Question
1. Consider the following partially incomplete Analysis of Variance.
Source SS df MS Fstats (0.01)
Treatments 12.5138 B C E F
Errors A 12 D
Total 15.8333 14
1.1. Complete the table.
1.2. Test for the difference among the treatment means at α = 0.01.
1.3. If it’s known that n_1=4; n_2=6; ( Y) ̅_1=8.875; Y ̅_2=6.7. Test for the difference between the two treatment means, by constructing a 99% confidence interval for the mean differences.
Source SS df MS Fstats (0.01)
Treatments 12.5138 B C E F
Errors A 12 D
Total 15.8333 14
1.1. Complete the table.
1.2. Test for the difference among the treatment means at α = 0.01.
1.3. If it’s known that n_1=4; n_2=6; ( Y) ̅_1=8.875; Y ̅_2=6.7. Test for the difference between the two treatment means, by constructing a 99% confidence interval for the mean differences.
Answers
GPT-4o mini
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
1. **Error Sum of Squares (SS)**:
\[
\text{Error SS} = \text{Total SS} - \text{Treatment SS} = 15.8333 - 12.5138 = 3.3195
\]
2. **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:
- \( \text{Treatment df} = 3 - 1 = 2 \)
- \( \text{Error df} = 15 - 3 = 12 \)
3. **Mean Squares (MS)**:
\[
\text{MS}_{\text{Treatments}} = \frac{\text{SS}_{\text{Treatments}}}{\text{df}_{\text{Treatments}}} = \frac{12.5138}{2} = 6.2569
\]
\[
\text{MS}_{\text{Error}} = \frac{\text{SS}_{\text{Error}}}{\text{df}_{\text{Error}}} = \frac{3.3195}{12} \approx 0.2766
\]
4. **F-statistic**:
\[
F = \frac{MS_{\text{Treatments}}}{MS_{\text{Error}}} = \frac{6.2569}{0.2766} \approx 22.63
\]
5. **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.
### 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
1. **Error Sum of Squares (SS)**:
\[
\text{Error SS} = \text{Total SS} - \text{Treatment SS} = 15.8333 - 12.5138 = 3.3195
\]
2. **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:
- \( \text{Treatment df} = 3 - 1 = 2 \)
- \( \text{Error df} = 15 - 3 = 12 \)
3. **Mean Squares (MS)**:
\[
\text{MS}_{\text{Treatments}} = \frac{\text{SS}_{\text{Treatments}}}{\text{df}_{\text{Treatments}}} = \frac{12.5138}{2} = 6.2569
\]
\[
\text{MS}_{\text{Error}} = \frac{\text{SS}_{\text{Error}}}{\text{df}_{\text{Error}}} = \frac{3.3195}{12} \approx 0.2766
\]
4. **F-statistic**:
\[
F = \frac{MS_{\text{Treatments}}}{MS_{\text{Error}}} = \frac{6.2569}{0.2766} \approx 22.63
\]
5. **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.