Let's break down the problem step-by-step.
1. Analysis of Variance Completion
1.1: Complete the ANOVA Table
Given Information:
- SS Total = 15.8333
- SS Treatments = 12.5138
- SS Errors = A
- df Total = 14
- df Treatments = B
- df Errors = D
- MS Treatments = E
- F-statistic = F
Step 1: Find Degrees of Freedom
For ANOVA:
- \( df_{Total} = n - 1 \) (Where \( n \) is the total number of observations)
- \( df_{Treatments} = k - 1 \) (Where \( k \) is the number of treatment groups)
- \( df_{Errors} = df_{Total} - df_{Treatments} \)
Due to \( df_{Total} = 14 \) and assuming there are 3 treatments, we can set:
- \( k = 3 \)
- \( df_{Treatments} = k - 1 = 3 - 1 = 2 \)
Thus:
- \( df_{Errors} = df_{Total} - df_{Treatments} = 14 - 2 = 12 \)
Step 2: Find Sum of Squares for Errors
Using the Total SS: \[ SS_{Errors} = SS_{Total} - SS_{Treatments} \] \[ A = 15.8333 - 12.5138 = 3.3195 \]
Step 3: Calculate Mean Squares
Mean Square (MS) is calculated as: \[ MS = \frac{SS}{df} \]
Calculate \( MS_{Treatments} \) and \( MS_{Errors} \): \[ MS_{Treatments} = \frac{SS_{Treatments}}{df_{Treatments}} = \frac{12.5138}{2} = 6.2569 \] \[ MS_{Errors} = \frac{SS_{Errors}}{df_{Errors}} = \frac{3.3195}{12} = 0.2766 \]
Step 4: Calculate F-statistic
\[ F = \frac{MS_{Treatments}}{MS_{Errors}} \] \[ F = \frac{6.2569}{0.2766} \approx 22.62 \]
Completed Table:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Source} & \text{SS} & \text{df} & \text{MS} & \text{F} & \text{(0.01)} \ \hline \text{Treatments} & 12.5138 & 2 & 6.2569 & 22.62 & E \ \text{Errors} & 3.3195 & 12 & 0.2766 & & \ \text{Total} & 15.8333 & 14 & & & \ \hline \end{array} \]
1.2: Test for Difference among Treatment Means
Using degrees of freedom \( df_{Treatments} = 2 \) and \( df_{Errors} = 12 \), we can determine the critical F-value at \( \alpha = 0.01 \). Typically from F-distribution tables, the critical value for \( F(2,12) \) at \( \alpha = 0.01 \) is approximately 5.676.
Since \( F \approx 22.62 \) > 5.676, we reject the null hypothesis. This indicates that there is a statistically significant difference among the treatment means.
1.3: 99% Confidence Interval for Mean Differences
To construct a confidence interval for the difference between two treatment means, we use the formula for the confidence interval:
\[ CI = ( \bar{Y}1 - \bar{Y}2 ) \pm t{(n_1 + n_2 - 2, 0.01/2)} \times \sqrt{ \frac{MS{Errors}}{n_1} + \frac{MS_{Errors}}{n_2} } \]
Here:
- \( n_1 = 4, n_2 = 6 \)
- \( \bar{Y}_1 = 8.875, \bar{Y}_2 = 6.7 \)
Calculate the difference of means: \[ \bar{Y}_1 - \bar{Y}_2 = 8.875 - 6.7 = 2.175 \]
Calculate \( t \)-value at \( \alpha = 0.01/2 = 0.005 \) for \( df = 8 \) (considering \( n_1+n_2-2 = 4+6-2 = 8 \)): From t-tables, \( t_{(8, 0.005)} \approx 4.501 \).
Calculate the standard error: \[ SE = \sqrt{ \frac{MS_{Errors}}{n_1} + \frac{MS_{Errors}}{n_2} } = \sqrt{ \frac{0.2766}{4} + \frac{0.2766}{6}} \] Calculating the numerical value, \[ = \sqrt{ 0.06915 + 0.0461 } = \sqrt{ 0.11525 } \approx 0.3395 \]
Now, construct the CI: \[ CI = 2.175 \pm 4.501 \cdot 0.3395 \]
Calculating the margin of error: \[ = 4.501 \cdot 0.3395 \approx 1.528 \] Thus the confidence interval is: \[ CI = 2.175 \pm 1.528 \Rightarrow (0.647, 3.703) \]
2. Complete the Table for the Study of Renal Dialysis Patients
Step 1: Group Total and Means
Group 1:
- Observations: 2.33, 1.50, 2.67, 1.17
- Calculate \( T_1 = 2.33 + 1.50 + 2.67 + 1.17 = 7.67 \)
- \( T_i^2/n \) can be calculated, but we need the number of observations \( n \) first.
Group 2:
- Observations: 2.83, 2.67, 2.17, 1.83, 1.67, 3.00
- \( T_2 = 2.83 + 2.67 + 2.17 + 1.83 + 1.67 + 3.00 = 14.17 \)
- Need \( n = 6 \) for number of observations.
Group 3:
- Observations: 2.17, 2.33, 2.50, 2.83, 2.67
- \( T_3 = 2.17 + 2.33 + 2.50 + 2.83 + 2.67 = 12.50 \)
- \( n = 5 \) for number of observations.
Total Calculation
- Sum of Observations \( n_{total} = 4 + 6 + 5 = 15 \)
Step 2: Compute \( \sum Y_{ij}^2 \)
Compute the sum of squares for each group: \[ \sum Y_i^2 = 2.33^2 + 1.50^2 + 2.67^2 + 1.17^2 + 2.83^2 + 2.67^2 + 2.17^2 + 1.83^2 + 1.67^2 + 3.00^2 + 2.17^2 + 2.33^2 + 2.50^2 + 2.83^2 + 2.67^2 \] Calculating the squares:
- Group 1: 5.4289 + 2.25 + 7.1289 + 1.3689 = 16.1767
- Group 2: 8.4489 + 7.1289 + 4.7089 + 3.3489 + 2.7889 + 9.0000 = 35.4235
- Group 3: 4.7089 + 5.4289 + 6.25 + 7.9889 + 7.1289 = 31.5056 \[ \sum Y_{ij}^2 = 16.1767 + 35.4235 + 31.5056 = 83.1058 \]
Step 4: Complete the Table:
\[ \begin{array}{|c|c|c|} \hline \text{Group} & \text{Observations (n)} & \frac{T_i^2}{n} \ \hline 1 & 4 & \frac{7.67^2}{4} = 14.67 \ 2 & 6 & \frac{14.17^2}{6} = 33.33 \ 3 & 5 & \frac{12.50^2}{5} = 31.25 \ \hline \end{array} \]
2.1: Test for Difference Among Groups
You will conduct ANOVA to compare these three groups.
- Calculating the Group Means and overall mean.
- Calculate Between Group and Within Group sums of squares.
- Conduct the necessary calculations for F-statistic.
If your F-statistic exceeds the critical value (for \(\alpha = 0.05\) in a group of 3, typically \(F(2,12)\)) from the F-distribution table, you reject the null hypothesis, indicating a difference among the groups.