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Sagot :
Alright, let's walk through each step in constructing the ANOVA table based on the given information. We'll fill out the ANOVA table and then interpret the results.
### Step-by-Step Solution:
#### Given Information:
- Total Sum of Squares (SST) = 76.83
- Sum of Squares for Treatment (SSTR) = 11.41
- Number of groups (C) = 4
- Number of observations per group ([tex]\( n_1 = n_2 = n_3 = n_4 \)[/tex]) = 15
#### Calculations:
1. Total Number of Observations [tex]\( (N) \)[/tex]
[tex]\[ N = n_1 + n_2 + n_3 + n_4 = 15 + 15 + 15 + 15 = 60 \][/tex]
2. Degrees of Freedom (df) Calculation:
- Total Degrees of Freedom [tex]\( (df_{\text{total}}) \)[/tex]
[tex]\[ df_{\text{total}} = N - 1 = 60 - 1 = 59 \][/tex]
- Between Groups Degrees of Freedom [tex]\( (df_{\text{between}}) \)[/tex]
[tex]\[ df_{\text{between}} = C - 1 = 4 - 1 = 3 \][/tex]
- Within Groups Degrees of Freedom [tex]\( (df_{\text{within}}) \)[/tex]
[tex]\[ df_{\text{within}} = df_{\text{total}} - df_{\text{between}} = 59 - 3 = 56 \][/tex]
3. Mean Squares Calculation:
- Mean Square Between Groups [tex]\( (MS_{\text{between}}) \)[/tex]
[tex]\[ MS_{\text{between}} = \frac{SSTR}{df_{\text{between}}} = \frac{11.41}{3} = 3.8033 \][/tex]
- Mean Square Within Groups [tex]\( (MS_{\text{within}}) \)[/tex]
[tex]\[ MS_{\text{within}} = \frac{SST - SSTR}{df_{\text{within}}} = \frac{76.83 - 11.41}{56} = \frac{65.42}{56} = 1.1682 \][/tex]
4. F-statistic Calculation:
[tex]\[ F = \frac{MS_{\text{between}}}{MS_{\text{within}}} = \frac{3.8033}{1.1682} = 3.2557 \][/tex]
5. Conclusion based on the p-value:
- Given p-value: 0.028
- Significance Level [tex]\( (\alpha) \)[/tex]: 0.05
Since the p-value (0.028) is less than the significance level (0.05), we reject the null hypothesis [tex]\( H_0 \)[/tex].
### ANOVA Table:
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline \text{Source of Variation} & \text{SS} & \text{df} & \text{MS} & F & p \text{-value} \\ \hline \text{Between Groups} & 11.41 & 3 & 3.8033 & 3.2557 & 0.028 \\ \hline \text{Within Groups} & 65.42 & 56 & 1.1682 & & \\ \hline \text{Total} & 76.83 & 59 & & & \\ \hline \end{array} \][/tex]
### Conclusion:
Based on the ANOVA table and the p-value:
- Reject [tex]\( H_0 \)[/tex]: we can conclude that some means differ.
### Step-by-Step Solution:
#### Given Information:
- Total Sum of Squares (SST) = 76.83
- Sum of Squares for Treatment (SSTR) = 11.41
- Number of groups (C) = 4
- Number of observations per group ([tex]\( n_1 = n_2 = n_3 = n_4 \)[/tex]) = 15
#### Calculations:
1. Total Number of Observations [tex]\( (N) \)[/tex]
[tex]\[ N = n_1 + n_2 + n_3 + n_4 = 15 + 15 + 15 + 15 = 60 \][/tex]
2. Degrees of Freedom (df) Calculation:
- Total Degrees of Freedom [tex]\( (df_{\text{total}}) \)[/tex]
[tex]\[ df_{\text{total}} = N - 1 = 60 - 1 = 59 \][/tex]
- Between Groups Degrees of Freedom [tex]\( (df_{\text{between}}) \)[/tex]
[tex]\[ df_{\text{between}} = C - 1 = 4 - 1 = 3 \][/tex]
- Within Groups Degrees of Freedom [tex]\( (df_{\text{within}}) \)[/tex]
[tex]\[ df_{\text{within}} = df_{\text{total}} - df_{\text{between}} = 59 - 3 = 56 \][/tex]
3. Mean Squares Calculation:
- Mean Square Between Groups [tex]\( (MS_{\text{between}}) \)[/tex]
[tex]\[ MS_{\text{between}} = \frac{SSTR}{df_{\text{between}}} = \frac{11.41}{3} = 3.8033 \][/tex]
- Mean Square Within Groups [tex]\( (MS_{\text{within}}) \)[/tex]
[tex]\[ MS_{\text{within}} = \frac{SST - SSTR}{df_{\text{within}}} = \frac{76.83 - 11.41}{56} = \frac{65.42}{56} = 1.1682 \][/tex]
4. F-statistic Calculation:
[tex]\[ F = \frac{MS_{\text{between}}}{MS_{\text{within}}} = \frac{3.8033}{1.1682} = 3.2557 \][/tex]
5. Conclusion based on the p-value:
- Given p-value: 0.028
- Significance Level [tex]\( (\alpha) \)[/tex]: 0.05
Since the p-value (0.028) is less than the significance level (0.05), we reject the null hypothesis [tex]\( H_0 \)[/tex].
### ANOVA Table:
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline \text{Source of Variation} & \text{SS} & \text{df} & \text{MS} & F & p \text{-value} \\ \hline \text{Between Groups} & 11.41 & 3 & 3.8033 & 3.2557 & 0.028 \\ \hline \text{Within Groups} & 65.42 & 56 & 1.1682 & & \\ \hline \text{Total} & 76.83 & 59 & & & \\ \hline \end{array} \][/tex]
### Conclusion:
Based on the ANOVA table and the p-value:
- Reject [tex]\( H_0 \)[/tex]: we can conclude that some means differ.
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