A researcher used a one-factor ANOVA for between-groups to test the effectivenes
ID: 3182881 • Letter: A
Question
A researcher used a one-factor ANOVA for between-groups to test the effectiveness of three teaching methods for autistic children. The experiment was conducted with three samples of n = 10 autistic children in each group. The results of the analysis are shown in the following summary table.
Source SS df MS
Between Treatments 50 ____ ____ F = 5.00
Within Treatments ____ ____
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 three 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.
Explanation / Answer
Part A)
Source
SS
df
MS
F
Between Treatments
50
3 - 1 = 2
25
5
Within Treatments
27 * 5 = 135
(3*10) -3 = 27
25 / 5 = 5
Total
135 + 50 = 185
(3*10) - 1 = 29
Source
SS
df
MS
F
Between Treatments
50
2
25
5
Within Treatments
135
27
5
Total
185
29
Part B)
Null and Alternative Hypothesis:
Null Hypothesis (H0): µ1 = µ2 = µ3; i.e, all the three population means are equal.
Alternative Hypothesis (H): At least one of the population mean differs.
For 2,27derees of freedom we get the critical F at 5% level of significance as 3.354
The null hypothesis can be rejected if calculated test statistics is greater or equal to the critical value of 3.354. Here the test statistics value is greater than the critical value (5 > 3.354); we reject the null hypothesis at 5% level of significance. The difference is statistically significant.
Eta^2 = 50 / 135 = 0.2703
Conclusion:
There was a significant effect of teaching method on autistic children at the p<.05 level for the three conditions [F(2, 27) = 5, p = 0.014]
P-value calculation in excel: =FDIST(5,2,27)
Source
SS
df
MS
F
Between Treatments
50
3 - 1 = 2
25
5
Within Treatments
27 * 5 = 135
(3*10) -3 = 27
25 / 5 = 5
Total
135 + 50 = 185
(3*10) - 1 = 29
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