statistics

1. In a study to investigate the relative effectiveness of six different instructional formats with respect to student achievement at the end of the semester, data were collected on a total of 90 students taught using one of the 6 formats described below:

Method 1: 3 hour lecture weekly

Method 2: 3 hour lecture weekly + weekly tutorial

Method 3: 3 hour lecture weekly + weekly pop quizzes

Method 4: Two 90-minute lectures weekly

Method 5: Two 90-minute lectures weekly + weekly tutorials

Method 6: Two 90-minute lectures weekly + weekly pop quizzes

Descriptive statistics for each group are given below.

Dependent Variable:Acht
Group	Mean	Std. Deviation	N
1	8.4667	1.45733	15
2	10.2667	1.79151	15
3	9.2267	1.82462	15
4	10.7333	2.11919	15
5	10.7800	1.68955	15
6	12.1267	1.99516	15
Total	10.2667	2.12999	90

An ANOVA table for testing the hypothesis of equal means was constructed and is shown below.

Tests of Between-Subjects Effects
Dependent Variable:Acht
Source	Type III Sum of Squares	df	Mean Square	F	Sig.
Corrected Model	123.937a	5	24.787	7.440	.000
Intercept	9486.400	1	9486.400	2847.520	.000
Group	123.937	5	24.787	7.440	.000
Error	279.843	84	3.331
Total	9890.180	90
Corrected Total	403.780	89
a. R Squared = .307 (Adjusted R Squared = .266)

(a) Based on the ANOVA table, what would you conclude about the equality of means? Show what you used to reach your conclusion.

The null hypothesis is, H0: μ1 = μ2 = μ3 = μ4 =μ5 = μ6, which is the means are equal of the methods that are used to teach the students.

The alternate hypothesis is H1: two or more means are unequal.

The p value obtained is 0.00 that is less that the alpha value of 0.05. The F value is 7.44 and with the low p value, it is obvious that the means are not equal. Therefore, the null hypothesis is rejected and alternate hypothesis is accepted.

(b) Determine Bonferroni, Tukey, and Scheffe critical values for testing all pairwise differences among the means.

The total number of pairs is, C = 6(6-1)/2 = 15

The Bonferroni adjustment =  = 0.05/15 = 0.0033

The critical value for Bonferroni corrections is 0.0033

The Tukey critical value is calculated using, q = q(α,r,dfW). In this case, α = 0.05, r = 6, df provided by the Anova table = 84.

The value from Tukey HSD table is 4.1246

The critical value for Scheffe test is calculated using, CV = (k-1)F = 5*7.44 = 37.2

(c) Construct 95% confidence intervals for Group 1 vs. Group 2 and Group 4 vs. Group 5 using each of the three procedures above, and assuming that you intend to test all pairwise comparisons (you need to use this in the Bonferroni contrasts). Explain what each contrast represents. Which procedure is the best for identifying differences?

Tukey Method	Critical q	Standard error	Mean difference	Confidence interval
Group 1 vs Group 2	4.1246	1.798	-1.8	9.21, -5.61
Group 4 vs. Group 5	4.1246	1.798	-0.0467	7.46, -7.36

Scheffe Method	Critical value	Confidence interval
Group 1 vs Group 2	3.33	5.13, -1.53
Group 4 vs. Group 5	3.33	3.37, --3.28

The contrast of the groups represents that the Group 1 vs Group 2 has more unequal mean in comparison to group 4 vs. Group 5. Therefore the group 1 and 2 caused most difference. The Tukey method is better for calculating as it provided wider confidence intervals thus, leading to inclusion of more results.

(d) Define a contrast to determine if students in the twice weekly lecture format have different mean achievement from those in the once weekly lecture format (averaging the means for each condition). Construct 95% confidence intervals using Scheffe and Bonferroni procedures. In using the Bonferroni procedure, assume that you also planned to test all pairwise comparisons. State your conclusion.

(e) Define a contrast to determine if the difference in mean achievement for students who got pop quizzes and students who attended tutorials is the same under the once weekly lecture format as under the twice weekly format. Think of this as (popquiz-tutorial) once weekly – (popquiz-tutorial) twice weekly. This is a difference between differences. Construct 95% confidence intervals using Scheffe and Bonferroni procedures. In using the Bonferroni procedure, assume that you also planned to test all pairwise comparisons. State your conclusion.