Question

A researcher is studying the effectiveness of different treatment options for jail inmates with mental illnesses. The researcher creates four groups: therapy plus medication; therapy alone; medication alone; and a control group (no intervention). The outcome measure is the number of disciplinary infractions each person commits within the 6 months following the start of the experiment. The data are displayed in the table. Using an alpha of .05, conduct a 5-step ANOVA hypothesis test to find out if there are differences between groups. If appropriate, calculate and interpret omega-squared.

The 5 step process must be shown.

Med. + Therapy	Medication Only	Therapy Only	Control Group
3	6	0	8
4	2	3	2
2	0	2	4
0	8	1	0
9	1	0	0
5	3	10	1
1	0	4	3
n1 = 7	n2 = 7	n3 = 7	n4 = 7

Answer 1

1.The following null and alternative hypotheses need to be tested:

Ho: μ₁ = μ₂ = μ₃ = μ₄

Ha: Not all means are equal.

2.Test Statistic:

using the data given:

Group 1	Group 2	Group 3	Group 4
	3	6	0	8
	4	2	3	2
	2	0	2	4
	0	8	1	0
	9	1	0	0
	5	3	10	1
	1	0	4	3
Sum =	24	20	20	18
Average =	3.429	2.857	2.857	2.571
∑i​Xij2​=	136	114	130	94
St. Dev. =	2.992	3.078	3.485	2.82
SS =	53.714285714286	56.857142857143	72.857142857143	47.714285714286
n =	7	7	7	7

The total sample size is N=28. Therefore, the total degrees of freedom are:

df_total​=28−1=27

Also, the between-groups degrees of freedom are df_between​=4−1=3, and the within-groups degrees of freedom are:

df_within​=df_total​−df_between​=27−3=24

∑​Xij​=24+20+20+18=82

Also, the sum of squared values is

∑​Xij2​=136+114+130+94=474

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

SS_within​=∑SS_withingroups​=53.714285714286+56.857142857143+72.857142857143+47.714285714286=231.143

MS_between​=​SS_between​​/df_between=2.714/3​=0.905

MS_within​=​SS_within​​/df_within=231.143/24​=9.631

F=MS_between/MS_within​​​=0.905​/9.631=0.094

3.REJECTION REGION/CRITICAl VALUE:

The significance level is α=0.05, and the degrees of freedom are df_1​=3 and df₂​=3, therefore, Fc​=3.009

4.Decision about the null hypothesis

Since it is observed that F=0.094≤Fc​=3.009, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p=0.9627, and since p=0.9627≥0.05, 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 4 population means are equal, at the α=0.05 significance level..

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