Math Problem Statement
calculate using actual mean method and direct method
Calculate Karl Pearson's coefficient of correlation between x and y using the actual mean method and direct method.
Solution
To calculate Karl Pearson's coefficient of correlation between and , we will proceed with both the actual mean method and the direct method.
Given dataTo calculate Karl Pearson's coefficient of correlation between and , we will proceed with both the actual mean method and the direct method.
Given
x & 4 & 6 & 8 & 10 & 12 \\ y & 2 & 3 & 4 & 6 & 10 \\ \end{array}$$ ### Step 1: Calculations for the Actual Mean Method 1. **Find the mean of $$x$$ and $$y$$:** $$\bar{x} = \frac{4 + 6 + 8 + 10 + 12}{5} = 8$$ $$\bar{y} = \frac{2 + 3 + 4 + 6 + 10}{5} = 5$$ 2. **Calculate deviations from the mean** for each value: - $$d_x = x - \bar{x}$$ - $$d_y = y - \bar{y}$$ $$d_x: -4, -2, 0, 2, 4$$ $$d_y: -3, -2, -1, 1, 5$$ 3. **Calculate $$d_x \times d_y$$, $$d_x^2$$, and $$d_y^2$$:** $$d_x \times d_y: 12, 4, 0, 2, 20$$ $$d_x^2: 16, 4, 0, 4, 16$$ $$d_y^2: 9, 4, 1, 1, 25$$ 4. **Sum up these values:** $$\sum d_x \times d_y = 12 + 4 + 0 + 2 + 20 = 38$$ $$\sum d_x^2 = 16 + 4 + 0 + 4 + 16 = 40$$ $$\sum d_y^2 = 9 + 4 + 1 + 1 + 25 = 40$$ 5. **Apply the correlation formula:** $$r = \frac{\sum d_x d_y}{\sqrt{\sum d_x^2 \cdot \sum d_y^2}}$$ Substituting the values: $$r = \frac{38}{\sqrt{40 \times 40}} = \frac{38}{40} = 0.95$$ ### Step 2: Calculations for the Direct Method In the direct method, we use the formula: $$r = \frac{n \sum xy - \sum x \sum y}{\sqrt{n \sum x^2 - (\sum x)^2} \cdot \sqrt{n \sum y^2 - (\sum y)^2}}$$ Where: - $$n = 5$$ (number of pairs) - $$\sum x = 4 + 6 + 8 + 10 + 12 = 40$$ - $$\sum y = 2 + 3 + 4 + 6 + 10 = 25$$ - $$\sum x^2 = 4^2 + 6^2 + 8^2 + 10^2 + 12^2 = 16 + 36 + 64 + 100 + 144 = 360$$ - $$\sum y^2 = 2^2 + 3^2 + 4^2 + 6^2 + 10^2 = 4 + 9 + 16 + 36 + 100 = 165$$ - $$\sum xy = (4 \times 2) + (6 \times 3) + (8 \times 4) + (10 \times 6) + (12 \times 10) = 8 + 18 + 32 + 60 + 120 = 238$$ Substitute these values into the formula: $$r = \frac{5 \times 238 - (40 \times 25)}{\sqrt{5 \times 360 - 40^2} \cdot \sqrt{5 \times 165 - 25^2}}$$ Calculating step-by-step: $$r = \frac{1190 - 1000}{\sqrt{1800 - 1600} \cdot \sqrt{825 - 625}}$$ $$r = \frac{190}{\sqrt{200} \cdot \sqrt{200}}$$ $$r = \frac{190}{200} = 0.95$$ ### Final Answer: The Karl Pearson’s coefficient of correlation $$r$$ between $$x$$ and $$y$$ is **0.95** using both methods. Would you like any further details on these steps, or do you have any questions? Here are 5 related questions you might find useful: 1. What does a correlation coefficient of 0.95 indicate about the relationship between $$x$$ and $$y$$? 2. How would you interpret a negative correlation coefficient? 3. Can you explain how the actual mean method and direct method are different conceptually? 4. What are the limitations of Pearson's correlation coefficient? 5. How would outliers affect the correlation coefficient? **Tip:** To confirm calculations, try breaking down each step and verifying intermediate sums and products before proceeding to the final formula.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Statistics
Correlation
Karl Pearson's coefficient of correlation
Formulas
r = Σ(dx * dy) / √(Σ(dx^2) * Σ(dy^2))
r = [nΣ(xy) - Σ(x)Σ(y)] / [√(nΣ(x^2) - (Σ(x))^2) * √(nΣ(y^2) - (Σ(y))^2)]
Theorems
Karl Pearson's correlation coefficient formula
Suitable Grade Level
Grades 11-12