Spearman Correlation
1. **State the problem:** We need to calculate the Spearman Correlation coefficient $r_s$ for the given data of hours studied and grades obtained by eight students.
2. **List the data:**
| Hours studied (x) | Grade (y) |
|-------------------|-----------|
| 11 | 79 |
| 8 | 56 |
| 13 | 72 |
| 18 | 95 |
| 8 | 70 |
| 5 | 54 |
| 2 | 33 |
| 5 | 44 |
3. **Assign ranks to x (hours studied):**
- Sort x values: 2, 5, 5, 8, 8, 11, 13, 18
- Ranks: 2 → 1
5 → (2+3)/2 = 2.5 (average rank for ties)
8 → (4+5)/2 = 4.5
11 → 6
13 → 7
18 → 8
So ranks for x are:
| x | Rank_x |
|---|--------|
| 11 | 6 |
| 8 | 4.5 |
| 13 | 7 |
| 18 | 8 |
| 8 | 4.5 |
| 5 | 2.5 |
| 2 | 1 |
| 5 | 2.5 |
4. **Assign ranks to y (grades):**
- Sort y values: 33, 44, 54, 56, 70, 72, 79, 95
- Ranks:
33 → 1
44 → 2
54 → 3
56 → 4
70 → 5
72 → 6
79 → 7
95 → 8
So ranks for y are:
| y | Rank_y |
|----|--------|
| 79 | 7 |
| 56 | 4 |
| 72 | 6 |
| 95 | 8 |
| 70 | 5 |
| 54 | 3 |
| 33 | 1 |
| 44 | 2 |
5. **Calculate differences $d = \text{Rank}_x - \text{Rank}_y$ and $d^2$:**
| Rank_x | Rank_y | $d$ | $d^2$ |
|--------|--------|-----|-------|
| 6 | 7 | -1 | 1 |
| 4.5 | 4 | 0.5 | 0.25 |
| 7 | 6 | 1 | 1 |
| 8 | 8 | 0 | 0 |
| 4.5 | 5 | -0.5| 0.25 |
| 2.5 | 3 | -0.5| 0.25 |
| 1 | 1 | 0 | 0 |
| 2.5 | 2 | 0.5 | 0.25 |
Sum of $d^2 = 1 + 0.25 + 1 + 0 + 0.25 + 0.25 + 0 + 0.25 = 3$.
6. **Apply the Spearman correlation formula:**
$$r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)}$$
where $n=8$.
Calculate denominator:
$$n(n^2 - 1) = 8(64 - 1) = 8 \times 63 = 504$$
Calculate numerator:
$$6 \sum d^2 = 6 \times 3 = 18$$
Calculate $r_s$:
$$r_s = 1 - \frac{18}{504} = 1 - 0.0357 = 0.9643$$
7. **Final answer:** The Spearman Correlation coefficient is approximately $r_s = 0.964$.
This indicates a very strong positive correlation between hours studied and grades obtained.