Question

1) The following data were collected from a repeated-measures study investigating the effects of 4 treatment conditions on test performance. Determine if there are any significant differences among the four treatments. State the null hypothesis. If you determine a significant treatment effect, use Tukey's HSD test (overall a = .05) to determine which treatments differ from which other treatments. Also, compute the percentage of variance explained by the treatment effect (12). Conclude with an appropriate summary describing what you found. Be sure that the statement includes/refers to some descriptive statistics and clearly states the outcome of any hypothesis and post-hoc tests you conducted. Participant Treatments A B C D 2 3 4 6 4 4 6 3 4 2 3 3 2 0 3 0 2 2 0
using the repeated measures anova

Answer 1

We use the following model: Observation of jthe replicate of the ith treatment

= general effect + effect due to the ith treatment + random error component

= $y_{ij} = \mu + \tau_i + e_{ij}$

The null hypothesis is there is no significant difference between treatment or symbollically

$H_0:\tau_1 = \tau_2 = \tau_3 = \tau_4$

$H_1: At \hspace{0.1cm} least \hspace{0.1cm} one \hspace{0.1cm} \tau_i \hspace{0.1cm} is \hspace{0.1cm} different$

Sources of Variation	df	Sum of Squares (SS)	Mean SS	F Value
Treatments	t - 1	SS of treatments	MST	MST / MSE
Error	n - t	SS of Error	MSE
Total	n - t	Total SS	-

Here t = 4, n = 16

Our given data

	Treatments
Participants	A	B	C	D
1	6	3	3	0
2	4	4	2	2
3	4	2	0	2
4	6	3	3	0

The ANOVA Table is given by

SUMMARY
Groups	Count	Sum	Average	Variance
A	4	20	5	1.333333
B	4	12	3	0.666667
C	4	8	2	2
D	4	4	1	1.333333
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	35	3	11.66667	8.75	0.00239	3.490295
Within Groups (error)	16	12	1.333333
Total	51	15

From the above ANOVA table we see F_Calculated > F_Crit

So we reject the Null hypothesis at 5% level of significance. In other words there are significant difference between treatments.

The Tukey Test for HSD is given by

$HSD = \frac{M_i - M_j}{\sqrt{\frac{MS_w}{n}}}, \hspace{0.2cm}M_i>M_j$

First we calculate ${\sqrt{\frac{MS_w}{n}}} = {\sqrt{\frac{1.333}{4}}} = 0.57727 \approx 0.577$

Now between Treatment A & B,

$HSD_{AB} = \frac{5-3}{0.577} = 3.466$

$HSD_{AC} = \frac{5-2}{0.577} = 5.199$

$HSD_{AD} = \frac{5-1}{0.577} = 6.932$

$HSD_{BC} = \frac{3-2}{0.577} = 1.733$

$HSD_{BD} = \frac{3-1}{0.577} = 3.466$

$HSD_{CD} = \frac{2-1}{0.577} = 1.733$

The Tukey's critical value at
g= 0,05
is at df

So from the above calculation we conclude that Treatment A & C and Treatment A & D are significantly different

The variance % explained by the treatment effect is 35/51 = 68%

Summary

It has been observed that there is a significant difference between the treatments as observed from ANOVA table. The Treatment A and C and Treatment A & D are significantly different but there is no significant difference between B, C and D. So it may conclude that Treatment A is better than B, C and D