In: Math
Group 1: 4.2, 4.2, 3.4
Group 2: 4.5, 2.1, 2.3
Group 3: 1.2, 0.3, -0.3, 2.3
Use the Bonferronni method to test each of the 3 possible hypotheses at the 3% significance level.
(a) Find the value of the test statistic for each of the 3 possible hypotheses.
(b) Which pairs of means are significantly different (using the Bonferronni method at the 3% significance level?
one way anova-
treatment | A | B | C | D | ||
count, ni = | 3 | 3 | 4 | |||
mean , x̅ i = | 3.933 | 2.97 | 0.875 | |||
std. dev., si = | 0.462 | 1.332 | 1.132 | |||
sample variances, si^2 = | 0.213 | 1.773 | 1.283 | |||
total sum | 11.8 | 8.9 | 3.5 | 24.2 | (grand sum) | |
grand mean , x̅̅ = | Σni*x̅i/Σni = | 2.42 | ||||
square of deviation of sample mean from grand mean,( x̅ - x̅̅)² | 2.290 | 0.299 | 2.387 | |||
TOTAL | ||||||
SS(between)= SSB = Σn( x̅ - x̅̅)² = | 6.871 | 0.897 | 9.548 | 17.31517 | ||
SS(within ) = SSW = Σ(n-1)s² = | 0.427 | 3.547 | 3.848 | 7.821 |
no. of treatment , k = 3
df between = k-1 = 2
N = Σn = 10
df within = N-k = 7
mean square between groups , MSB = SSB/k-1 =
8.6576
mean square within groups , MSW = SSW/N-k =
1.1173
F-stat = MSB/MSW = 7.7489
SS | df | MS | F | p-value | |
Between: | 17.315 | 2 | 8.6576 | 7.749 | 0.0168 |
Within: | 7.821 | 7 | 1.1173 | ||
Total: | 25.136 | 9 |
============
The Bonferroni correction sets the significance cut-off at α/n = 0.03/3 = 0.01
Level of significance | 0.0100 |
no. of treatments,k | 3 |
DF error =N-k= | 7 |
MSE | 1.117 |
t-critical value,t(α/2,df) | 3.4995 |
test statistic= mean difference ±√(MSE(1/ni+1/nj))
population mean difference | test statistic | |
µ1-µ2 | 1.12 | |
µ1-µ3 | 3.79 | |
µ2-µ3 | 2.59 |
b)
t critical value=t(0.01,7) = 3.50
if test stat > 3.50, mean are significantly different
population mean difference | Test stat | t -critical value | result | ||||
µ1-µ2 | 1.12 | 3.50 | means are not different | ||||
µ1-µ3 | 3.79 | 3.50 | means are different | ||||
µ2-µ3 | 2.59 | 3.50 | means are not different |