Question

The dataset gives the dried weight in pounds of three groups of 10 batches of plants, where each group of 10 batches received a different treatment. The Weight variable gives the weight of the batch and the Groups variable gives the treatment received.

A.) Conduct an analysis of variance to test the hypothesis of no difference in the weights of the plants under different treatments.

B.) Plot and analyze the residuals from the study and comment on model adequacy (All tests are to be performed at α = 0.05)

Control	Group 1	Group 2
4.17	4.81	6.31
5.58	4.17	5.12
5.18	4.41	5.54
6.11	3.59	5.5
4.5	5.87	5.37
4.61	3.83	5.29
5.17	6.03	4.92
4.53	4.89	6.15
5.33	4.32	5.8
5.14	4.69	5.26

Answer 1

The total sample size is N = 30. Therefore, the total degrees of freedom are:

Also, the between-groups degrees of freedom are

And the within-groups degrees of freedom are:

First, we need to compute the total sum of values and the grand mean. The following is obtained

Also, the sum of squared values is

Based on the above calculations, the total sum of squares is computed as follows

The within sum of squares is computed as shown in the calculation below:

The between sum of squares is computed directly as shown in the calculation below:

Now that sum of squares are computed, we can proceed with computing the mean sum of squares:

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows:

(1) Null and Alternative Hypotheses

(2) Critical value

At = 0.05 and df1 = 2 and df2 = 27, the critical value F_c​=3.354

(3) Test Statistics

(4) Decision about the null hypothesis

Since it is observed that F = 4.846 > F_c = 3.354, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0159, and since p = 0.0159 < 0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 3 population means are equal, at 0.05 significance level.