Question 1170416: (i) When do you apply the analysis of variance technique?
(ii) The three samples given below have been obtained from three normal populations with equal variances. Test the hypothesis that the population means are equal at 5% level of significance.
Sample 1
6
8
5
12
9
Sample 2
5
3
8
7
7
Sample 3
10
7
11
10
12
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
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