statistics

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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

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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.

15

15

15

15

15

10.2667

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.

Dependent Variable:Acht

5

1

9486.400

.000

Group

5

24.787

7.440

.000

Total

90

 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.

15

15

15

15

15

10.2667

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.

Dependent Variable:Acht

5

1

9486.400

.000

Group

5

24.787

7.440

.000

Total

90

 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?

4.1246

1.798

 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

Confidence interval

Group 1 vs Group 2

Group 4 vs. Group 5

3.33

 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.

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