Question 1188071
You're describing a situation where you need to perform a Kruskal-Wallis test. This is a non-parametric test used to compare three or more groups when the data is not assumed to be normally distributed.  Here's how to perform the test:

**1. State the Hypotheses:**

*   **Null Hypothesis (H0):** There is no significant difference in union loyalty scores among the three unions.
*   **Alternative Hypothesis (H1):** There is a significant difference in union loyalty scores among the three unions.

**2. Organize the Data and Rank:**

You'll need the actual data values for each of the 6 members from the three unions.  Let's assume you have the following data:

| Union 1 | Union 2 | Union 3 |
|---|---|---|
| 85 | 72 | 92 |
| 90 | 68 | 88 |
| 78 | 75 | 95 |
| 82 | 65 | 89 |
| 87 | 70 | 91 |
| 80 | 73 | 93 |

Now, combine all the scores and rank them from lowest to highest.  Give tied scores the average of the ranks they would have received.

| Score | Rank |
|---|---|
| 65 | 1 |
| 68 | 2 |
| 70 | 3 |
| 72 | 4 |
| 73 | 5 |
| 75 | 6 |
| 78 | 7 |
| 80 | 8 |
| 82 | 9 |
| 85 | 10 |
| 87 | 11 |
| 88 | 12 |
| 89 | 13 |
| 90 | 14 |
| 91 | 15 |
| 92 | 16 |
| 93 | 17 |
| 95 | 18 |

**3. Calculate the Rank Sums:**

Add up the ranks for each union:

*   Union 1: 10 + 14 + 7 + 9 + 11 + 8 = 59
*   Union 2: 4 + 2 + 3 + 1 + 5 + 6 = 21
*   Union 3: 16 + 12 + 18 + 13 + 15 + 17 = 91

**4. Calculate the Kruskal-Wallis H Statistic:**

H = (12 / (N(N+1))) * Σ(Ri² / ni) - 3(N+1)

Where:

*   N = Total number of observations (6 + 6 + 6 = 18)
*   ni = Number of observations in each group (6)
*   Ri = Sum of ranks for each group

H = (12 / (18 * 19)) * [(59² / 6) + (21² / 6) + (91² / 6)] - 3(18 + 1)
H = (12 / 342) * [580.17 + 73.5 + 1380.17] - 57
H ≈ 0.0351 * 2033.84 - 57
H ≈ 71.49 - 57
H ≈ 14.49

**5. Determine the Degrees of Freedom:**

df = Number of groups - 1 = 3 - 1 = 2

**6. Find the Critical Value:**

Consult a chi-square distribution table or use a calculator. For α = 0.05 and df = 2, the critical value is approximately 5.991.

**7. Make a Decision:**

Our calculated H statistic (14.49) is *greater* than the critical value (5.991).  Therefore, we *reject* the null hypothesis.

**8. Conclusion:**

There is sufficient evidence at the 0.05 significance level to conclude that there is a significant difference in union loyalty scores among the three unions.