Consider the following data with some missing values:

| Treatment | I | II | III |
|———–|—-|—-|—–|
| A | 12 | 14 | 12 |
| B | 10 | y | 8 |
| C | x | 15 | 10 |

**Objective:** Obtain the estimates of the missing values using Yates method and analyze the given data using a suitable technique.

### Solution:

1. **Estimate Missing Value \(x\):**
– Convert the two missing plots problem into one missing plot problem by taking the average of the values in Block I, where \(x\) is missing.
– Average for Block I: \((10 + 12) / 2 = 11\)
– Estimate of \(x\): \(x_1 = 11\)

2. **Form the following table with \(x_1 = 11\):**
\[
\begin{array}{|c|c|c|c|c|}
\hline \text{Treatments} & \text{I} & \text{II} & \text{III} & \text{Total} \\
\hline \mathbf{A} & 12 & 14 & 12 & T_A = 38 \\
\hline \mathbf{B} & 10 & y & 8 & T_B = 18 + y \\
\hline \mathbf{C} & 11 & 15 & 10 & T_C = 36 \\
\hline \text{Total} & B_1 = 33 & B_2 = 29 + y & B_3 = 30 & G = 92 + y \\
\hline
\end{array}
\]

– Values: \(p = 3, q = 3, B_2′ = 29, T_B’ = 18, G’ = 92\)

3. **Estimate Missing Value \(y\) using the Missing Estimation Formula:**
\[
\hat{Y} = \frac{pT_B’ + qB_2′ – G’}{(q-1)(p-1)} = \frac{3 \times 18 + 3 \times 29 – 92}{4} = \frac{54 + 87 – 92}{4} = \frac{49}{4} = 12.25 \approx 12
\]
– Estimate of \(y\): \(y_1 = 12\)

4. **Form the following table with \(y_1 = 12\):**
\[
\begin{array}{|c|c|c|c|c|}
\hline \text{Treatments} & \text{I} & \text{II} & \text{III} & \text{Total} \\
\hline \mathbf{A} & 12 & 14 & 12 & T_A = 38 \\
\hline \mathbf{B} & 10 & 12 & 8 & T_B = 30 \\
\hline \mathbf{C} & x & 15 & 10 & T_C = 25 + x \\
\hline \text{Total} & B_1 = 22 + x & B_2 = 41 & B_3 = 30 & G = 93 + x \\
\hline
\end{array}
\]

– Values: \(p = 3, q = 3, B_1′ = 22, T_C’ = 25, G’ = 93\)

5. **Estimate Missing Value \(x\) again using the Missing Estimation Formula:**
\[
\hat{x} = \frac{3 \times 25 + 3 \times 22 – 93}{4} = \frac{75 + 66 – 93}{4} = \frac{48}{4} = 12
\]
– Estimate of \(x\): \(x_2 = 12\)

6. **Validate Estimate of \(y\) with \(x_2 = 12\):**
\[
\hat{y} = \frac{3 \times 18 + 3 \times 29 – 93}{4} = \frac{54 + 87 – 93}{4} = \frac{47}{4} = 11.75 \approx 12
\]
– Second estimate of \(y\) (\(y_2\)) is not significantly different from \(y_1\).

7. **Final Estimated Values:**
– \(\hat{x} = 12\)
– \(\hat{y} = 12\)

8. **Form the table with both estimated values of \(x\) and \(y\):**
\[
\begin{array}{|c|c|c|c|c|}
\hline \text{Treatments} & \text{I} & \text{II} & \text{III} & \text{Total} \\
\hline \mathbf{A} & 12 & 14 & 12 & T_A = 38 \\
\hline \mathbf{B} & 10 & 12 & 8 & T_B = 30 \\
\hline \mathbf{C} & 12 & 15 & 10 & T_C = 37 \\
\hline \text{Total} & B_1 = 34 & B_2 = 41 & B_3 = 30 & G = 105 \\
\hline
\end{array}
\]

### ANOVA Analysis:

1. **Correction Factor (CF):**
\[
CF = \frac{(105)^2}{9} = \frac{11025}{9} = 1225
\]

2. **Raw Sum of Squares (RSS):**
\[
RSS = (12)^2 + (10)^2 + \ldots + (8)^2 + (10)^2 = 1261
\]

3. **Total Sum of Squares (TSS):**
\[
TSS = RSS – CF = 1261 – 1225 = 36
\]

4. **Treatment Sum of Squares (SST):**
\[
\begin{aligned}
SST &= \frac{(38)^2 + (30)^2 + (37)^2}{3} – CF \\
&= \frac{1444 + 900 + 1369}{3} – 1225 \\
&= \frac{3713}{3} – 1225 = 1237.67 – 1225 = 12.67
\end{aligned}
\]

5. **Block Sum of Squares (SSB):**
\[
\begin{aligned}
SSB &= \frac{(34)^2 + (41)^2 + (30)^2}{3} – CF \\
&= \frac{1156 + 1681 + 900}{3} – 1225 \\
&= 1245.67 – 1225 = 20.67
\end{aligned}
\]

6. **Error Sum of Squares (SSE):**
\[
SSE = TSS – SST – SSB = 36 – 12.67 – 20.67 = 2.66
\]

### ANOVA Table:
\[
\begin{array}{|c|c|c|c|c|c|}
\hline \text{Source of Variation} & \text{DF} & \text{SS} & \text{MSS} & \text{Variance Ratio} & \text{Tabulated} \\
\hline \text{Treatments} & 2 & 12.67 & 6.34 & 4.77 & 9.55 \\
\hline \text{Blocks} & 2 & 20.67 & 10.34 & 7.77 & 9.55 \\
\hline \text{Error} & 2 & 2.66 & 1.33 & & \\
\hline \text{Total} & 6 & & & & \\
\hline
\end{array}
\]

### Conclusion:
In the case of both treatments and blocks, the calculated value of \(F\) is less than the tabulated value of \(F\) at a 5% level of significance, indicating that treatment and block means are not significantly different.