Question

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...

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?

Solutions

Expert Solution

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

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