Correlation Coefficient

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Correlation Coefficient · Grade 11 · Worksheet 2

1. A scatter plot shows the relationship between the number of hours students spend practicing piano each week (x) and their performance scores on a standardized music assessment (y) for 25 students. The data points form a linear pattern with a correlation coefficient of r = -0.72. The regression equation is ŷ = -2.8x + 92. If a student practices for 6 hours per week, what performance score would you predict using the least squares regression line? Answer: 75.2 Solution: Identify the regression equation: ŷ = -2.8x + 92 Substitute x = 6 (hours practiced) into the equation: ŷ = -2.8(6) + 92 Calculate -2.8 × 6 = -16.8 Add this result to 92: -16.8 + 92 = 75.2 The predicted performance score is 75.2 The answer is 75.2.
   Full step-by-step solution

   Step 1: Identify the regression equation: ŷ = -2.8x + 92 Step 2: Substitute x = 6 (hours practiced) into the equation: ŷ = -2.8(6) + 92 Step 3: Calculate -2.8 × 6 = -16.8 Step 4: Add this result to 92: -16.8 + 92 = 75.2 Step 5: The predicted performance score is 75.2 The answer is 75.2.

2. Noah, a sports scientist, is studying the relationship between the number of hours of strength training per week (x) and the maximum bench press weight in kilograms (y) for 11 athletes. He collects the following summary statistics: the mean of x is 6 hours, the mean of y is 81 kg, the standard deviation of x is 2 hours, the standard deviation of y is 11 kg, and the sum of the products of z-scores (Σzₓzᵧ) is 8.6. Compute the Pearson correlation coefficient r, and interpret the strength and direction of the linear relationship between weekly strength training hours and maximum bench press weight. Answer: r = 0.86, indicating a strong positive linear relationship between weekly strength training hours and maximum bench press weight. Solution: Recall the formula for the Pearson correlation coefficient using the sum of products of z-scores: r = (Σzₓzᵧ) / (n - 1). Here, Σzₓzᵧ = 8.6 and n = 11.
   Full step-by-step solution

   Step 1: Recall the formula for the Pearson correlation coefficient using the sum of products of z-scores: r = (Σzₓzᵧ) / (n - 1). Here, Σzₓzᵧ = 8.6 and n = 11. Step 2: Substitute the values into the formula: r = 8.6 / (11 - 1) = 8.6 / 10. Step 3: Calculate the quotient: 8.6 ÷ 10 = 0.86. Step 4: Interpret the direction: Since r = 0.86 is positive, there is a positive linear relationship. This means that as the number of hours of strength training per week increases, the maximum bench press weight tends to increase. Step 5: Interpret the strength: The absolute value |r| = 0.86 is greater than 0.7. Using standard guidelines (|r| < 0.3 is weak, 0.3 ≤ |r| < 0.7 is moderate, |r| ≥ 0.7 is strong), this indicates a strong linear relationship. Final Answer: r = 0.86, indicating a strong positive linear relationship between weekly strength training hours and maximum bench press weight.

3. Given the dataset: (2, 5), (4, 11), (6, 17), (8, 23), compute the Pearson correlation coefficient r = ? Answer: 1 Solution: List the data points: (2,5), (4,11), (6,17), (8,23) Calculate the means: x̄ = (2+4+6+8)/4 = 20/4 = 5, ȳ = (5+11+17+23)/4 = 56/4 = 14 Calculate the sums needed for the formula: ∑(x-x̄)(y-ȳ), ∑(x-x̄)², ∑(y-ȳ)² For (2,5): x-x̄ = -3, y-ȳ = -9, product = 27, (x-x̄)² = 9, (y-ȳ)² = 81 For (4,11): x-x̄…
   Full step-by-step solution

   Step 1: List the data points: (2,5), (4,11), (6,17), (8,23) Step 2: Calculate the means: x̄ = (2+4+6+8)/4 = 20/4 = 5, ȳ = (5+11+17+23)/4 = 56/4 = 14 Step 3: Calculate the sums needed for the formula: ∑(x-x̄)(y-ȳ), ∑(x-x̄)², ∑(y-ȳ)² Step 4: Calculate deviations: For (2,5): x-x̄ = -3, y-ȳ = -9, product = 27, (x-x̄)² = 9, (y-ȳ)² = 81 For (4,11): x-x̄ = -1, y-ȳ = -3, product = 3, (x-x̄)² = 1, (y-ȳ)² = 9 For (6,17): x-x̄ = 1, y-ȳ = 3, product = 3, (x-x̄)² = 1, (y-ȳ)² = 9 For (8,23): x-x̄ = 3, y-ȳ = 9, product = 27, (x-x̄)² = 9, (y-ȳ)² = 81 Step 5: Sum the products: ∑(x-x̄)(y-ȳ) = 27+3+3+27 = 60 Step 6: Sum the squared deviations: ∑(x-x̄)² = 9+1+1+9 = 20, ∑(y-ȳ)² = 81+9+9+81 = 180 Step 7: Apply the formula: r = ∑(x-x̄)(y-ȳ) / √[∑(x-x̄)² × ∑(y-ȳ)²] = 60 / √(20 × 180) = 60 / √3600 = 60 / 60 = 1 The answer is 1.

4. Aroha, a climate researcher, is investigating the relationship between the number of days of extreme heat (above 35°C) per year and the annual yield of kiwifruit (in tonnes per hectare) for 15 orchards in a specific region of New Zealand. She collects the data and calculates the Pearson correlation coefficient, finding r = 0.63. Interpret this correlation coefficient in the context of the study, describing both the strength and direction of the linear relationship between extreme heat days and kiwifruit yield. Answer: There is a moderate positive linear relationship between the number of extreme heat days and kiwifruit yield; as extreme heat days increase, kiwifruit yield tends to increase moderately. Solution: Interpret the direction. The correlation coefficient r = 0.63 is positive, meaning that as the number of extreme heat days per year increases, the annual kiwifruit yield also tends to increase.
   Full step-by-step solution

   Step 1: Interpret the direction. The correlation coefficient r = 0.63 is positive, meaning that as the number of extreme heat days per year increases, the annual kiwifruit yield also tends to increase. This indicates a positive linear relationship. Step 2: Interpret the strength. The absolute value of r is |0.63| = 0.63. Using standard guidelines for correlation strength: - 0.00 to 0.30: weak correlation - 0.30 to 0.70: moderate correlation - 0.70 to 1.00: strong correlation Since 0.63 falls within the moderate range (0.30 to 0.70), this indicates a moderate strength relationship. Step 3: Combine direction and strength. The relationship is a moderate positive linear relationship. In context: orchards experiencing more extreme heat days tend to have moderately higher kiwifruit yields, but the relationship is not extremely strong, meaning other factors also influence yield significantly. Final answer: There is a moderate positive linear relationship between the number of extreme heat days and kiwifruit yield; as extreme heat days increase, kiwifruit yield tends to increase moderately.

5. A research team is studying the relationship between study hours and exam scores for Grade 11 students. They collected data from 30 students and calculated a correlation coefficient of r = 0.72. The team wants to determine if this correlation is statistically significant at the α = 0.05 level. If the critical value for a two-tailed test with 28 degrees of freedom is approximately 0.361, what conclusion should the researchers draw about the relationship between study hours and exam scores? Answer: The correlation is statistically significant because the calculated correlation coefficient (0.72) is greater than the critical value (0.361). Solution: - Sample size n = 30 - Correlation coefficient r = 0.72 - Significance level α = 0.05 - Degrees of freedom df = n - 2 = 28 - Critical value from table = 0.361 We need to check if the correlation is statistically significant.
   Full step-by-step solution

   Step 1: Understand the problem We are given: - Sample size n = 30 - Correlation coefficient r = 0.72 - Significance level α = 0.05 - Two-tailed test - Degrees of freedom df = n - 2 = 28 - Critical value from table = 0.361 We need to check if the correlation is statistically significant. --- Step 2: Recall the decision rule for correlation significance For a two-tailed test, we compare the absolute value of the calculated r with the critical value from the table. If |r| > critical value, then the correlation is statistically significant at the given α level. --- Step 3: Apply the numbers Calculated r = 0.72 Critical value = 0.361 Since |0.72| = 0.72 and 0.72 > 0.361, the calculated correlation coefficient exceeds the critical value. --- Step 4: Draw the conclusion Because the calculated r is greater than the critical value, we reject the null hypothesis (which states that there is no correlation in the population). Thus, the correlation between study hours and exam scores is statistically significant at α = 0.05. --- Final Answer: The correlation is statistically significant because the calculated correlation coefficient (0.72) is greater than the critical value (0.361).

6. Given the data points (1, 2), (2, 4), (3, 6), (4, 8), compute the Pearson correlation coefficient r = ? Answer: 1 Solution: We have data points: (1, 2), (2, 4), (3, 6), (4, 8).
   Full step-by-step solution

   Let's compute the Pearson correlation coefficient r step by step. We have data points: (1, 2), (2, 4), (3, 6), (4, 8). --- **Step 1: List the x and y values** x: 1, 2, 3, 4 y: 2, 4, 6, 8 n = 4 (number of pairs) --- **Step 2: Compute the means** mean_x = (1 + 2 + 3 + 4) / 4 = 10 / 4 = 2.5 mean_y = (2 + 4 + 6 + 8) / 4 = 20 / 4 = 5 --- **Step 3: Compute deviations and products** We will use the formula: r = [ sum( (x_i - mean_x) * (y_i - mean_y) ) ] / [ sqrt( sum( (x_i - mean_x)^2 ) * sum( (y_i - mean_y)^2 ) ) ] First, compute (x_i - mean_x) and (y_i - mean_y): For (1, 2): dx = 1 - 2.5 = -1.5 dy = 2 - 5 = -3 product = (-1.5)*(-3) = 4.5 dx^2 = 2.25 dy^2 = 9 For (2, 4): dx = 2 - 2.5 = -0.5 dy = 4 - 5 = -1 product = (-0.5)*(-1) = 0.5 dx^2 = 0.25 dy^2 = 1 For (3, 6): dx = 3 - 2.5 = 0.5 dy = 6 - 5 = 1 product = 0.5 * 1 = 0.5 dx^2 = 0.25 dy^2 = 1 For (4, 8): dx = 4 - 2.5 = 1.5 dy = 8 - 5 = 3 product = 1.5 * 3 = 4.5 dx^2 = 2.25 dy^2 = 9 --- **Step 4: Sum the necessary values** sum of products = 4.5 + 0.5 + 0.5 + 4.5 = 10 sum of dx^2 = 2.25 + 0.25 + 0.25 + 2.25 = 5 sum of dy^2 = 9 + 1 + 1 + 9 = 20 --- **Step 5: Compute r** r = 10 / sqrt(5 * 20) = 10 / sqrt(100) = 10 / 10 = 1 --- **Step 6: Interpretation** The Pearson correlation coefficient r = 1 means there is a perfect positive linear relationship between x and y. Indeed, y = 2x exactly for all points. --- **Final Answer:** 1