Hypothesis :
H0 : μ1 = μ2 = μ3
Ha : atleast one μ is different
Treatment → | A | B | C |
Input Data → | 11.0 12.0 10.0 13.0 14.0 |
14.0 18.0 17.0 15.0 13.0 16.0 12.0 |
17.0 16.0 18.0 19.0 12.0 14.0 |
Treatment → | A | B | C | Pooled Total |
observations N | 5 | 7 | 6 | 18 |
sum ∑xi∑xi | 60.0000 | 105.0000 | 96.0000 | 261.0000 |
mean ¯xx¯ | 12.0000 | 15.0000 | 16.0000 | 14.5000 |
sum of squares ∑x2i∑xi2 | 730.0000 | 1,603.0000 | 1,570.0000 | 3,903.0000 |
sample variance s2s2 | 2.5000 | 4.6667 | 6.8000 | 6.9706 |
sample std. dev. ss | 1.5811 | 2.1602 | 2.6077 | 2.6402 |
std. dev. of mean SE¯xSEx¯ | 0.7071 | 0.8165 | 1.0646 | 0.6223 |
source | sum of squares SS |
degrees of freedom νν |
mean square MS |
F statistic | p-value |
treatment | 46.5000 | 2 | 23.2500 | 4.8438 | 0.0238 |
error | 72.0000 | 15 | 4.8000 | ||
total | 118.5000 | 17 |
p value = 0.0238 < 0.05, So reject H0, Not all means are equal
post-hoc Tukey HSD Test Calculator results:
k=3 treatments
degrees of freedom for the error term ν=15
Critical values of the Studentized Range Q statistic:
At α=0.01,k=3,ν=15 Qcritical= 4.8341
At α=0.05,k=3,ν=15 Qcritical= 3.6719
Q1,2,3,4 Answers
Tukey HSD results
treatments pair |
Tukey HSD Q statistic |
Tukey HSD p-value |
Tukey HSD inferfence |
A vs B | 3.3072 | 0.0807676 | insignificant |
A vs C | 4.2640 | 0.0222569 | * p<0.05 |
B vs C | 1.1602 | 0.6859788 | insignificant |
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