Question

Minimum number of replications required, when the coefficient of the variation for the plot values is given to be 12%, for an observed difference of 10% among the sample means to be significant at 5% level, is

• A. 5
• B. 7
• C. 8
• D. 11

Hint:

Sample size formulas involving t-distributions are iterative because the t-value depends on the sample size you are trying to find. In exams, you can often use an approximation (like \(t \approx 2\) or \(t \approx 1.96\)) and see which option is closest, or you might need to guess a reasonable df, find r, and then check if the t-value for that df is consistent.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question asks for the minimum number of replications (r) needed to detect a certain difference between treatment means as statistically significant. This is a sample size or power analysis question in the context of ANOVA. We use the formula for the Least Significant Difference (LSD) or Critical Difference (CD).

Step 2: Key Formula or Approach:
We want the observed difference to be at least as large as the Critical Difference (CD). Observed Difference \( \ge \) CD. The CD formula is \( \text{CD} = t_{\alpha/2, \text{df}} \times \sqrt{\text{MSE} \left( \frac{1}{r_1} + \frac{1}{r_2} \right)} \). Assuming equal replications \(r\), this becomes \( \text{CD} = t_{\alpha/2, \text{df}} \times \sqrt{\frac{2\text{MSE}}{r}} \). The Coefficient of Variation (CV) is given by \( \text{CV} = \frac{\sqrt{\text{MSE}}}{\bar{y}} \times 100% \), where \(\bar{y}\) is the grand mean. The observed difference is also given as a percentage of the mean.

Step 3: Detailed Explanation:
Let \(d\) be the observed difference between two sample means, and \(\bar{y}\) be the grand mean. We are given \(d / \bar{y} = 10%\) or \(d = 0.10 \bar{y}\). The Coefficient of Variation is \(CV = \frac{\sqrt{MSE}}{\bar{y}} = 12%\) or \( \sqrt{MSE} = 0.12 \bar{y} \). Squaring this gives \( MSE = (0.12 \bar{y})^2 = 0.0144 \bar{y}^2 \). The condition for significance is that the difference \(d\) must be greater than or equal to the LSD. \[ d \ge t_{\alpha/2, \text{df}} \times \sqrt{\frac{2 \text{MSE}}{r}} \] Substituting the expressions for \(d\) and MSE: \[ 0.10 \bar{y} \ge t_{\alpha/2, \text{df}} \times \sqrt{\frac{2(0.0144 \bar{y}^2)}{r}} \] The \(\bar{y}\) term can be simplified: \[ 0.10 \bar{y} \ge t_{\alpha/2, \text{df}} \times \bar{y} \sqrt{\frac{0.0288}{r}} \] Cancel \(\bar{y}\) from both sides: \[ 0.10 \ge t_{\alpha/2, \text{df}} \times \sqrt{\frac{0.0288}{r}} \] We need to solve for r. Squaring both sides: \[ 0.01 \ge t^2_{\alpha/2, \text{df}} \times \frac{0.0288}{r} \] \[ r \ge \frac{t^2_{\alpha/2, \text{df}} \times 0.0288}{0.01} = 2.88 \times t^2_{\alpha/2, \text{df}} \] The value of \(t\) depends on the degrees of freedom for error, which in turn depends on \(r\) and the number of treatments (\(t\)). Let's assume there are at least two treatments. The error df is \(t(r-1)\) for a CRD. This makes direct solving difficult. For a large number of df, we can approximate \(t_{0.025}\) with the Z-value, 1.96. \[ r \ge 2.88 \times (1.96)^2 \approx 2.88 \times 3.8416 = 11.06 \] This suggests \(r=12\). But let's check the options. Often for such problems, the formula is approximated or given in a different form. Let's use \(t \approx 2\). \[ r \ge 2.88 \times (2)^2 = 2.88 \times 4 = 11.52 \] This gives \(r=12\). Still not matching the options. Let's re-examine the formula. Maybe it's a one-tailed test? For \(\alpha=0.05\) one-tailed, \(t\) is smaller. \(t_{0.05} \approx 1.645\). \[ r \ge 2.88 \times (1.645)^2 \approx 2.88 \times 2.706 = 7.79 \] This gives \(r=8\). This matches option (C). It is plausible that the "significance test" is a one-sided comparison.
Step 4: Final Answer:
Assuming a one-sided test is intended, the minimum number of replications is 8.