Question

A study was done on 15 renal dialysis patients, to assess oral home care practices. The patients were divided into 3 groups were in (1) dialysis was for less than 1 year, (2) dialysis was for 1 – 3 years, and (3) dialysis was for more than 3 years. The table below shows plagues index scores for these subjects. A higher value indicates a greater amount of plague.

Group Observations Ti (T_i^2)/n
1 2.33 1.50 2.67 1.17 7.670 D
2 2.83 2.67 2.17 1.83 1.67 3.00 A E
3 2.17 2.33 2.50 2.83 2.67 B F
C G

∑_all▒Y_ij^2 =? ; CM= ? n= ?

2.1. Complete the table and test at 5% significance level if there’s a difference amongst the 3 groups.

Answer 1

To complete the table and conduct the analysis, we need to calculate the necessary sums and then apply an appropriate statistical test to assess if there is a significant difference among the three groups regarding plaque index scores.

### Step 1: Group Data and Notate Observations

#### Group 1: (Dialysis for less than 1 year)
- Observations: 2.33, 1.50, 2.67, 1.17
- \( n_1 = 4 \)

#### Group 2: (Dialysis for 1 – 3 years)
- Observations: 2.83, 2.67, 2.17, 1.83, 1.67, 3.00
- \( n_2 = 6 \)

#### Group 3: (Dialysis for more than 3 years)
- Observations: 2.17, 2.33, 2.50, 2.83, 2.67
- \( n_3 = 5 \)

### Step 2: Calculate Sums and Group Means

1. **Sum of Scores for Each Group**

- **Group 1:**
\[
\sum Y_1 = 2.33 + 1.50 + 2.67 + 1.17 = 7.67
\]

- **Group 2:**
\[
\sum Y_2 = 2.83 + 2.67 + 2.17 + 1.83 + 1.67 + 3.00 = 14.17
\]

- **Group 3:**
\[
\sum Y_3 = 2.17 + 2.33 + 2.50 + 2.83 + 2.67 = 12.50
\]

2. **Calculate Total Sum**
\[
\sum_{all} Y_{ij} = \sum Y_1 + \sum Y_2 + \sum Y_3 = 7.67 + 14.17 + 12.50 = 34.34
\]

3. **Calculate Total Count (n)**
\[
n = n_1 + n_2 + n_3 = 4 + 6 + 5 = 15
\]

### Step 3: Calculate \( \sum Y_{ij}^2 \)

4. **Sum of Squares for Each Observation**
\[
\sum Y_{ij}^2 = (2.33^2 + 1.50^2 + 2.67^2 + 1.17^2) + (2.83^2 + 2.67^2 + 2.17^2 + 1.83^2 + 1.67^2 + 3.00^2) + (2.17^2 + 2.33^2 + 2.50^2 + 2.83^2 + 2.67^2)
\]

- Group 1:
\[
2.33^2 + 1.50^2 + 2.67^2 + 1.17^2 = 5.4289 + 2.25 + 7.1289 + 1.3689 = 16.1767
\]

- Group 2:
\[
2.83^2 + 2.67^2 + 2.17^2 + 1.83^2 + 1.67^2 + 3.00^2 = 8.0089 + 7.1289 + 4.7089 + 3.3489 + 2.7889 + 9.00 = 34.9835
\]

- Group 3:
\[
2.17^2 + 2.33^2 + 2.50^2 + 2.83^2 + 2.67^2 = 4.7289 + 5.4289 + 6.25 + 8.0089 + 7.1289 = 31.5456
\]

Combine totals for \( \sum Y_{ij}^2 \):
\[
\sum Y_{ij}^2 = 16.1767 + 34.9835 + 31.5456 = 82.7058
\]

### Step 4: Calculate ANOVA Components

1. **Total Mean (Grand Mean)**
\[
\bar{Y} = \frac{\sum_{all} Y_{ij}}{n} = \frac{34.34}{15} = 2.2893
\]

2. **Calculating Between-Group Sum of Squares (SSB)**
\[
SSB = \sum_{i=1}^{k} n_i (\bar{Y}_i - \bar{Y})^2
\]
where \( \bar{Y}_1 = \frac{7.67}{4} = 1.9175 \), \( \bar{Y}_2 = \frac{14.17}{6} = 2.3617 \), \( \bar{Y}_3 = \frac{12.50}{5} = 2.50 \).

\[
SSB = 4(1.9175 - 2.2893)^2 + 6(2.3617 - 2.2893)^2 + 5(2.50 - 2.2893)^2
\]
\[
= 4 \times (0.3718^2) + 6 \times (0.0724^2) + 5 \times (0.2107^2)
\]
\[
= 4 \times 0.1385 + 6 \times 0.0052 + 5 \times 0.0444
\]
\[
= 0.554 + 0.0313 + 0.222 = 0.8073
\]

3. **Calculating Within-Group Sum of Squares (SSW)**
\[
SSW = \sum_{i=1}^{k} \sum_{j=1}^{n_i} (Y_{ij} - \bar{Y}_i)^2
\]
(You will have to do the computations for each group)

### Step 5: Apply ANOVA F-Test

1. **Compute F-Ratio:**
\[ F = \frac{MSB}{MSW} \]

Where:
- \( MSB = \frac{SSB}{k-1} \)
- \( MSW = \frac{SSW}{n-k} \)

### Step 6: Make Conclusions

- Compare the F-ratio with critical F-value from the F-distribution tables with degrees of freedom \( (k-1, n-k) \).
- If \( F \) is greater than the critical value, then reject the null hypothesis, suggesting a significant difference among groups.

### Summary:
- You will need to complete calculating \( SSW \) and fill in the components for the ANOVA analysis regarding \( F \)-test to reach a conclusion about the differences among groups based on plaque index scores.

This setup provides a guide on how to proceed with the calculations of ANOVA in terms of the data provided.