Question

The following data were collected from a repeated-measures study. Perform an ANOVA. Use a .05 alpha level. (10 POINTS) Treatment A B G = 36 EX? = 192 4 4 8 T= 12 T= 24 SS = 4 SS 8 1) State the null hypothesis. 2) Calculate necessary statistics (USE AN ANOVA even though K=2, and a=.05). 3) Make a decision about your null hypothesis and explain what that decision means. 4) Make a conclusion about your IV and DV that includes an APA format summary of your statistical analysis. 224

Answer 1

The Hypothesis:

H0:$\mu_1 = mu_2$: There is no difference between the means of the.2 treatments.

Ha: $\mu_1 \neq mu_2$ : There is no difference between the means of the.2 treatments.

The ANOVA table is as below

Tthe p value is calculated for F = 9 for df1 = 1 and df2 = 6

The Fcritical is calculated at $\alpha$ = 0.05 for df1 = 1 and df2 = 6

Source	SS	DF	Mean Square	F	Fcv	p
Between	18.00	1	18.00	9.00	5.9874	0.0240
Within/Error	12.00	6	2.00
Total	30.00	7

The Decision Rule:

If Ftest is > F critical, Then Reject H0.

Also if p-value is < α, Then reject H0.

The Decision:

Since Ftest (9) is > F critical (5.9874), We Reject H0.

Also since p-value (0.0240) is < $\alpha$ (0.05), We reject H0.

The Conclusion: There is sufficient evidence at the 95% level of significance to conclude that there is a difference in the means of the 2 treatments.

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Calculations For the ANOVA Table:

Overall Mean = (12 + 24) / 8 = 4.5

SS treatment = SUM n* ($\bar{x_i}$ - overall mean)² = 4 * (3 - 4.5)² + 4 * (6 - 4.5)² = 18

df1 = k - 1 = 2 - 1 = 1

MSTR = SS treatment/df1 = 18 / 1 = 18

SSerror = SUM (Sum of Squares) = 4 + 8 = 12

df2 = N - k = 8 - 2 = 6

Therefore MS error = SSerror/df2 = 12 / 6 = 2

F = MSTR/MSE = 18 / 2 = 9

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