Question

A researcher used a one-factor ANOVA for between-groups to test the effectiveness of four teaching methods for autistic children. The experiment was conducted with four samples of n = 12 autistic children in each group.

The results of the analysis are shown in the following summary table:

Source	SS	df	MS
Between Treatments (F=5.00)	?	?	?
Within Treatments	88	?	?
Total	?	?

A. Fill in all missing values in the table. Show your work (i.e., all computational steps for finding the missing values). Hint: start with the df values.

B. Do these data indicate any significant differences among the four teaching methods (assume p < .05)?

Your answer should include:

- the null hypothesis H0 & the alternative hypothesis H1

- the critical F value used for the decision about H0

- if the difference is statistically significant, compute the effect size, ?2

- the conclusion in APA style format.

*NOTE: This is a Psychology Statistics question

Answer 1

Given data:

Effectiveness of teaching methods = 4

experiment was conducted with four samples of n = 12

REQURIED:

A.Missing values in the table?

B.Data indicate any significant differences among the four teaching methods?

SOLUTION:

A:

Df for treatment    = 4 – 1

=       3   (Given Four teaching methods)

Df for total            = 48 – 1

= 47    (Given n=12 in each group,

No of groups = 4

12 * 4 = 48)

Df   for within treatment = 47 - 3

=    44

Ms for with in treatment   = 88 /44

= 2

MS treatment =    10        (F = between treatment MS / within treatment MS

5 = treatment MS / 2

Ms treatment = 5 * 2

= 10)

SS Between treatment = 3 * 10

= 30           (MS = SS/ Df)

10 = SS/ 3

SS = 10 *3

SS   = 30 )

TOTAL SS = 30 + 88

= 118

Therefore missing valules are shown in the above Table

B:

H0 : All the teaching methods are same

                             H1 : All the teaching methods are same

Therefore p < 0.05

H0 is rejected at 5% level of significant

Critical value = F

                           F = 3,44,0.05

                               = 2.82 (approximately)

Calculate F value = 5.00

    F value   > F 3, 44, 0.05 = 2.82

There fore H0 is rejected at 5%

HENCE THERE IS SIGNIFICANT DIFFERENCE BETWEEN THE TEACHING METHODS