Question

The conclusion of a​ one-way ANOVA procedure for the data shown in the table is to reject the null hypothesis that the means are all equal. Determine which means are different using alpha α equals=.05		Sample 1	Sample 2	Sample 3
		88	1515	2020
		1515	1414	2424
		77	1717	2222
		99	1313	1919

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alphaαequals=0.050.05.

Let

x overbarx1​,

x overbarx2​,

and

x overbarx3

be the means for samples​ 1, 2, and​ 3, respectively. Find the absolute values of the differences between the means.

StartAbsoluteValue x overbar 1 minus x overbar 2 EndAbsoluteValuex1−x2equals=55

​,

StartAbsoluteValue x overbar 1 minus x overbar 3 EndAbsoluteValuex1−x3equals=11.511.5

​, and

StartAbsoluteValue x overbar 2 minus x overbar 3 EndAbsoluteValuex2−x3equals=6.56.5

Find the​ Tukey-Kramer critical range CR. Note that since the samples are all the same​ size,

CR​1,2equals=CR​1,3equals=CR​2,3equals=CR.

CRequals=19.67519.675

​(Round to two decimal places as​ needed.)Conclude that the means for

▼

populations 1 and 3 and populations 2 and 3

populations 1 and 2 and populations 1 and 3

populations 1 and 3

populations 1 and 2 and populations 2 and 3

all the populations

populations 1 and 2

none of the populations

populations 2 and 3

are different.

Answer 1

Applying one way ANOVA: (use excel: data: data analysis: one way ANOVA: select Array):

Difference	Absolute Value	Conclusion
x1-x2	5.00	not significant difference
x3-x1	11.50	significant difference
x3-x2	6.50	significant difference

critical q with 0.05 level and at k=3 and N-k=9 degree of freedom=						3.95
Tukey's critical range for group i and j =                  (q/√2)*(sp*√(1/ni+1/nj)         =						5.19

populations 1 and 3 and populations 2 and 3 are different