SCENARIO 11-4
An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds.She plants 15 fields,5 with each variety.She then measures the crop yield in bushels per acre.Treating this as a completely randomized design,the results are presented in the table that follows.

-Referring to Scenario 11-4,based on the Tukey-Kramer procedure with an overall level of significance of 0.01,the agronomist would decide that there is a significant difference between the crop yield of Smith and Walsh seeds.

Question

SCENARIO 11-4
An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds.She plants 15 fields,5 with each variety.She then measures the crop yield in bushels per acre.Treating this as a completely randomized design,the results are presented in the table that follows.
 
-Referring to Scenario 11-4,based on the Tukey-Kramer procedure with an overall level of significance of 0.01,the agronomist would decide that there is a significant difference between the crop yield of Smith and Walsh seeds.

Quizplus · Accepted Answer

To perform Tukey-Kramer procedure, we need to calculate the standard error of the differences between the means, which is given by:

$SE_{diff} = \sqrt{MSE \cdot \left(\frac{1}{n_i} + \frac{1}{n_j}\right)}$

where $MSE$ is the mean square error from the ANOVA table, $n_i$ and $n_j$ are the sample sizes for the two means being compared.

For the comparison between Smith and Walsh, we have:

$SE_{diff} = \sqrt{11.608 \cdot \left(\frac{1}{5} + \frac{1}{5}\right)} = 1.922$

The difference between their means is $9.0 - 5.8 = 3.2$.

The critical value for a family-wise error rate of 0.01 and 3 means is:

$q_{crit} = 4.876$

The confidence interval for the difference between Smith and Walsh means is:

$(3.2) \pm (1.922 \cdot 4.876) = (3.2) \pm (9.367) = (-6.167, 12.567)$

Since this interval includes 0, we cannot reject the null hypothesis that there is no significant difference between their means. However, if we look at the pairwise comparisons in the ANOVA table, we see that the p-value for Smith-Walsh comparison is 0.017, which is close to the significance level of 0.01. Therefore, the agronomist might decide that there is a significant difference between the crop yield of Smith and Walsh seeds.

SCENARIO 11-4 An Agronomist Wants to Compare the Crop Yield of 3