Question

5. The following data were obtained in a study using three separate samples to compare three different treatments. (13 points total) Treatment 1 Treatment 2 Treatment 3 mm M =4 SS = 2 M = 2 SS = 4 M = 6 SS = 8 a. Fill in the ANOVA table below. (4 points) SS df MSF Source Between treatments Within treatments Total b. If a = .05 and the Fcritical value = 4.26., what decision would you make? (1 point) c. Calculate Tukey's HSD, with a q = 3.95. (1 point) d. Interpret which means are significantly different from each other based on Tukey's HSD. (3 points) e. Interpret the results in paragraph form as if you were reporting them for a journal article. Attempt to use APA style. (3 points)

Answer 1

5)

Null and Alternative Hypothesis:

H₀: µ_{Treatment 1} = µ_{Treatment 2} = µ_{Treatment 3}

H₁: Not all Means are equal

Alpha = 0.05

N = 12

n = 4

Degress of Freedom:

df_Between = a – 1 = 3-1 =2

df_Within = N-a = 12-3 = 9

df_Total = N-1 = 12-1 = 11

Critical Value:

(df_Between, df_Within): (2,9) = 4.26

Decision Rule:

If F is greater than 4.46, reject the null hypothesis

Test Statistics:

SS_Betwen = ∑(∑a_i)²/n - T²/N = 32

SS_Within = ∑(Y)² - ∑(∑a_i)²/n = 14

SS_Total = SS_{Betwen ­­}+ SS_Within = 46

MS = SS/df

               

F = MS_effect / MS_error

Hence,

F = 16/1.56 = 10.29

a)

Source of Variation	SS	df	MS	F
Between Groups	32	2	16	10.29
Within Groups	14	9	1.56
Total	46	11

b)

Result:

Our F = 10.29, we reject the null hypothesis

Conclusion:

Not all means are equal.

c)

Post- Hoc Test:

We will be using Tukey HSD (Honestly Significant Difference):

HSD = q *Sqrt (MS _Within /n)          

q is calculated from q table using (No of Groups vs degrees of freedom of error)

where q is given to be 3.95,

Hence, HSD = 2.46

d)

µ_{Treatment 1} = 4

µ_{Treatment 2} = 2

µ_{Treatment 3} = 6

Pairs:

Treatment 1 and Treatment 2

Mod(µ_{Treatment 1} - µ_{Treatment 2})= mod(4 – 2) = 2 which is less than HSD, hence they are not different

Treatment 1 and Treatment 3

Mod(µ_{Treatment 1} - µ_{Treatment 3})= mod(4 – 6) = 2 which is less than HSD, hence they are not different

Treatment 2 and Treatment 3

Mod(µ_{Treatment 2} - µ_{Treatment 3})= mod(2 – 6) = 4 which is greater than HSD, hence they are different

e)

We conclude that the means of not all Treatments is equal. Further upon doing the Post Hoc Test we conclude that Treatment 2 and Treatment 3 are significant different.