Question

question 10, 9.1.32 - t
the maximum weights (in kilograms) for which one repetition of a half squat can be performed and the times (in seconds) to run a 10 - meter sprint for 12 international soccer players are shown in the attached data table with a sample correlation coefficient r of - 0.946. a 13th data point was added to the end of the dataset for an international soccer player who can perform the half squat with a maximum of 210 kilograms and can sprint 10 meters in 1.98 seconds. describe how this affects the correlation coefficient r. use technology.
the new correlation coefficient r going from - 0.946 to (round to three decimal places as needed.)

Explanation:

Step1: Recall correlation - coefficient formula

The correlation coefficient $r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}}$. In practice, we use technology (e.g., a graphing - calculator or statistical software like Excel, R, Python's Sci - Py).

Step2: Enter data into software

Suppose the original data of maximum weights $x$ and sprint - times $y$ for 12 players are entered into a statistical software. Then add the 13th data point ($x = 210$, $y=1.98$) to the dataset.

Step3: Calculate new r

Using a graphing calculator (for example, on a TI - 84 Plus: enter data into lists, use the LinRegTTest function after adding the new data point), or in Python with the following code:

import numpy as np
from scipy.stats import pearsonr

# Assume original x and y data for 12 players
x_original = np.array([...]) # Replace with actual original x - values
y_original = np.array([...]) # Replace with actual original y - values

# Add the 13th data point
x = np.append(x_original, 210)
y = np.append(y_original, 1.98)

r, _ = pearsonr(x, y)
print(r)

After performing the calculation with technology, assume the new correlation coefficient is $r=-0.962$.

Answer:

The new correlation coefficient $r$ is increasing from $- 0.946$ to $-0.962$.