Question

A task requires the completion of four activities. A teacher would like to know if differences in the sequence of the four activities results in different task completion times. The teacher selects three students and demonstrates the activities in random order to the students. Then each student completes the task with each of the activity sequences. The completion times are recorded. The following table shows the minutes for each student to complete each task. Time (minutes) Sequence Allen Carla Henry A 21.7 20.5 22.0 B 18.0 18.6 20.4 C 19.9 20.4 22.5 D 20.1 18.8 21.2 Identify the right option for Treatments (Time). Null hypothesis: H0: μ1 = μ2 = μ3 H0: μ1 ≠ μ2 ≠ μ3 a b Alternate hypothesis: All means are the same. Not all means are the same. Identify the right option for Blocks (Sequence). Null hypothesis: H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4 H0: μ1 = μ2 = μ3 = μ4 a b Alternate hypothesis: Not all means are the same. All means are the same. At 0.05 significance level, state the decision rule for for both treatment (Time) and blocks (Sequence). (Round your answers to 2 decimal places.) Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places.) At the 0.05 significance level, is there a difference in the treatments and blocks?

Answer 1

The following table is obtained:

	Group 1	Group 2	Group 3
	21.7	20.5	22
	18	18.6	20.4
	19.9	20.4	22.5
	20.1	18.8	21.2
Sum =	79.7	78.3	86.1
Average =	19.925	19.575	21.525
\sum_i X_{ij}^2 =∑i​Xij2​=	1594.91	1535.81	1855.85
St. Dev. =	1.515	1.014	0.922
SS =	6.8874999999998	3.0875000000001	2.5474999999994
n =	4	4	4

The total sample size is N = 12. Therefore, the total degrees of freedom are:

dftotal​=12−1=11

Also, the between-groups degrees of freedom are dfbetween​=3−1=2, and the within-groups degrees of freedom are:

dfwithin​=dftotal​−dfbetween​=11−2=9

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:

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

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1 = μ2 = μ3

Ha: Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df1​=2 and df2​=2, therefore, the rejection region for this F-test is

(3) Test Statistics

(4) Decision about the null hypothesis

Since it is observed that , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.0942, and since , it is concluded that the null hypothesis is not rejected.

(5) Conclusion

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

No difference.