Question 1170416
Let's address each part of the question.

**(i) When do you apply the analysis of variance (ANOVA) technique?**

The analysis of variance (ANOVA) technique is used to test the hypothesis that the means of two or more populations are equal. Specifically, you would apply ANOVA when:

* **You have more than two groups (or populations) to compare:** If you only have two groups, a t-test is appropriate. ANOVA is designed for situations where you need to compare three or more groups.
* **You want to determine if there are statistically significant differences between the group means:** ANOVA helps you determine if the observed differences between the sample means are likely due to real differences in the population means or simply due to random chance.
* **The dependent variable is continuous:** The variable you are measuring should be a continuous variable (e.g., height, weight, test scores).
* **The independent variable is categorical:** The independent variable represents the groups or categories you are comparing (e.g., different treatment groups, different teaching methods).
* **The assumptions of ANOVA are met:** These assumptions include:
    * Normality: The data within each group should be approximately normally distributed.
    * Homogeneity of variances: The variances of the populations should be equal.
    * Independence: The observations should be independent of each other.

**(ii) Testing the hypothesis of equal population means**

We will perform a one-way ANOVA test.

**1. Set up the hypotheses:**

* Null hypothesis (H₀): μ₁ = μ₂ = μ₃ (The population means are equal).
* Alternative hypothesis (H₁): At least one mean is different.

**2. Calculate the sample means and variances:**

* Sample 1 (x₁): 6, 8, 5, 12, 9
    * n₁ = 5
    * Mean (x̄₁) = (6 + 8 + 5 + 12 + 9) / 5 = 40 / 5 = 8
    * Variance (s₁²) = Σ(x₁ᵢ - x̄₁)² / (n₁ - 1) = (4 + 0 + 9 + 16 + 1) / 4 = 30 / 4 = 7.5
* Sample 2 (x₂): 5, 3, 8, 7, 7
    * n₂ = 5
    * Mean (x̄₂) = (5 + 3 + 8 + 7 + 7) / 5 = 30 / 5 = 6
    * Variance (s₂²) = Σ(x₂ᵢ - x̄₂)² / (n₂ - 1) = (1 + 9 + 4 + 1 + 1) / 4 = 16 / 4 = 4
* Sample 3 (x₃): 10, 7, 11, 10, 12
    * n₃ = 5
    * Mean (x̄₃) = (10 + 7 + 11 + 10 + 12) / 5 = 50 / 5 = 10
    * Variance (s₃²) = Σ(x₃ᵢ - x̄₃)² / (n₃ - 1) = (0 + 9 + 1 + 0 + 4) / 4 = 14 / 4 = 3.5

**3. Calculate the overall mean (x̄):**

* x̄ = Σ(Σxᵢⱼ) / N = (40 + 30 + 50) / 15 = 120 / 15 = 8

**4. Calculate the sum of squares between groups (SSB):**

* SSB = Σnᵢ(x̄ᵢ - x̄)² = 5(8 - 8)² + 5(6 - 8)² + 5(10 - 8)² = 0 + 20 + 20 = 40

**5. Calculate the sum of squares within groups (SSW):**

* SSW = Σ(nᵢ - 1)sᵢ² = 4(7.5) + 4(4) + 4(3.5) = 30 + 16 + 14 = 60

**6. Calculate the degrees of freedom:**

* Degrees of freedom between groups (dfB) = k - 1 = 3 - 1 = 2 (where k is the number of groups)
* Degrees of freedom within groups (dfW) = N - k = 15 - 3 = 12 (where N is the total number of observations)

**7. Calculate the mean squares:**

* Mean square between groups (MSB) = SSB / dfB = 40 / 2 = 20
* Mean square within groups (MSW) = SSW / dfW = 60 / 12 = 5

**8. Calculate the F-statistic:**

* F = MSB / MSW = 20 / 5 = 4

**9. Find the critical F-value:**

* Using an F-distribution table or calculator, with dfB = 2 and dfW = 12, and α = 0.05, the critical F-value is approximately 3.89.

**10. Make a decision:**

* Since the calculated F-statistic (4) is greater than the critical F-value (3.89), we reject the null hypothesis.

**11. Conclusion:**

* There is sufficient evidence to conclude that the population means are not all equal at the 5% level of significance.

Therefore, we reject the null hypothesis.