Step 1: Concept Overview:
The problem requires finding the minimum number of replicates (\(r\)) to achieve statistical significance for a difference between treatment means. This is essentially a sample size/power analysis problem within ANOVA, utilizing the Least Significant Difference (LSD) or Critical Difference (CD) formula.
Step 2: Core Formula and Approach:
The goal is for the observed difference to equal or exceed the Critical Difference (CD): Observed Difference \( \ge \) CD. The CD is calculated as \( \text{CD} = t_{\alpha/2, \text{df}} \times \sqrt{\text{MSE} \left( \frac{1}{r_1} + \frac{1}{r_2} \right)} \). With equal replications \(r\), this simplifies to \( \text{CD} = t_{\alpha/2, \text{df}} \times \sqrt{\frac{2\text{MSE}}{r}} \). The Coefficient of Variation (CV) is \( \text{CV} = \frac{\sqrt{\text{MSE}}}{\bar{y}} \times 100% \), where \(\bar{y}\) is the grand mean. The observed difference is expressed as a percentage of the mean.
Step 3: Detailed Solution:
Let \(d\) represent the observed difference between sample means, and \(\bar{y}\) the grand mean. We're given \(d / \bar{y} = 10%\), thus \(d = 0.10 \bar{y}\). The Coefficient of Variation is \(CV = \frac{\sqrt{MSE}}{\bar{y}} = 12%\), implying \( \sqrt{MSE} = 0.12 \bar{y} \). Squaring this gives \( MSE = (0.12 \bar{y})^2 = 0.0144 \bar{y}^2 \).
The significance criterion is \(d\) must be greater than or equal to the LSD:
\[ d \ge t_{\alpha/2, \text{df}} \times \sqrt{\frac{2 \text{MSE}}{r}} \]
Substituting for \(d\) and MSE:
\[ 0.10 \bar{y} \ge t_{\alpha/2, \text{df}} \times \sqrt{\frac{2(0.0144 \bar{y}^2)}{r}} \]
Simplifying the \(\bar{y}\) term:
\[ 0.10 \bar{y} \ge t_{\alpha/2, \text{df}} \times \bar{y} \sqrt{\frac{0.0288}{r}} \]
Canceling \(\bar{y}\) from both sides:
\[ 0.10 \ge t_{\alpha/2, \text{df}} \times \sqrt{\frac{0.0288}{r}} \]
Solving for \(r\):
\[ 0.01 \ge t^2_{\alpha/2, \text{df}} \times \frac{0.0288}{r} \]
\[ r \ge \frac{t^2_{\alpha/2, \text{df}} \times 0.0288}{0.01} = 2.88 \times t^2_{\alpha/2, \text{df}} \]
The value of \(t\) depends on the error degrees of freedom, which depends on \(r\) and the number of treatments (\(t\)). Assuming at least two treatments, the error df for a CRD is \(t(r-1)\). This makes solving directly difficult. Approximating \(t_{0.025}\) with the Z-value, 1.96 (for large df):
\[ r \ge 2.88 \times (1.96)^2 \approx 2.88 \times 3.8416 = 11.06 \]
This suggests \(r=12\).
Approximating \(t \approx 2\):
\[ r \ge 2.88 \times (2)^2 = 2.88 \times 4 = 11.52 \]
Again, \(r=12\).
Re-examining and assuming a one-tailed test (\(\alpha=0.05\), \(t_{0.05} \approx 1.645\)):
\[ r \ge 2.88 \times (1.645)^2 \approx 2.88 \times 2.706 = 7.79 \]
This results in \(r=8\), which aligns with option (C), suggesting a possible one-sided comparison.
Step 4: Conclusion:
Assuming a one-sided test, the minimum required number of replications is 8.