Question

Instruction: In this assignment, you are required to perform two tasks- (1) Test the hypotheses in the two-way ANOVA manually. Here, show all the necessary steps and compute effect sizes and run post hoc (when necessary). (2) Using the same data, run the analyses using SPSS. Here, first copy and paste the output tables into word document and then report the important results in the standard ANOVA summary table and interpret the results. Include effect sizes and plots.

1. A statistics professor conducts an experiment to compare the effectiveness of two methods of teaching his course. Method I is the usual way he teaches the course: lectures, homework assignments, and a final exam. Method II is the same as method I, except that students receiving method II get 1 additional hour per week in which they solve illustrative problems under the guidance of the professor. The professor is also interested in how the methods affect students of differing mathematical abilities, so volunteers for the experiment are subdivided according to mathematical ability into superior, average, and poor groups. Five students from each group are randomly assigned to method I and 5 students from each group to method II. At the end of the course, all 30 students take the same final exam. The following final exam scores resulted:

		Mathematical ability
		Superior (1)	Average (2)	Poor (3)
Teaching method	Method I (1)	39               41	43                36	30                33
		48            42	40              35	29              36
		44	42	37
	Method II (2)	49               47	38               46	37                41
		47               48	45            44	34              33
		43	42	40

Answer 1

Solution:

Here, we have to use the two-way analysis of variance or two way ANOVA F test for checking the given claims. The null and alternative hypotheses for this two way ANOVA test are given as below:

Hypothesis 1

Null hypothesis: H₀: There is no any significant difference in the average final exam scores due to three different types of mathematical abilities.

Alternative hypothesis: H_a: There is a significant difference in the average final exam scores due to three different types of mathematical abilities.

Hypothesis 2

Null hypothesis: H₀: There is no any significant difference in the average final exam scores due to two different methods of teaching.

Alternative hypothesis: H_a: There is a significant difference in the average final exam scores due to two different methods of teaching.

Hypothesis 3

Null hypothesis: H₀: There is no any significant interaction between the two variables mathematical ability and teaching method.

Alternative hypothesis: H_a­: There is a significant interaction between the two variables mathematical ability and teaching method.

We assume a 5% level of significance for this test.

From the given data, a two way ANOVA table for this test is given as below:

Two-way ANOVA: Final exam score versus Mathematical Abi, Teaching method Analysis of Variance for Final ex Source Mathemat Teaching Interaction 2 Error Total DF MS 489.8 116.0 244.9 22.75 0.000 116.0 10.78 0.003 0.00 0.997 24 10.8 258.4 864.3 Multiple Comparisons Final exam score 95% Confidence Interval Mean Difference (l- 3.7000 9.8000 3.7000 6.1000 9.8000 6.1000 Si Lower Bound Upper Bound 7.3646 13.4646 .0354 9.7646 6.1354 2.4354 0354 6.1354 7.3646 2.4354 13.4646 9.7646 Superior Average Poor Superior Poor Superior Average 1.46742 1.46742 1.46742 1.46742 1.46742 1.46742 048 048 001 Average Poor 001 Based on observed means The error term is Mean Square(Error)-10.767 t. The mean difference is significant at the 0.05 level

The p-value for the factor A mathematical ability is given as 0.00 which is less than alpha value 0.05, so we reject the null hypothesis that there is no any significant difference in the average final exam scores due to three different types of mathematical abilities. There is sufficient evidence to conclude that there is a significant difference in the average final exam scores due to three different types of mathematical abilities.

The p-value for the factor B teaching method is given as 0.003 which is less than alpha value 0.05, so we reject the null hypothesis that there is no any significant difference in the average final exam scores due to two different methods of teaching. There is sufficient evidence to conclude that there is a significant difference in the average final exam scores due to two different methods of teaching.

The p-value for the interaction between mathematical ability and teaching method is given as 0.997 which is greater than the alpha value 0.5, so we do not reject the null hypothesis that there is no any significant interaction between the two variables mathematical ability and teaching method. There is insufficient evidence to conclude that there is a significant interaction between the two variables mathematical ability and teaching method.

We find out the significant differences between the three mathematical abilities and two teaching methods, so we need to conduct post hoc tests for this test.

From given Tukey post hoc comparison matrix, it is observed that there is a significant difference between all pairs of comparison. Note, all p-values are less than alpha value 0.05.