Question

Tukey's Post test. Alpha = .05. Group I: 0, 2, 2, 0, 1 Group II: 4, 6, 1, 5, 4 Group III: 1, 3, 0, 1, 0. Step by Step how to solve

Answer 1

Group I: 0, 2, 2, 0, 1

Group II: 4, 6, 1, 5, 4

Group III: 1, 3, 0, 1, 0

Mean of Group I = (0 + 2 + 2 + 0 + 1) /5 = 1

Mean of Group II = (4 + 6 + 1 + 5 + 4) /5 = 4

Mean of Group III = (1 + 3 + 0 + 1 + 0) /5 = 1

SSE =

= (0 - 1)^2 + (2 - 1)^2 + (2 - 1)^2 + (0 - 1)^2 + (1 - 1)^2

+ (4 - 4)^2 + (6- 4)^2 + (1 - 4)^2 + (5 - 4)^2 + (4- 4)^2

+ (1 - 1)^2 + (3 - 1)^2 + (0 - 1)^2 + (1 - 1)^2 + ( 0 - 1)^2  

= 24

DF for SSE = Number of observations - Number of groups = 15 - 3 = 12

MSE = SSE / DF for SSE = 24 / 12 = 2

Tukey critical value at = 0.05 is calculated using the below formula

where is a critical value of the studentized range for , the number of treatments or groups r, and the within-groups degrees of freedom and n is the sample size for each treatment.

From Anova table,

MSE = 2 , = 12

n = 5, r = 3

From studentized range table, = 3.773

So,

The groups are significantly different for which the means differ by more than 2.386

Mean Difference of Group II and Group I = 4 -1 = 3

Mean Difference of Group II and Group III = 4 -1 = 3

Mean Difference of Group I and Group III = 1 -1 = 0

Hence Group II is significantly different from Group I and Group III