Consider the following data with some missing values :

\begin{tabular}{|c|c|c|c|}

\hline \multirow{2}{*}{ Treatment } & & Block & \\

\cline { 2 – 4 } & I & II & III \\

\hline A & 12 & 14 & 12 \\

B & 10 & \(y\) & 8 \\

C & \(x\) & 15 & 10 \\

\hline

\end{tabular}

Obtain the estimates of the missing values using Yates method. Also analyse the given data using suitable technique.

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.