation coefficient \( r \) is calculated using the formula:

\[ r = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y} \]

Where:

  • \( \text{Cov}(X,Y) = -70.809 \)
  • \( \sigma_X = 11.852 \)
  • \( \sigma_Y = 6.179 \)

Substituting the values:

\[ r = \frac{-70.809}{11.852 \cdot 6.179} \approx -0.967 \]

Thus, the correlation coefficient is:

\[ \boxed{r = -0.967} \]

Step 2: Find the Equation of the Regression Line

To find the regression line, we first calculate the means of \( x \) and \( y \):

\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = 30.85 \] \[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = 29.25 \]

Next, we calculate the numerator and denominator for the slope \( \beta \):

Numerator for \( \beta \):

\[ \sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 6722.64 - 8 \cdot 30.85 \cdot 29.25 = -495.66 \]

Denominator for \( \beta \):

\[ \sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 8596.48 - 8 \cdot 30.85^2 = 983.32 \]

Now, we can calculate the slope \( \beta \):

\[ \beta = \frac{-495.66}{983.32} \approx -0.5 \]

Next, we calculate the intercept \( \alpha \):

\[ \alpha = \bar{y} - \beta \bar{x} = 29.25 - (-0.5 \cdot 30.85) \approx 44.8 \]

Thus, the equation of the regression line is:

\[ y = 44.8 - 0.5x \]

Step 3: Predict the Finishing Time for a Given VO2 Max

To predict the finishing time for a runner with \( \mathrm{VO}_2 \) max of \( 25.86 \, \mathrm{ml/kg/min} \):

\[ \text{Predicted Time} = \alpha + \beta \cdot \text{VO2 Max} = 44.8 - 0.5 \cdot 25.86 \]

Calculating this gives:

\[ \text{Predicted Time} = 44.8 - 12.93 \approx 31.87 \]

Thus, the predicted finishing time is:

\[ \boxed{31.87 \text{ minutes}} \]

Final Answer

  • Correlation Coefficient: \( \boxed{r = -0.967} \)
  • Regression Line: \( \boxed{y = 44.8 - 0.5x} \)
  • Predicted 5K Finishing Time: \( \boxed{31.87 \text{ minutes}} \)
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