Unit 2: Solved Numerical Problems
This section contains 15+ fully solved problems covering correlation and regression with detailed step-by-step solutions.
Section A: Karl Pearson’s Correlation Coefficient
Problem 1: Correlation (Direct Method)
Step 1: Create calculation table
| X | Y | X² | Y² | XY |
|---|---|---|---|---|
| 10 | 50 | 100 | 2500 | 500 |
| 15 | 60 | 225 | 3600 | 900 |
| 20 | 65 | 400 | 4225 | 1300 |
| 25 | 75 | 625 | 5625 | 1875 |
| 30 | 85 | 900 | 7225 | 2550 |
| ΣX=100 | ΣY=335 | ΣX²=2250 | ΣY²=23175 | ΣXY=7125 |
Step 2: Note the values
- n = 5
- ΣX = 100, ΣY = 335
- ΣX² = 2250, ΣY² = 23175
- ΣXY = 7125
Step 3: Apply the formula \(r = \frac{n\sum XY - \sum X \sum Y}{\sqrt{[n\sum X^2 - (\sum X)^2][n\sum Y^2 - (\sum Y)^2]}}\)
Step 4: Calculate numerator \(n\sum XY - \sum X \sum Y = 5(7125) - (100)(335)\) \(= 35625 - 33500 = 2125\)
Step 5: Calculate denominator \(n\sum X^2 - (\sum X)^2 = 5(2250) - (100)^2 = 11250 - 10000 = 1250\) \(n\sum Y^2 - (\sum Y)^2 = 5(23175) - (335)^2 = 115875 - 112225 = 3650\)
\[\sqrt{1250 \times 3650} = \sqrt{4562500} = 2136.0\]Step 6: Calculate r \(r = \frac{2125}{2136.0} = 0.995\)
Answer: r = 0.995 (Very strong positive correlation)
Interpretation: There is a very strong positive linear relationship between advertising spend and sales. As advertising increases, sales increase proportionally.
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Question: Calculate Karl Pearson’s correlation coefficient for the following data:
| X (Advertising Spend in ‘000) | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|
| Y (Sales in ‘000) | 50 | 60 | 65 | 75 | 85 |
### Problem 2: Correlation (Deviation Method)
Click to reveal solution
**Step 1:** Calculate means $$\bar{X} = \frac{75}{5} = 15, \quad \bar{Y} = \frac{120}{5} = 24$$ **Step 2:** Create deviation table | X | Y | x = X - 15 | y = Y - 24 | x² | y² | xy | | --------- | --- | ---------- | ---------- | ------- | ------- | ------- | | 5 | 12 | -10 | -12 | 100 | 144 | 120 | | 10 | 18 | -5 | -6 | 25 | 36 | 30 | | 15 | 25 | 0 | 1 | 0 | 1 | 0 | | 20 | 30 | 5 | 6 | 25 | 36 | 30 | | 25 | 35 | 10 | 11 | 100 | 121 | 110 | | **Total** | | **0** | **0** | **250** | **338** | **290** | **Step 3:** Apply formula $$r = \frac{\sum xy}{\sqrt{\sum x^2 \times \sum y^2}} = \frac{290}{\sqrt{250 \times 338}}$$ $$= \frac{290}{\sqrt{84500}} = \frac{290}{290.69} = 0.998$$ **Answer:** r = **0.998** (Very strong positive correlation)Question: Calculate correlation coefficient using deviation method:
| X | 5 | 10 | 15 | 20 | 25 |
|---|---|---|---|---|---|
| Y | 12 | 18 | 25 | 30 | 35 |
### Problem 3: Correlation (Assumed Mean Method)
Click to reveal solution
**Step 1:** Create table with assumed deviations | X | Y | dx = X - 30 | dy = Y - 50 | dx² | dy² | dx.dy | | --------- | --- | ----------- | ----------- | ------- | ------- | ------- | | 20 | 40 | -10 | -10 | 100 | 100 | 100 | | 25 | 45 | -5 | -5 | 25 | 25 | 25 | | 30 | 52 | 0 | 2 | 0 | 4 | 0 | | 35 | 55 | 5 | 5 | 25 | 25 | 25 | | 40 | 58 | 10 | 8 | 100 | 64 | 80 | | **Total** | | **0** | **0** | **250** | **218** | **230** | **Step 2:** Apply formula (same as deviation method since Σdx = Σdy = 0) $$r = \frac{\sum dx.dy}{\sqrt{\sum dx^2 \times \sum dy^2}} = \frac{230}{\sqrt{250 \times 218}}$$ $$= \frac{230}{\sqrt{54500}} = \frac{230}{233.45} = 0.985$$ **Answer:** r = **0.985**Question: Calculate r using assumed mean method with Ax = 30 and Ay = 50:
| X | 20 | 25 | 30 | 35 | 40 |
|---|---|---|---|---|---|
| Y | 40 | 45 | 52 | 55 | 58 |
### Problem 4: Negative Correlation
Click to reveal solution
**Step 1:** Create calculation table | X | Y | X² | Y² | XY | | ------- | ------- | -------- | --------- | -------- | | 10 | 100 | 100 | 10000 | 1000 | | 15 | 85 | 225 | 7225 | 1275 | | 20 | 70 | 400 | 4900 | 1400 | | 25 | 55 | 625 | 3025 | 1375 | | 30 | 40 | 900 | 1600 | 1200 | | **100** | **350** | **2250** | **26750** | **6250** | **Step 2:** Apply formula $$r = \frac{5(6250) - (100)(350)}{\sqrt{[5(2250) - 10000][5(26750) - 122500]}}$$ $$= \frac{31250 - 35000}{\sqrt{[11250 - 10000][133750 - 122500]}}$$ $$= \frac{-3750}{\sqrt{1250 \times 11250}} = \frac{-3750}{\sqrt{14062500}}$$ $$= \frac{-3750}{3750} = -1.00$$ **Answer:** r = **-1.00** (Perfect negative correlation) **Interpretation:** Price and demand have a perfect inverse relationship. As price increases, demand decreases proportionally.Question: Calculate correlation for price and demand:
| Price (Rs.) | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|
| Demand (Units) | 100 | 85 | 70 | 55 | 40 |
### Problem 5: Correlation from Grouped Data
Click to reveal solution
**Step 1:** Assign mid-values and step deviations - X mid-values: 5, 15, 25; let ux = (X - 15)/10 - Y mid-values: 15, 25, 35; let uy = (Y - 25)/10 **Step 2:** Create calculation table | Cell | f | ux | uy | fux | fuy | fux² | fuy² | fuxuy | | --------- | ------ | --- | --- | ----- | ----- | ------ | ------ | ------ | | (5,15) | 8 | -1 | -1 | -8 | -8 | 8 | 8 | 8 | | (5,25) | 5 | -1 | 0 | -5 | 0 | 5 | 0 | 0 | | (5,35) | 2 | -1 | 1 | -2 | 2 | 2 | 2 | -2 | | (15,15) | 4 | 0 | -1 | 0 | -4 | 0 | 4 | 0 | | (15,25) | 12 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | | (15,35) | 6 | 0 | 1 | 0 | 6 | 0 | 6 | 0 | | (25,15) | 3 | 1 | -1 | 3 | -3 | 3 | 3 | -3 | | (25,25) | 8 | 1 | 0 | 8 | 0 | 8 | 0 | 0 | | (25,35) | 12 | 1 | 1 | 12 | 12 | 12 | 12 | 12 | | **Total** | **60** | | | **8** | **5** | **38** | **35** | **15** | **Step 3:** Apply formula $$r = \frac{n\sum fuxuy - \sum fux \times \sum fuy}{\sqrt{[n\sum fux^2 - (\sum fux)^2][n\sum fuy^2 - (\sum fuy)^2]}}$$ $$r = \frac{60(15) - (8)(5)}{\sqrt{[60(38) - 64][60(35) - 25]}}$$ $$= \frac{900 - 40}{\sqrt{[2280 - 64][2100 - 25]}} = \frac{860}{\sqrt{2216 \times 2075}}$$ $$= \frac{860}{\sqrt{4598200}} = \frac{860}{2144.3} = 0.401$$ **Answer:** r = **0.401** (Moderate positive correlation)Question: Find correlation coefficient:
| X \ Y | 10-20 | 20-30 | 30-40 | Total |
|---|---|---|---|---|
| 0-10 | 8 | 5 | 2 | 15 |
| 10-20 | 4 | 12 | 6 | 22 |
| 20-30 | 3 | 8 | 12 | 23 |
| Total | 15 | 25 | 20 | 60 |
## Section B: Spearman's Rank Correlation
### Problem 6: Rank Correlation (No Ties)
Click to reveal solution
**Step 1:** Create difference table | Participant | R₁ | R₂ | d = R₁ - R₂ | d² | | ----------- | --- | --- | ----------- | ------ | | A | 1 | 2 | -1 | 1 | | B | 3 | 1 | 2 | 4 | | C | 2 | 3 | -1 | 1 | | D | 5 | 6 | -1 | 1 | | E | 4 | 5 | -1 | 1 | | F | 7 | 8 | -1 | 1 | | G | 6 | 4 | 2 | 4 | | H | 8 | 7 | 1 | 1 | | **Total** | | | | **14** | **Step 2:** Apply Spearman's formula $$\rho = 1 - \frac{6\sum d^2}{n(n^2 - 1)}$$ **Step 3:** Substitute values $$\rho = 1 - \frac{6 \times 14}{8(64 - 1)} = 1 - \frac{84}{8 \times 63}$$ $$= 1 - \frac{84}{504} = 1 - 0.167 = 0.833$$ **Answer:** ρ = **0.833** (Strong positive correlation) **Interpretation:** The two judges show strong agreement in their rankings.Question: Two judges ranked 8 participants in a competition:
| Participant | A | B | C | D | E | F | G | H |
|---|---|---|---|---|---|---|---|---|
| Judge 1 | 1 | 3 | 2 | 5 | 4 | 7 | 6 | 8 |
| Judge 2 | 2 | 1 | 3 | 6 | 5 | 8 | 4 | 7 |
Calculate Spearman’s rank correlation coefficient.
### Problem 7: Rank Correlation with Tied Ranks
**Question:** Calculate Spearman's correlation for:
| Student | Test 1 | Test 2 |
| ------- | ------ | ------ |
| A | 85 | 90 |
| B | 70 | 72 |
| C | 85 | 85 |
| D | 60 | 65 |
| E | 75 | 72 |
| F | 90 | 88 |
---
### Problem 8: Converting Raw Data to Ranks
**Question:** Calculate Spearman's correlation:
| X (Training Hours) | 10 | 15 | 8 | 20 | 12 | 18 |
| --------------------- | --- | --- | --- | --- | --- | --- |
| Y (Performance Score) | 70 | 82 | 65 | 90 | 75 | 85 |
---
## Section C: Simple Linear Regression
### Problem 9: Regression Line of Y on X
**Question:** Find the regression equation of Y on X:
| X (Experience in years) | 2 | 4 | 6 | 8 | 10 |
| ----------------------- | --- | --- | --- | --- | --- |
| Y (Salary in lakhs) | 3 | 5 | 6 | 8 | 10 |
---
### Problem 10: Regression Line of X on Y
Click to reveal solution
**Step 1:** Assign ranks (average for ties) **Test 1 ranking:** - 90 → Rank 1 - 85 (two values) → Average of 2,3 = 2.5 - 75 → Rank 4 - 70 → Rank 5 - 60 → Rank 6 **Test 2 ranking:** - 90 → Rank 1 - 88 → Rank 2 - 85 → Rank 3 - 72 (two values) → Average of 4,5 = 4.5 - 65 → Rank 6 **Step 2:** Create table | Student | Test 1 | R₁ | Test 2 | R₂ | d | d² | | --------- | ------ | --- | ------ | --- | ---- | -------- | | A | 85 | 2.5 | 90 | 1 | 1.5 | 2.25 | | B | 70 | 5 | 72 | 4.5 | 0.5 | 0.25 | | C | 85 | 2.5 | 85 | 3 | -0.5 | 0.25 | | D | 60 | 6 | 65 | 6 | 0 | 0 | | E | 75 | 4 | 72 | 4.5 | -0.5 | 0.25 | | F | 90 | 1 | 88 | 2 | -1 | 1 | | **Total** | | | | | | **4.00** | **Step 3:** Apply formula with correction for ties $$\rho = 1 - \frac{6[\sum d^2 + \frac{m_1(m_1^2-1)}{12} + \frac{m_2(m_2^2-1)}{12}]}{n(n^2 - 1)}$$ Where m₁ = number of tied items in Test 1 = 2, m₂ = 2 $$\rho = 1 - \frac{6[4 + \frac{2(4-1)}{12} + \frac{2(4-1)}{12}]}{6(36 - 1)}$$ $$= 1 - \frac{6[4 + 0.5 + 0.5]}{6 \times 35} = 1 - \frac{6 \times 5}{210}$$ $$= 1 - \frac{30}{210} = 1 - 0.143 = 0.857$$ **Answer:** ρ = **0.857** (Strong positive correlation)Click to reveal solution
**Step 1:** Assign ranks | X | Rank of X | Y | Rank of Y | | --- | --------- | --- | --------- | | 10 | 4 | 70 | 4 | | 15 | 3 | 82 | 3 | | 8 | 6 | 65 | 6 | | 20 | 1 | 90 | 1 | | 12 | 4 | 75 | 4 | | 18 | 2 | 85 | 2 | Wait, let me re-rank correctly: **Ranking X (highest = 1):** 20→1, 18→2, 15→3, 12→4, 10→5, 8→6 **Ranking Y (highest = 1):** 90→1, 85→2, 82→3, 75→4, 70→5, 65→6 **Step 2:** Create difference table | X | R₁ | Y | R₂ | d | d² | | --------- | --- | --- | --- | --- | ----- | | 10 | 5 | 70 | 5 | 0 | 0 | | 15 | 3 | 82 | 3 | 0 | 0 | | 8 | 6 | 65 | 6 | 0 | 0 | | 20 | 1 | 90 | 1 | 0 | 0 | | 12 | 4 | 75 | 4 | 0 | 0 | | 18 | 2 | 85 | 2 | 0 | 0 | | **Total** | | | | | **0** | **Step 3:** Calculate $$\rho = 1 - \frac{6 \times 0}{6(36 - 1)} = 1 - 0 = 1.00$$ **Answer:** ρ = **1.00** (Perfect positive correlation) **Interpretation:** Training hours and performance have a perfect positive rank relationship.Click to reveal solution
**Step 1:** Create calculation table | X | Y | X² | XY | | --------- | --------- | ----------- | ----------- | | 2 | 3 | 4 | 6 | | 4 | 5 | 16 | 20 | | 6 | 6 | 36 | 36 | | 8 | 8 | 64 | 64 | | 10 | 10 | 100 | 100 | | **ΣX=30** | **ΣY=32** | **ΣX²=220** | **ΣXY=226** | **Step 2:** Calculate means $$\bar{X} = \frac{30}{5} = 6, \quad \bar{Y} = \frac{32}{5} = 6.4$$ **Step 3:** Calculate slope (b) $$b = \frac{n\sum XY - \sum X \sum Y}{n\sum X^2 - (\sum X)^2}$$ $$b = \frac{5(226) - (30)(32)}{5(220) - (30)^2} = \frac{1130 - 960}{1100 - 900}$$ $$= \frac{170}{200} = 0.85$$ **Step 4:** Calculate intercept (a) $$a = \bar{Y} - b\bar{X} = 6.4 - 0.85(6) = 6.4 - 5.1 = 1.3$$ **Step 5:** Write regression equation $$\hat{Y} = a + bX = 1.3 + 0.85X$$ **Answer:** The regression equation is **Ŷ = 1.3 + 0.85X** **Interpretation:** - For each additional year of experience, salary increases by Rs. 0.85 lakhs - A person with 0 years experience would have salary of Rs. 1.3 lakhs (base salary) **Prediction Example:** Expected salary for 5 years experience: $$\hat{Y} = 1.3 + 0.85(5) = 1.3 + 4.25 = 5.55 \text{ lakhs}$$Click to reveal solution
**Step 1:** Calculate additional values needed | X | Y | Y² | XY | | ------ | ------ | ------- | ------- | | 2 | 3 | 9 | 6 | | 4 | 5 | 25 | 20 | | 6 | 6 | 36 | 36 | | 8 | 8 | 64 | 64 | | 10 | 10 | 100 | 100 | | **30** | **32** | **234** | **226** | **Step 2:** Calculate slope (b') $$b' = \frac{n\sum XY - \sum X \sum Y}{n\sum Y^2 - (\sum Y)^2}$$ $$b' = \frac{5(226) - (30)(32)}{5(234) - (32)^2} = \frac{1130 - 960}{1170 - 1024}$$ $$= \frac{170}{146} = 1.164$$ **Step 3:** Calculate intercept (a') $$a' = \bar{X} - b'\bar{Y} = 6 - 1.164(6.4) = 6 - 7.45 = -1.45$$ **Step 4:** Write regression equation $$\hat{X} = -1.45 + 1.164Y$$ **Answer:** The regression equation of X on Y is **X̂ = -1.45 + 1.164Y** **Prediction Example:** Expected experience for salary of 7 lakhs: $$\hat{X} = -1.45 + 1.164(7) = -1.45 + 8.15 = 6.7 \text{ years}$$Question: Using the data from Problem 9, find the regression equation of X on Y.
### Problem 11: Complete Regression Analysis
**Question:** For the following data, find:
a) Both regression equations
b) Correlation coefficient from regression coefficients
c) Predicted Y when X = 50
| X | 20 | 30 | 40 | 50 | 60 |
| --- | --- | --- | --- | --- | --- |
| Y | 15 | 25 | 35 | 40 | 50 |
---
### Problem 12: Regression Through Origin
**Question:** If regression line passes through origin, find b for:
| X | 2 | 4 | 6 | 8 | 10 |
| --- | --- | --- | --- | --- | --- |
| Y | 5 | 11 | 16 | 22 | 26 |
---
### Problem 13: Finding Missing Values Using Regression
**Question:** Given:
- Regression of Y on X: Y = 0.5X + 10
- Regression of X on Y: X = 0.8Y + 5
- Variance of X = 16
Find: (a) Means of X and Y, (b) Correlation coefficient, (c) Standard deviation of Y
---
### Problem 14: Standard Error of Estimate
**Question:** Calculate the standard error of estimate for the regression of Y on X:
| X | 1 | 2 | 3 | 4 | 5 |
| --- | --- | --- | --- | --- | --- |
| Y | 4 | 5 | 7 | 8 | 11 |
---
### Problem 15: Coefficient of Determination
Click to reveal solution
**Step 1:** Calculate necessary sums | X | Y | X² | Y² | XY | | ------- | ------- | -------- | -------- | -------- | | 20 | 15 | 400 | 225 | 300 | | 30 | 25 | 900 | 625 | 750 | | 40 | 35 | 1600 | 1225 | 1400 | | 50 | 40 | 2500 | 1600 | 2000 | | 60 | 50 | 3600 | 2500 | 3000 | | **200** | **165** | **9000** | **6175** | **7450** | $$\bar{X} = 40, \quad \bar{Y} = 33$$ **Step 2:** Calculate regression coefficient of Y on X (byx) $$b_{yx} = \frac{5(7450) - (200)(165)}{5(9000) - (200)^2} = \frac{37250 - 33000}{45000 - 40000}$$ $$= \frac{4250}{5000} = 0.85$$ **Step 3:** Regression equation of Y on X $$a = 33 - 0.85(40) = 33 - 34 = -1$$ $$\hat{Y} = -1 + 0.85X$$ **Step 4:** Calculate regression coefficient of X on Y (bxy) $$b_{xy} = \frac{5(7450) - (200)(165)}{5(6175) - (165)^2} = \frac{4250}{30875 - 27225}$$ $$= \frac{4250}{3650} = 1.164$$ **Step 5:** Regression equation of X on Y $$a' = 40 - 1.164(33) = 40 - 38.41 = 1.59$$ $$\hat{X} = 1.59 + 1.164Y$$ **Step 6:** Calculate r from regression coefficients $$r = \pm\sqrt{b_{yx} \times b_{xy}} = \pm\sqrt{0.85 \times 1.164}$$ $$= \pm\sqrt{0.989} = \pm 0.995$$ Since both regression coefficients are positive, r is positive. $$r = +0.995$$ **Step 7:** Predict Y when X = 50 $$\hat{Y} = -1 + 0.85(50) = -1 + 42.5 = 41.5$$ **Answers:** - a) Y on X: **Ŷ = -1 + 0.85X**; X on Y: **X̂ = 1.59 + 1.164Y** - b) r = **0.995** - c) When X = 50, Y = **41.5**Click to reveal solution
When regression passes through origin (a = 0): **Step 1:** Formula simplifies to $$b = \frac{\sum XY}{\sum X^2}$$ **Step 2:** Calculate | X | Y | X² | XY | | ------ | ------ | ------- | ------- | | 2 | 5 | 4 | 10 | | 4 | 11 | 16 | 44 | | 6 | 16 | 36 | 96 | | 8 | 22 | 64 | 176 | | 10 | 26 | 100 | 260 | | **30** | **80** | **220** | **586** | $$b = \frac{586}{220} = 2.664$$ **Answer:** Regression equation is **Y = 2.664X**Click to reveal solution
**Part (a): Finding Means** Both regression lines pass through (X̄, Ȳ): From Y on X: Ȳ = 0.5X̄ + 10 ... (1) From X on Y: X̄ = 0.8Ȳ + 5 ... (2) Substituting (1) in (2): $$X̄ = 0.8(0.5X̄ + 10) + 5 = 0.4X̄ + 8 + 5$$ $$X̄ - 0.4X̄ = 13$$ $$0.6X̄ = 13$$ $$X̄ = 21.67$$ From (1): Ȳ = 0.5(21.67) + 10 = 20.83 **Part (b): Correlation Coefficient** $$r = \pm\sqrt{b_{yx} \times b_{xy}} = \pm\sqrt{0.5 \times 0.8} = \pm\sqrt{0.4} = \pm 0.632$$ Since both coefficients are positive, r = **+0.632** **Part (c): Standard Deviation of Y** We know: $b_{yx} = r \times \frac{\sigma_Y}{\sigma_X}$ Given: $\sigma_X = \sqrt{16} = 4$ $$0.5 = 0.632 \times \frac{\sigma_Y}{4}$$ $$\sigma_Y = \frac{0.5 \times 4}{0.632} = \frac{2}{0.632} = 3.16$$ **Answers:** - (a) X̄ = **21.67**, Ȳ = **20.83** - (b) r = **0.632** - (c) σY = **3.16**Click to reveal solution
**Step 1:** Find regression equation first | X | Y | X² | XY | | ------ | ------ | ------ | ------- | | 1 | 4 | 1 | 4 | | 2 | 5 | 4 | 10 | | 3 | 7 | 9 | 21 | | 4 | 8 | 16 | 32 | | 5 | 11 | 25 | 55 | | **15** | **35** | **55** | **122** | $$b = \frac{5(122) - 15(35)}{5(55) - 225} = \frac{610 - 525}{275 - 225} = \frac{85}{50} = 1.7$$ $$a = 7 - 1.7(3) = 7 - 5.1 = 1.9$$ Equation: Ŷ = 1.9 + 1.7X **Step 2:** Calculate predicted values and errors | X | Y | Ŷ = 1.9 + 1.7X | Y - Ŷ | (Y - Ŷ)² | | --------- | --- | -------------- | ----- | -------- | | 1 | 4 | 3.6 | 0.4 | 0.16 | | 2 | 5 | 5.3 | -0.3 | 0.09 | | 3 | 7 | 7.0 | 0 | 0 | | 4 | 8 | 8.7 | -0.7 | 0.49 | | 5 | 11 | 10.4 | 0.6 | 0.36 | | **Total** | | | | **1.10** | **Step 3:** Calculate standard error $$S_{yx} = \sqrt{\frac{\sum(Y - \hat{Y})^2}{n-2}} = \sqrt{\frac{1.10}{3}} = \sqrt{0.367} = 0.606$$ **Answer:** Standard error of estimate = **0.606**Click to reveal solution
**Step 1:** Calculate total variation (SST) | Y | Y - Ȳ | (Y - Ȳ)² | | --------- | ----- | -------- | | 4 | -3 | 9 | | 5 | -2 | 4 | | 7 | 0 | 0 | | 8 | 1 | 1 | | 11 | 4 | 16 | | **Total** | | **30** | SST (Total Sum of Squares) = 30 **Step 2:** Calculate explained variation (SSR) | Ŷ | Ŷ - Ȳ | (Ŷ - Ȳ)² | | --------- | ----- | --------- | | 3.6 | -3.4 | 11.56 | | 5.3 | -1.7 | 2.89 | | 7.0 | 0 | 0 | | 8.7 | 1.7 | 2.89 | | 10.4 | 3.4 | 11.56 | | **Total** | | **28.90** | SSR (Regression Sum of Squares) = 28.90 **Step 3:** Calculate unexplained variation (SSE) SSE = SST - SSR = 30 - 28.90 = 1.10 (same as Σ(Y - Ŷ)²) **Step 4:** Calculate R² $$R^2 = \frac{SSR}{SST} = \frac{28.90}{30} = 0.963$$ Or: $R^2 = 1 - \frac{SSE}{SST} = 1 - \frac{1.10}{30} = 0.963$ **Answers:** - a) R² = **0.963** or **96.3%** - b) Explained variation = **28.90**, Unexplained = **1.10** **Interpretation:** 96.3% of the variation in Y is explained by the linear relationship with X.Question: From Problem 14, calculate: a) Coefficient of determination (R²) b) Explained and unexplained variation
## Practice Problems
1. Calculate Karl Pearson's r for: X = {5, 8, 12, 15, 20}, Y = {25, 30, 38, 42, 50}
2. Find Spearman's rank correlation:
| Student | Maths | Physics |
|---------|-------|---------|
| A | 78 | 84 |
| B | 65 | 62 |
| C | 82 | 78 |
| D | 55 | 58 |
| E | 90 | 92 |
3. Find both regression equations for: X = {10, 20, 30, 40, 50}, Y = {25, 32, 45, 52, 60}
4. If byx = 0.6 and bxy = 0.4, find r.
5. Given regression equations Y = 2X + 5 and X = 0.4Y + 2, find means of X and Y.
---
## Summary of Formulas
| Measure | Formula |
| ---------------------------- | ---------------------------------------------------------------------------------------------- |
| **Karl Pearson's r** | $r = \frac{n\sum XY - \sum X \sum Y}{\sqrt{[n\sum X^2 - (\sum X)^2][n\sum Y^2 - (\sum Y)^2]}}$ |
| **Spearman's ρ** | $\rho = 1 - \frac{6\sum d^2}{n(n^2-1)}$ |
| **Regression slope (b)** | $b = \frac{n\sum XY - \sum X \sum Y}{n\sum X^2 - (\sum X)^2}$ |
| **Regression intercept (a)** | $a = \bar{Y} - b\bar{X}$ |
| **r from regression** | $r = \pm\sqrt{b_{yx} \times b_{xy}}$ |
| **Standard Error** | $S_{yx} = \sqrt{\frac{\sum(Y-\hat{Y})^2}{n-2}}$ |
| **R²** | $R^2 = \frac{SSR}{SST} = r^2$ |
