A researcher wanted to investigate whether students ...



Use the ANOVA – Example 1 data from the Web site and the following scenario:

You are to investigate whether students’ scores on a reading midterm examination differ as a function of the method of direction/instruction they received. For this study, 45 students have been randomly selected to participate. The students were then randomly assigned to one of three treatment groups – to make 15 members in each group. All three groups are given the same required readings, but class time was spent differently in each group. Group 1 (Discussion) spent class time in directed discussions of the assigned readings. Group 2 (No Class) was excused from any obligations to attend class for the first half of the semester, but its members were given additional text materials they could use to help them understand the assigned readings. Group 3 (Lecture) was instructed using the professor’s ordinary lecture methods. For this exercise, use an a priori alpha level of significance of α = 0.05 for each statistical analysis necessary to answer the following questions.

1. What would the null hypothesis be for this study? Show/write the appropriate symbols or expression in words.

H0: (discussion = (no class = (lecture or H0: (1 = (2 = (3

The group means for the three instruction styles are equal.

2. What would the research/alternative hypothesis be for this study? Show/write the appropriate symbols or expression in words.

Ha: (i ( (k for some i, k or Ha: (i ( (j for some i, j

At least one of the group means significantly differs from the other group means.

-or- At least two of the group means are significantly different from each other.

3. Prior to examining whether group means differ, you need to test the assumption underlying the one-way ANOVA.

a. Was the assumption of independence met for these data? Indicate how you made this determination.

YES, the assumption of independence was met. This is indicated by the three groups of children being independent of each other. Children were randomly selected and then randomly assigned to one of the three treatment groups.

b. Was the assumption of normality met for these data? Indicate how you made this determination (using an alpha level of .01 for the Shapiro-Wilks test).

YES, the assumption of normality is met for this set of data. This is indicated by the fact that none of the standardized skewness values exceeded +3.29, nor were any of the probability values less than (or equal to) the .01 alpha level set for the Shapiro-Wilks test.

Discussion: [pic]

No Class: [pic]

Lecture: [pic]

Discussion’s p (1.000), No Class’ p (.952), and Lecture’s p (.361) each > α (.01)

c. Was the assumption of homogeneity of variance met for these data? Indicate how you made this determination.

Yes, the assumption of homogeneity is met.

This is indicated by the Levene’s Test of Homogeneity of Variances, F(2, 42) = 1.560, p = .222. With an alpha level of .05, p (.222) > ( (.05), which indicates non-significance, the null hypothesis (no variance difference) is retained – as such, indicating that the assumption of homogeneity of variance is met.

4. The next question that needs to be answered is whether all of the groups are the same on their reading midterm examination. This is answered by conducting a One-way Analysis of Variance using method of direction/instruction (group) as the independent variable and students’ scores on a reading midterm examination (test) as the dependent variable. What is your conclusion (at this point) from this analysis? Indicate how you came to your conclusion.

Since the obtained F ratio (48.362) was significant at the .05 alpha level (p = .000, that is p < .001) reported as F(2, 42) = 48.362, p < 0.001, we can concluded that at least two of the three instruction styles differ significantly on their average scores on a reading midterm examination. That is, p (.000) < ( (.05).

However, beyond that, post hoc follow-up procedures will need to be conducted to test the difference between all unique pairwise comparisons.

5. Calculate the measure of association and interpret its meaning.

[pic]

Therefore, we conclude that approximately 68% of the total variance in student’s average scores on a reading midterm examination is accounted for by the three levels of direction/instruction style.

-or-

This indicates that the independent variable (three levels of direction/instruction style) accounts for approximately 68% of the total variance in the dependent variable (average scores on a reading midterm examination) for this sample.

6. Assuming that you found a significant F, how do the pairs of groups differ? Indicate which post hoc procedure you used and why. Indicate your findings from the post hoc analysis. That is, how did you determine the pair to be significant or not (this must go beyond the * as an indication)? Be sure to discuss all unique pairwise comparisons.

Tukey’s post hoc procedure is used since the homogeneity of variance assumption was met…

Using an a priori alpha level of .05

Discussion vs. No Class (Mean difference = 18.200) is significant, p (.000) < ( (.05)

Discussion vs. Lecture (Mean difference = 7.933) is significant, p (000) < ( (.05)

No Class vs. Lecture (Mean difference = 10.267) is significant, p (.000) < ( (.05)

That is, all three unique pairwise comparisons were found to be significant at the alpha = .05 level of significance.

7. Then for all significant pairwise comparisons, calculate and report the effect size.

Using [pic] where [pic]

ES for Discussion vs. No Class = [pic]= 3.581

ES for Discussion vs. Lecture = [pic]= 1.561

ES for No Class vs. Lecture = [pic]= 2.020

8. Complete the three tables below that present your statistical results from these analyses. Use the values reported in the SPSS output – you do not need to round (Note that in reporting your data, APA suggests being consistent in your reporting of values, e.g., all rounded to two or three decimals).

See following pages…

Results

A One-way Analysis of Variance (ANOVA) was used to examine the question of whether students experiencing different types of direction/instruction have different scores on a reading midterm examination. The independent variable represented the different types of direction/instruction styles with three groups being represented: 1) discussion; 2) no class; and 3) lecture. The dependent variable was the average score that students made on a reading midterm examination. See Table 1 for the means and standard deviations for each of the three groups.

Table 1

Means and Standard Deviations of Midterm Reading Examination Scores by Treatment Group

|Treatment Group |n |Mean |SD |

|Discussion |15 |88.330 |6.149 |

|No Class |15 |70.130 |4.941 |

|Lecture |15 |80.400 |3.906 |

|Total |45 |79.620 |9.023 |

An alpha level of 0.05 was used for all analyses. The test for homogeneity of variance was not significant [Levene’s F (2, 42) = 1.560, p = .222] indicating that this assumption underlying the application of ANOVA was met. The one-way ANOVA of student’s average score on their reading midterm examination (see Table 2) revealed a statistically significant main effect [F (2, 42) = 48.362, p < .001] indicating that not all three styles of direction/instruction had the same average score on a reading midterm examination. Omega squared ((2 = .678) indicated that approximately 68% of the total variation in average score on students’ reading midterm examination is attributable to differences between the three styles of direction/instruction.

Table 2

Analysis of Variance for Midterm Reading Examination Scores by Treatment Group

|Source |SS |df |MS |F |p |

|Between |2497.911 |2 |1248.956 |48.362 |.000 |

|Within |1084.667 |42 |25.825 | | |

|Total |3582.578 |44 | | | |

Post hoc comparisons, using the Tukey post hoc procedure, were used to determine which pairs of the three style’s means differed significantly. These results are given in Table 3 and indicate that students using the discussion style (M = 88.330, SD = 6.149) had a significantly higher average score on their reading midterm examination than students using the lecture style (M = 80.400, SD = 3.906) and students using the no class style (M = 70.130, SD = 4.941). The effect sizes for these two significant effects were 1.561 and 3.581, respectively. Additionally, students using the lecture style had a significantly higher average score on their reading midterm examination than students using the no class style, with an effect size of 2.020.

Table 3

Post Hoc Results for Midterm Reading Examination Scores by Treatment Group

|Treatment Group |Mean |Mean Differences ([pic]) |

| | |(Effect Sizes are indicated in parentheses) |

| | |1 |2 |3 |

|1. Discussion |88.330 |0.00 | | |

|2. No Class |70.130 |18.200*** |0.00 | |

| | |(3.581) | | |

|3. Lecture |80.400 |7.930*** |10.270*** |0.00 |

| | |(1.561) |(2.020) | |

*Significant at α = 0.05, ** Significant at α = 0.01, *** Significant at α = 0.001

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