Question

find the correlation coefficient, r, of the data described below.
sally is a professional dancer booked for a major event. although she has a rigorous practice schedule, she sometimes has to end a practice session when she begins to experience stiffness. she wonders if stretching more before starting a session will allow her to practice for a longer time.
over the next few days sally varies the number of minutes she stretches before the start of her session, x, and notes the number of minutes she can practice before experiencing stiffness, y.
minutes spent stretching minutes spent practicing
0.45 199
11.90 199
19.34 220
23.42 266
28.21 260

Explanation:

Step 1: List the data points

Let \( x \) be the minutes spent stretching and \( y \) be the minutes spent practicing. The data points are:
\( (0.45, 199) \), \( (11.90, 199) \), \( (19.34, 220) \), \( (23.42, 266) \), \( (28.21, 260) \)

Step 2: Calculate necessary sums

First, we calculate \( \sum x \), \( \sum y \), \( \sum xy \), \( \sum x^2 \), and \( \sum y^2 \).

• \( \sum x = 0.45 + 11.90 + 19.34 + 23.42 + 28.21 = 83.32 \)
• \( \sum y = 199 + 199 + 220 + 266 + 260 = 1144 \)
• \( \sum xy = (0.45 \times 199) + (11.90 \times 199) + (19.34 \times 220) + (23.42 \times 266) + (28.21 \times 260) \)
• \( 0.45 \times 199 = 89.55 \)
• \( 11.90 \times 199 = 2368.1 \)
• \( 19.34 \times 220 = 4254.8 \)
• \( 23.42 \times 266 = 6239.72 \)
• \( 28.21 \times 260 = 7334.6 \)
• \( \sum xy = 89.55 + 2368.1 + 4254.8 + 6239.72 + 7334.6 = 20286.77 \)
• \( \sum x^2 = (0.45)^2 + (11.90)^2 + (19.34)^2 + (23.42)^2 + (28.21)^2 \)
• \( 0.45^2 = 0.2025 \)
• \( 11.90^2 = 141.61 \)
• \( 19.34^2 = 374.0356 \)
• \( 23.42^2 = 548.4964 \)
• \( 28.21^2 = 795.8041 \)
• \( \sum x^2 = 0.2025 + 141.61 + 374.0356 + 548.4964 + 795.8041 = 1860.1486 \)
• \( \sum y^2 = (199)^2 + (199)^2 + (220)^2 + (266)^2 + (260)^2 \)
• \( 199^2 = 39601 \) (twice, so \( 2 \times 39601 = 79202 \))
• \( 220^2 = 48400 \)
• \( 266^2 = 70756 \)
• \( 260^2 = 67600 \)
• \( \sum y^2 = 79202 + 48400 + 70756 + 67600 = 265958 \)

Step 3: Use the correlation coefficient formula

The formula for the correlation coefficient \( r \) is:

$$ r = \frac{n\sum xy - \sum x \sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} $$

where \( n = 5 \) (number of data points).

First, calculate the numerator:
\( n\sum xy - \sum x \sum y = 5 \times 20286.77 - 83.32 \times 1144 \)
\( = 101433.85 - 95318.08 = 6115.77 \)

Next, calculate the denominator:

• First part inside the square root: \( n\sum x^2 - (\sum x)^2 = 5 \times 1860.1486 - (83.32)^2 \)

\( = 9300.743 - 6942.2224 = 2358.5206 \)

• Second part inside the square root: \( n\sum y^2 - (\sum y)^2 = 5 \times 265958 - (1144)^2 \)

\( = 1329790 - 1308736 = 21054 \)

• Multiply these two parts: \( 2358.5206 \times 21054 \approx 4965639.27 \)
• Take the square root: \( \sqrt{4965639.27} \approx 2228.37 \)

Now, calculate \( r \):
\( r = \frac{6115.77}{2228.37} \approx 0.89 \) (rounded to two decimal places)

Answer:

The correlation coefficient \( r \) is approximately \( 0.89 \).