## Answers

**SOLUTION:**

(2) Null hypothesis (H): There is no significant difference between the mean effects due to the three treatments ie. 4 Alternative hypothesis (H): At least two of the three treatments differ significantly in their mean effects. i.e Given a 0.05 (3) The decision rule is, reject H, if calculated F> F-table value (4) Number of treatments (k) 3 (5) From the given data, N 24 Grand (G)72 Also Given that x =393 105 ニ=35 3 k Ex 393 =196.6 2 -1 F taldel 196.6 35 = 5.614 Since F-cal >The critical value of the test is, F-1) = Foospan reject our H = 4.4199 So we At least two of the three treatments differ significantly in their mean effects

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