In the Tukey test, if $\bar { X } _ { 2 } = 8$ for a group size $n _ { 2 } = 8 , \bar { X } _ { 4 } = 12$ for a group size $n _ { 4 }$ and the within-group variance $s _ { w } ^ { 2 } = 24$, then the $q$ test value for comparing group 2 to group 4 is
A) -0.17
B) -0.01
C) -0.82
D) -2.31

Question

Quizplus · Accepted Answer

The Tukey test $q$ value is calculated using the formula: $$q = \frac{{|\bar{X}_i - \bar{X}_j|}}{{\sqrt{\frac{s_w^2}{n}}}}$$ where $\bar{X}_i$ and $\bar{X}_j$ are the means of the two groups being compared, $s_w^2$ is the within-group variance, and $n$ is the size of each group (assuming equal sizes for simplicity in this formula). Given $\bar{X}_2 = 8$, $\bar{X}_4 = 12$, $s_w^2 = 24$, and $n_2 = n_4 = 8$, the calculation is: $$q = \frac{|8 - 12|}{\sqrt{\frac{24}{8}}} = \frac{4}{\sqrt{3}} \approx 2.31$$ Since the question does not specify the direction of the comparison (i.e., whether it's $2-4$ or $4-2$), and typically $q$ values are reported as positive, the answer is $2.31$, which corresponds to choice D.

In the Tukey Test, If Xˉ2=8\bar { X } _ { 2 } = 8Xˉ2​=8

In the Tukey Test, If $\bar { X } _ { 2 } = 8$