ANOVA Computations.

1. Determine the critical value of F for the following.

a. df_between = 5, df_within = 72, α = .05

2.35

b. df_between = 3, df_within = 46, α = .01

4.24

2. Use the following one-way ANOVA source table to answer the following questions.

Source of Variance	SS	df	Mean Square	F observed
Between Groups	40	4	10	1.79
Within Groups (Error)	140	25	5.60
Total	180	29

a. What is the mean square for the independent variable effects? 10

b. What is the mean square for the error variance? 5.6

c. What is the observed F-statistic? 1.79

d. How many total subjects were used in this study? 30

e. What is the total amount of variability in this design? S² =6.21, SS_TOTAL = 180

f. What is the critical value of F at α = .05? 2.76

g. What is your decision regarding Ho in this study?

Fail to reject the Null Hypothesis.

h. What is your overall conclusion about the differences between the means?

There is no significant difference between the means.

3. Poly Nomial believed that the magnitude of reinforcement would have an effect on altering the disruptive behavior of students in a class. Three groups of students were randomly selected from a class. One group received a small reinforcer at the end of the day for appropriate behavior. The second and third groups received medium and large rewards. At the end of the first week of this study the number of inappropriate behaviors were counted by independent, trained observers. Their results are summarized below.

	Small Reinforcer	Medium Reinforcer	Large Reinforcer	Total
	12	10	8
	14	9	10
	13	12	11
	12	13	9
	16	10	9
	15	11	7
	14	12	10
	13	9	7
Sum	109	86	71	266
Mean	13.625	10.75	8.875	11.083
N	8	8	8	24
SS	13.875	15.5	14.875	135.833

Source	SS	df	MS	F
Between treatments				21.70
Within treatments	44.25
Total	135.833

Note: The Total Sum of Squares comes from the Total in the upper table, and the Within Treatments data comes from the sum of the Within Groups listed in the bottom row of the upper table.

a. Complete the ANOVA summary table using the information in the table.

b. What is the null hypothesis for this study?

The means for the three groups are equal.

Or

There is no difference between the three means.

Or

(Statistically) M_Small = M_Medium = M_Large

c. What is your conclusion based on the F ratio?

There is a significant difference between the three means (F(2,21) = 21.70, p <.05. The reinforcement had a significant effect.

d. Compute Tukey’s HSD

HSD_.05 = 3.58 X SQRT(2.11/8) = 1.84

e. What is your conclusion based on the TUKEY test?

The Medium reinforcer was significantly more effective than the small reinforcer and the large reinforcement was significantly more effective than the medium reinforcer.

f. What is Eta²? How do you interpret this?

Eta^2 =	91.582	= .674
	135.833

Sixty-seven percent of the variability in the dependent variable (number of inappropriate behaviors) can be accounted for by knowing the reinforcement level. This appears to be a strong effect.