SOLUTION: Given the following information, complete the ANOVA table; then conduct the F-test with α=0.05. SST=581.6 SSE=470.4 k=3 n1=n2=n3=10

Question 1177242: Given the following information, complete the ANOVA table; then conduct the F-test with α=0.05.
SST=581.6 SSE=470.4 k=3 n1=n2=n3=10
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to complete the ANOVA table and conduct the F-test:
**1. Calculate SS(between):**
* SS(between) = SST - SSE
* SS(between) = 581.6 - 470.4 = 111.2
**2. Calculate Degrees of Freedom (df):**
* df(between) = k - 1 = 3 - 1 = 2
* df(within) = N - k = (n1 + n2 + n3) - k = (10 + 10 + 10) - 3 = 30 - 3 = 27
* df(total) = N - 1 = 30 - 1 = 29
**3. Calculate Mean Squares (MS):**
* MS(between) = SS(between) / df(between) = 111.2 / 2 = 55.6
* MS(within) = SSE / df(within) = 470.4 / 27 = 17.4222 (approximately)
**4. Calculate the F-statistic:**
* F = MS(between) / MS(within) = 55.6 / 17.4222 = 3.1913 (approximately)
**5. Complete the ANOVA Table:**
| Source | SS | df | MS | F |
|---------------|---------|-----|----------|---------|
| Between Groups| 111.2 | 2 | 55.6 | 3.1913 |
| Within Groups | 470.4 | 27 | 17.4222 | |
| Total | 581.6 | 29 | | |
**6. Conduct the F-test:**
* **Significance level (α):** 0.05
* **Degrees of freedom:** df(between) = 2, df(within) = 27
* **Find the critical F-value:** Using an F-distribution table or calculator, with α = 0.05, df1 = 2, and df2 = 27, the critical F-value is approximately 3.35.
* **Compare the calculated F-statistic with the critical F-value:** 3.1913 < 3.35
**7. Make a Decision:**
* Since the calculated F-statistic (3.1913) is less than the critical F-value (3.35), we fail to reject the null hypothesis.
**8. Conclusion:**
* At the 0.05 significance level, there is not enough evidence to conclude that there is a significant difference between the means of the groups.
**Final Answer:**
The ANOVA table is completed as shown above. The F-test indicates that we fail to reject the null hypothesis. There is not sufficient evidence to say that the means of the k groups are different.

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