Question

A researcher is interested in whether or not there is a difference in student’s performance based on the study methods they use. Participants are assigned one of three different study methods (read the text and re-read it, read and answer prepared questions, and read then create your own questions). There are 6 people in each group and the researcher measures the participants’ performance on a test. Perform a hypothesis test and show all four steps. Hints: Use a two tailed test with an alpha of .05.

Read and Reread	Read and answer prepared questions	Read, then create your own questions
T = 30 n = 6 SS = 24	T = 54 n = 6 SS = 34	T = 60    n = 6   SS = 30	G = 144 ΣX2 = 1324

Answer 1

(1) Hypothesis

H0: The means of the three groups are Equal.

Ha: Not all the means are equal.

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(2) We use a One way ANOVA. with df between = 2, and df within = 15

The critical value at = 0.05, is 3.68

The Rejection rule is that if F test is > 3.68, Then Reject H0.

If p value is < 0.05, Then Reject H0.

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(3) The ANOVA Table is Below, showing SS, MS and F

Source	SS	DF	Mean Square	F	Fcv	p
Between	84.00	2	42.00	7.16	3.682	0.0066
Within/Error	88.00	15	5.87
Total	172.00	17

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(4) Since F test is > F critical, Reject H0

Since p value is < 0.05, Reject H0.

There is sufficient evidence at the 95% level of significance to conclude that the mean of at least one group is different from the others.

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Calculations For the ANOVA Table:

Overall Mean = (144 / 18) = 8

SS treatment = SUM [n* ( Individual Mean - overall mean)²] = 6 * (5 - 8)² + 6 * (9 - 8)² + 6 * (10 - 8)² = 84

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment/df1 = 84 / 2 = 42

SS error = SUM (Sum of Squares) = 24 + 34 + 30 = 88

df2 = N - k = 18 - 3 = 15

Therefore MS error = SS error/df2 = 88 / 15 = 5.87

F = MSTR / MSE = 42 / 5.87 = 7.16

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