Correlation Coefficient

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Answer Key & Explanations

Correlation Coefficient · Grade 11 · Worksheet 1

1. Mason, a hydrologist, is studying the relationship between the depth of a river (in meters) at a specific point and the water flow velocity (in meters per second) at that same point. He collects data from 12 different river locations and computes the following summary statistics: the mean depth is 2.7 meters, the mean velocity is 1.2 meters per second, the standard deviation of depth is 0.7 meters, the standard deviation of velocity is 0.4 meters per second, and the sum of the products of the z-scores (Σzₓzᵧ) is 7.7. Compute the Pearson correlation coefficient r, and interpret the strength and direction of the linear relationship between river depth and water flow velocity. Answer: r = 0.70, indicating a strong positive linear relationship between river depth and water flow velocity. Solution: Recall the formula for the Pearson correlation coefficient using the sum of products of z-scores: r = (Σzₓzᵧ) / (n - 1). Here, Σzₓzᵧ = 7.7 and n = 12.
   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 = 7.7 and n = 12. Step 2: Substitute the values into the formula: r = 7.7 / (12 - 1) = 7.7 / 11. Step 3: Calculate the quotient: 7.7 ÷ 11 = 0.7 exactly (since 7.7/11 = 77/110 = 7/10 = 0.7). Step 4: Interpret the direction: Since r = 0.7 is positive, there is a positive linear relationship. This means that as river depth increases, water flow velocity also tends to increase. Step 5: Interpret the strength: The absolute value |r| = 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 = 0.70, indicating a strong positive linear relationship between river depth and water flow velocity.

2. A scatter plot shows the relationship between the number of hours per week Aroha spends on creative writing (x) and her score on a vocabulary test (y) for 7 weeks. The data points are: (3, 51), (5, 59), (7, 67), (9, 75), (11, 83), (13, 91), (15, 99). Calculate the Pearson correlation coefficient r for this data, and interpret its strength and direction. Answer: 1.000 Solution: Compute the sums. n = 7.
   Full step-by-step solution

   Step 1: Compute the sums. n = 7. Σx = 3 + 5 + 7 + 9 + 11 + 13 + 15 = 63 Σy = 51 + 59 + 67 + 75 + 83 + 91 + 99 = 525 Σxy = 3*51 + 5*59 + 7*67 + 9*75 + 11*83 + 13*91 + 15*99 = 153 + 295 + 469 + 675 + 913 + 1183 + 1485 = 5173 Σx² = 3² + 5² + 7² + 9² + 11² + 13² + 15² = 9 + 25 + 49 + 81 + 121 + 169 + 225 = 679 Σy² = 51² + 59² + 67² + 75² + 83² + 91² + 99² = 2601 + 3481 + 4489 + 5625 + 6889 + 8281 + 9801 = 41167 Step 2: Use the Pearson formula: r = [n(Σxy) - (Σx)(Σy)] / sqrt{[n(Σx²) - (Σx)²][n(Σy²) - (Σy)²]} Numerator: n(Σxy) - (Σx)(Σy) = 7(5173) - (63)(525) = 36211 - 33075 = 3136 Denominator part 1: n(Σx²) - (Σx)² = 7(679) - (63)² = 4753 - 3969 = 784 Denominator part 2: n(Σy²) - (Σy)² = 7(41167) - (525)² = 288169 - 275625 = 12544 Product: 784 * 12544 = 9834496 Square root: sqrt(9834496) = 3136 Step 3: r = 3136 / 3136 = 1.000 Interpretation: r = 1.000 indicates a perfect positive linear relationship. As Aroha's creative writing hours increase, her vocabulary test score increases exactly along a straight line, with no scatter around the line.

3. Isabella, a marine biologist, is studying the relationship between the surface water temperature (in degrees Celsius) and the number of sea stars observed per square meter on a rocky shore. She collects data from 12 different tide pools and records the following summary statistics: the mean temperature is 17°C, the mean sea star count is 22 per m², the standard deviation of temperature is 3°C, the standard deviation of sea star count is 7 per m², and the sum of the products of the z-scores (Σzₓzᵧ) is 8.8. Compute the Pearson correlation coefficient r, and interpret the strength and direction of the linear relationship between surface water temperature and sea star abundance. Answer: r = 0.80, indicating a strong positive linear relationship between surface water temperature and sea star abundance. Solution: Recall the formula for the Pearson correlation coefficient using the sum of products of z-scores: r = (Σzₓzᵧ) / (n - 1). Here, Σzₓzᵧ = 8.8 and n = 12.
   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 = 8.8 and n = 12. Step 2: Substitute the values into the formula: r = 8.8 / (12 - 1) = 8.8 / 11. Step 3: Calculate the quotient: 8.8 ÷ 11 = 0.8. Step 4: Interpret the direction: Since r = 0.8 is positive, there is a positive linear relationship. This means that as surface water temperature increases, the number of sea stars per square meter also tends to increase. Step 5: Interpret the strength: The absolute value |r| = 0.8 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 = 0.80, indicating a strong positive linear relationship between surface water temperature and sea star abundance.

4. Given ∑x = 55, ∑y = 66, ∑xy = 902, ∑x² = 605, ∑y² = 968, n = 5, compute r = ? Answer: 1.0 Solution: Write the Pearson correlation formula: r = [n∑xy - (∑x)(∑y)] / sqrt([n∑x² - (∑x)²][n∑y² - (∑y)²]) Calculate the numerator: n∑xy - (∑x)(∑y) = 5×902 - 55×66 = 4510 - 3630 = 880 Calculate the first part of denominator: n∑x² - (∑x)² = 5×605 - 55² = 3025 - 3025 = 0 Calculate the second part of…
   Full step-by-step solution

   Step 1: Write the Pearson correlation formula: r = [n∑xy - (∑x)(∑y)] / sqrt([n∑x² - (∑x)²][n∑y² - (∑y)²]) Step 2: Calculate the numerator: n∑xy - (∑x)(∑y) = 5×902 - 55×66 = 4510 - 3630 = 880 Step 3: Calculate the first part of denominator: n∑x² - (∑x)² = 5×605 - 55² = 3025 - 3025 = 0 Step 4: Calculate the second part of denominator: n∑y² - (∑y)² = 5×968 - 66² = 4840 - 4356 = 484 Step 5: Calculate the full denominator: sqrt(0 × 484) = sqrt(0) = 0 Step 6: Since the denominator is 0, this indicates a perfect linear relationship. The correlation coefficient for a perfect positive linear relationship is 1.0. The answer is 1.0.

5. Mere, a kaitiaki (environmental guardian), is studying the relationship between the number of native trees planted per hectare (x) and the annual rainfall infiltration rate in millimeters per hour (y) across 11 different restoration sites. She calculates the following summary statistics: the mean number of trees planted per hectare is 240, the mean infiltration rate is 18 mm/hr, the standard deviation of tree count is 45 trees, the standard deviation of infiltration rate is 4 mm/hr, and the sum of the products of the z-scores (Σzₓzᵧ) is -7.2. Compute the Pearson correlation coefficient r, and interpret the strength and direction of the linear relationship between the number of native trees planted per hectare and the annual rainfall infiltration rate. Answer: r = -0.72, indicating a strong negative linear relationship between the number of native trees planted per hectare and the annual rainfall infiltration rate. Solution: Recall the formula for the Pearson correlation coefficient using the sum of the products of z-scores: r = (Σzₓzᵧ) / (n - 1). Here, Σzₓzᵧ = -7.2 and n = 11.
   Full step-by-step solution

   Step 1: Recall the formula for the Pearson correlation coefficient using the sum of the products of z-scores: r = (Σzₓzᵧ) / (n - 1). Here, Σzₓzᵧ = -7.2 and n = 11. Step 2: Substitute the values into the formula: r = (-7.2) / (11 - 1) = -7.2 / 10. Step 3: Calculate the quotient: -7.2 ÷ 10 = -0.72. Step 4: Interpret the direction: Since r = -0.72 is negative, there is a negative linear relationship. This means that as the number of native trees planted per hectare increases, the annual rainfall infiltration rate tends to decrease. Step 5: Interpret the strength: The absolute value |r| = 0.72 is greater than or equal to 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 = -0.72, indicating a strong negative linear relationship between the number of native trees planted per hectare and the annual rainfall infiltration rate.