Question

ample Data Item 1 2 3 4 5 Group 1 Group 2 Group 3 13 19 16 13 18 14 13 19 16 13 19 15 12 18 15 Print Done

Answer 1

[a] Option D is correct. becuase this is null hypothesis used in one-way anova.

[b] Option A is correct if p value is < 0.05 rejet null hypothesis, Otherwise do not reject.

  

T=1 T=2 T=3

N 5 5 5

$\sum T_{i}$ 64 93 76

Mean(T) 12.8 18.6 15.2

ANOVA TABLE

Source SS df MS F ratio

Between treatment 84.933 2 42.4667 F=106.1667

Within treatments 4.8 12 0.4

Total 89.733 14

Here from one way anova F ratio is 106.167 & P-value is 0.0093. which is <0.05 SO we reject h0: m1=m2=m3

Group 1 and Group 3 are significantly different by Tukey's HSD

N1:5 df1=4 M1=12.8 var_1= 0.2

N2:5 df1=4 M2=15.2 var_2= 0.7

T-value Calculation: (M1-M2)/sqrt(var_1+var_1) = - 5.66 and p-value is 0.000239.