To complete your Analysis of Variance table, we first need to solve for the missing values using the formulas associated with ANOVA.
1.1 Complete the ANOVA Table
The ANOVA table gives us the following information:
- Treatment Sum of Squares (SS) = 12.5138
- Total Sum of Squares (Total SS) = 15.8333
- Error Sum of Squares (Error SS) = Total SS - Treatment SS
- Degrees of Freedom (df) = The number of groups minus one for treatments and the total number of observations minus the number of groups for errors.
Calculating Missing Values
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Error Sum of Squares (SS): \[ \text{Error SS} = \text{Total SS} - \text{Treatment SS} = 15.8333 - 12.5138 = 3.3195 \]
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Degrees of Freedom (df): Let \( k \) be the number of treatments, and \( N \) be the total number of observations:
- Assuming there are \( k \) treatments, we have:
- \( \text{Treatment df} = k - 1 \)
- To find \( \text{Error df} \): \( \text{Error df} = N - k \)
- \( N - 1 = 14 \rightarrow N = 15 \) total samples.
Since \( 12.5138 \) as total SS and we've assumed three treatments for this example, let's say \( k=3 \) (which will be confirmed later). Thus:
- \( \text{Treatment df} = 3 - 1 = 2 \)
- \( \text{Error df} = 15 - 3 = 12 \)
- Assuming there are \( k \) treatments, we have:
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Mean Squares (MS): \[ \text{MS}{\text{Treatments}} = \frac{\text{SS}{\text{Treatments}}}{\text{df}{\text{Treatments}}} = \frac{12.5138}{2} = 6.2569 \] \[ \text{MS}{\text{Error}} = \frac{\text{SS}{\text{Error}}}{\text{df}{\text{Error}}} = \frac{3.3195}{12} \approx 0.2766 \]
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F-statistic: \[ F = \frac{MS_{\text{Treatments}}}{MS_{\text{Error}}} = \frac{6.2569}{0.2766} \approx 22.63 \]
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Look up the Critical F Statistic for \( df_{1} = 2 \) and \( df_{2} = 12 \) at \( \alpha = 0.01 \) (from the F-table). For \( 2, 12 \) degrees of freedom, \( F_{0.01}(2, 12) \approx 5.19 \), but you can look this up based on the latest F-table.
Completed ANOVA Table
Source SS df MS Fstats (0.01)
Treatments 12.5138 2 6.2569 22.635 5.19
Errors 3.3195 12 0.2766
Total 15.8333 14
1.2 Test for Difference among Treatment Means at \( \alpha = 0.01 \)
Since the calculated F-statistic (22.635) is greater than the critical F value (5.19), we reject the null hypothesis.
Conclusion: There is significant evidence at the \( \alpha = 0.01 \) level that not all treatment means are equal.
1.3 Constructing a 99% Confidence Interval for the Mean Difference
To construct the confidence interval for the difference between two treatment means, we use the following formula:
\[ \text{Confidence Interval} = (\bar{Y_1} - \bar{Y_2}) \pm t_{\alpha/2} \cdot SE \] Where \( SE \) is the standard error of the difference between means, calculated as:
\[ SE = \sqrt{\frac{MS_{Error}}{n_1} + \frac{MS_{Error}}{n_2}} \]
Given:
- \( n_1 = 4, n_2 = 6 \)
- \( \bar{Y_1} = 8.875, \bar{Y_2} = 6.7 \)
- \( MS_{Error} = 0.2766 \)
Calculate SE:
\[ SE = \sqrt{\frac{0.2766}{4} + \frac{0.2766}{6}} = \sqrt{0.06915 + 0.0461} = \sqrt{0.11525} \approx 0.3394 \]
Next, find \( t_{\alpha/2} \) for \( df = 12 \) at a 99% confidence level (\( \alpha = 0.01 \)):
Assuming \( t_{0.005}(12) \approx 3.055 \) (can also verify from a t-table).
Now compute the mean difference:
\[ \bar{Y_1} - \bar{Y_2} = 8.875 - 6.7 = 2.175 \]
Construct the confidence interval:
\[ 2.175 \pm 3.055 \cdot 0.3394 \] \[ = 2.175 \pm 1.0373 \] \[ = [1.1377, 3.2123] \]
Final Conclusion
The 99% confidence interval for the difference in treatment means is approximately \( [1.1377, 3.2123] \). Hence, we can conclude that there is a significant difference between the two treatment means based on the confidence interval not including zero.
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