USA: +1(669)289-8683 
Sign in
hello i need it done by tomorrow 100% correct solution required.
thanx
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 |
|||||||
| 3 |
9.2267 |
1.82462 |
|||||||
| 4 |
10.7333 |
2.11919 |
|||||||
| 5 |
10.7800 |
1.6 89 55 |
|||||||
| 6 |
12.1267 |
1.99516 |
|||||||
|
Total |
2.12999 |
90 |
An ANOVA table for testing the hypothesis of equal means was constructed and is shown below.
|
Tests of Between-Subjects Effects |
||||||||
|
Source |
Type III Sum of Squares |
df |
Mean Square |
F |
Sig. |
|||
|
Corrected Model |
123.937 a |
24.787 |
7.440 |
.000 |
||||
|
Intercept |
9486.400 |
2 84 7.520 |
||||||
| 123.937 | ||||||||
|
Error |
279.843 |
84 |
3.331 |
|||||
|
9890.180 |
||||||||
|
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.
(b) Determine Bonferroni, Tukey, and Scheffe critical values for testing all pairwise differences among the means.
(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?
(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.
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 |
|||||||
| 3 |
9.2267 |
1.82462 |
|||||||
| 4 |
10.7333 |
2.11919 |
|||||||
| 5 |
10.7800 |
1.6 89 55 |
|||||||
| 6 |
12.1267 |
1.99516 |
|||||||
|
Total |
2.12999 |
90 |
An ANOVA table for testing the hypothesis of equal means was constructed and is shown below.
|
Tests of Between-Subjects Effects |
||||||||
|
Source |
Type III Sum of Squares |
df |
Mean Square |
F |
Sig. |
|||
|
Corrected Model |
123.937 a |
24.787 |
7.440 |
.000 |
||||
|
Intercept |
9486.400 |
2 84 7.520 |
||||||
| 123.937 | ||||||||
|
Error |
279.843 |
84 |
3.33 1 |
|||||
|
9890.180 |
||||||||
|
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 |
-0.0467 |
7.46, -7.36 |
|
Scheffe Method |
Critical value |
|
| 3.33 |
5.13, -1.53 |
|
|
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.
