A group of 72 participants was equally divided into four groups. A Tukey HSD test produced a value that led to the conclusion that Mean 1 was significantly larger than Mean 2, p

Question

A group of 72 participants was equally divided into four groups. A Tukey HSD test produced a value that led to the conclusion that Mean 1 was significantly larger than Mean 2, p < .05. Which of the following situations would lead to such a conclusion?
A) $\bar { X } _ { 1 } = 7 , \bar { X } _ { 2 } = 0 , M S _ { \text {error } } = 80$
B) $\bar { X } _ { 1 } = 24 , \bar { X } _ { 2 } = 13 , M S _ { \text {error } } = 180$
C) both of the descriptive alternatives are correct
D) neither of the descriptive alternatives is correct.

Quizplus · Accepted Answer

The Tukey HSD test requires the mean difference to be large relative to the error term (MS error) to conclude significance. In both options A and B, the mean differences are not large enough relative to the given MS error to achieve significance at p < .05.

A Group of 72 Participants Was Equally Divided into Four Xˉ1=7,Xˉ2=0,MSerror =80\bar { X } _ { 1 } = 7 , \bar { X } _ { 2 } = 0 , M S _ { \text {error } } = 80Xˉ1​=7,Xˉ2​=0,MSerror ​=80

A Group of 72 Participants Was Equally Divided into Four $\bar { X } _ { 1 } = 7 , \bar { X } _ { 2 } = 0 , M S _ { \text {error } } = 80$