Question

In: Statistics and Probability

In a completely randomized design, six experimental units were used for each of the three levels...

  1. In a completely randomized design, six experimental units were used for each of the three levels of the factor. (2 points each; 6 points total)

Source of Variation

Sum of Squares

Degrees of Freedom

Mean Square

F

Treatment

Error

432076.5

Total

675643.3

  1. Complete the ANOVA table.
  2. Find the critical value at the 0.05 level of significance from the F table for testing whether the population means for the three levels of the factors are different.
  3. Use the critical value approach and α = 0.05 to test whether the population means for the three levels of the factors are the same.

Solutions

Expert Solution

ANSWER::

Here ,

Number of treatments k = 3

Each treatments contains 2 points

So , N = 3 * 2 = 6

degrees of freedom for treatments = k - 1 = 3 - 1 = 2

degrees of freedom for errors = n - k  = 6 - 3 = 3

degrees of freedom for Total = N - 1 = 6 - 1 = 5

Treatment sum of squares = Total sum of squares - Error sum of squares  

= 675643.3 - 432076.5

= 243566.8

Now ,

Mean square = Sum of squares/Degrees of freedom

F = Mean square(treatment) / Mean square(error)

a)

Source of Variation Sum of Squares Degrees of Freedom Mean Square F ratio
Treatment 243566.8 2 121783.4 0.8456
Error 432076.5 3 144025.5
Total 675643.3 5

b)

= 0.05

Critical value of the test

= F0.05 , df1 , df2

= F0.05,2,3

= 9.552  

(Use f distribution table)

c)

Since F = 0.8456 < F0.05,2,3 , we do not reject the null hypothesis.

We conclude that

the population means for the three levels of the factors are not significantly different.

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