Question

In an effort to evaluate different methods of teaching a foreign language, a researcher examined
the performance of students learning Spanish under four different teaching methods. Group 1 learns in a
traditional lecture/lab format; group 2 is taught from a programmed text; group 3 is taught from podcasts; and
group 4 is taught from videos of people and life in Spain. After several weeks of instruction, all students were
given a Spanish quiz; these scores were recorded:
Group 1 Group 2 Group 3 Group 4
17 14 18 27
18 22 27 30
20 16. 24 25
17 15 18 26
16 16 20 32
21 20 22 27
18 18 19 31
15 14 25 28

a. Describe the statistical test to be done.

b. State the null and alternative hypotheses.
c. What is the critical value of the test statistic? What is the decision rule for rejecting Ho?
CV: ______________________ Reject Ho if F ≥ _________________
d. Paste in SPSS output below, circling the observed value of the test statistic. Also show your work for eta2
, and a

Tukey test, if necessary.

e. Write out an APA Style conclusion about the research project, using the context of the study.

Answer 1

a) Since there are four groups we have to do comparison between the means of the data and hence we perform one factor Analysis of Variance (ANOVA).

b) Null Hypothesis

H_o :

H₁ : Atleast on mean is different.

c)

the critical value of the test statistic? What is the decision rule for rejecting Ho?
CV: 2.946 Reject Ho if F ≥ Critical value.

d)

One-way ANOVA: Group 1, Group 2, Group 3, Group 4

Method

Null hypothesis	All means are equal
Alternative hypothesis	Not all means are equal
Significance level	α = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor	Levels	Values
Factor	4	Group 1, Group 2, Group 3, Group 4

Analysis of Variance

Source	DF	Adj SS	Adj MS	F-Value	P-Value
Factor	3	643.7	214.583	28.37	0.000
Error	28	211.8	7.563
Total	31	855.5

Model Summary

S	R-sq	R-sq(adj)	R-sq(pred)
2.75	75.25%	72.60%	67.67%

Means

Factor	N	Mean	StDev	95% CI
Group 1	8	17.750	1.982	(15.758, 19.742)
Group 2	8	16.88	2.90	(14.88, 18.87)
Group 3	8	21.63	3.42	(19.63, 23.62)
Group 4	8	28.250	2.493	(26.258, 30.242)

Pooled StDev = 2.75

Tukey Pairwise Comparisons

Grouping Information Using the Tukey Method and 95% Confidence

Factor	N	Mean	Grouping
Group 4	8	28.250	A
Group 3	8	21.63		B
Group 1	8	17.750			C
Group 2	8	16.88			C

Means that do not share a letter are significantly different.

Tukey Simultaneous Tests for Differences of Means

Difference of Levels	Difference of Means	SE of Difference	95% CI	T-Value	Adjusted P-Value
Group 2 - Group 1	-0.88	1.38	(-4.63, 2.88)	-0.64	0.919
Group 3 - Group 1	3.88	1.38	(0.12, 7.63)	2.82	0.041
Group 4 - Group 1	10.50	1.38	(6.75, 14.25)	7.64	0.000
Group 3 - Group 2	4.75	1.38	(1.00, 8.50)	3.45	0.009
Group 4 - Group 2	11.38	1.38	(7.62, 15.13)	8.27	0.000
Group 4 - Group 3	6.63	1.38	(2.87, 10.38)	4.82	0.000

e) Since CV: 2.946 is less than the calculated test statistic F (28.37) Reject Ho , hence we conclude that all the means are not equal.

From tukey's results we can see that all the pairs are significantly different except group 2 and group1 because they share zero in their confidence interval.