Ask Statistics and Probability Expert

Please write up and interpret the results for the following repeated measures ANOVA, using the Activity 6.sav data set. Score_0 through Score _12 is the repeated measure (7 levels) and gender is a fixed factor. Discuss especially both main effects and the presence/absence of an interaction between the two.

All of the relevant data is given below.

 

Within-Subjects Factors

Measure:   MEASURE_1 

Score

Dependent Variable

1

Score_0

2

Score_2

3

Score_4

4

Score_6

5

Score_8

6

Score_10

7

Score_12

Between-Subjects Factors

 

Value Label

N

Gender

F

Female

8

M

Male

4

Descriptive Statistics

 

Gender

Mean

Std. Deviation

N

Pre-test score

Female

28.25

8.172

8

Male

32.25

19.432

4

Total

29.58

12.221

12

Week 2 score

Female

29.75

6.319

8

Male

39.75

13.889

4

Total

33.08

10.113

12

Week 4 score

Female

33.63

5.181

8

Male

39.00

16.432

4

Total

35.42

9.885

12

Week 6 score

Female

35.88

6.556

8

Male

35.25

17.802

4

Total

35.67

10.671

12

Week 8 score

Female

39.38

5.370

8

Male

41.00

16.633

4

Total

39.92

9.718

12

Week 10 score

Female

44.88

5.743

8

Male

47.25

13.961

4

Total

45.67

8.690

12

Week 12 score

Female

48.38

8.518

8

Male

53.25

13.793

4

Total

50.00

10.189

12

Multivariate Testsa

Effect

Value

F

Hypothesis df

Error df

Sig.

Score

Pillai's Trace

.961

20.439b

6.000

5.000

.002

Wilks' Lambda

.039

20.439b

6.000

5.000

.002

Hotelling's Trace

24.526

20.439b

6.000

5.000

.002

Roy's Largest Root

24.526

20.439b

6.000

5.000

.002

Score * Gender

Pillai's Trace

.491

.804b

6.000

5.000

.607

Wilks' Lambda

.509

.804b

6.000

5.000

.607

Hotelling's Trace

.965

.804b

6.000

5.000

.607

Roy's Largest Root

.965

.804b

6.000

5.000

.607

a. Design: Intercept + Gender

 Within Subjects Design: Score

b. Exact statistic

 

Mauchly's Test of Sphericitya

 

Measure:   MEASURE_1 

 

Within Subjects Effect

Mauchly's W

Approx. Chi-Square

df

Sig.

Epsilonb

 

Greenhouse-Geisser

Huynh-Feldt

Lower-bound

 

Score

.001

56.876

20

.000

.441

.674

.167

 

Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix.

 

a. Design: Intercept + Gender

 Within Subjects Design: Score

 

b. May be used to adjust the degrees of freedom for the averaged tests of significance. Corrected tests are displayed in the Tests of Within-Subjects Effects table.

 

                             

Tests of Within-Subjects Effects

Measure:   MEASURE_1 

Source

Type III Sum of Squares

df

Mean Square

F

Sig.

Score

Sphericity Assumed

3246.536

6

541.089

20.609

.000

Greenhouse-Geisser

3246.536

2.646

1227.164

20.609

.000

Huynh-Feldt

3246.536

4.045

802.659

20.609

.000

Lower-bound

3246.536

1.000

3246.536

20.609

.001

Score * Gender

Sphericity Assumed

182.155

6

30.359

1.156

.342

Greenhouse-Geisser

182.155

2.646

68.853

1.156

.341

Huynh-Feldt

182.155

4.045

45.035

1.156

.344

Lower-bound

182.155

1.000

182.155

1.156

.307

Error(Score)

Sphericity Assumed

1575.321

60

26.255

 

 

Greenhouse-Geisser

1575.321

26.456

59.546

 

 

Huynh-Feldt

1575.321

40.447

38.948

 

 

Lower-bound

1575.321

10.000

157.532

 

 

 

Tests of Within-Subjects Contrasts

Measure:   MEASURE_1 

Source

Score

Type III Sum of Squares

df

Mean Square

F

Sig.

Score

Linear

2962.680

1

2962.680

46.905

.000

Quadratic

143.040

1

143.040

4.305

.065

Cubic

51.361

1

51.361

2.242

.165

Order 4

73.724

1

73.724

2.765

.127

Order 5

3.584

1

3.584

.405

.539

Order 6

12.147

1

12.147

4.444

.061

Score * Gender

Linear

25.537

1

25.537

.404

.539

Quadratic

21.254

1

21.254

.640

.442

Cubic

66.694

1

66.694

2.911

.119

Order 4

55.767

1

55.767

2.092

.179

Order 5

5.060

1

5.060

.572

.467

Order 6

7.841

1

7.841

2.869

.121

Error(Score)

Linear

631.638

10

63.164

 

 

Quadratic

332.272

10

33.227

 

 

Cubic

229.083

10

22.908

 

 

Order 4

266.594

10

26.659

 

 

Order 5

88.403

10

8.840

 

 

Order 6

27.330

10

2.733

 

 

Tests of Between-Subjects Effects

Measure:   MEASURE_1 

Transformed Variable:   Average 

Source

Type III Sum of Squares

df

Mean Square

F

Sig.

Intercept

114349.339

1

114349.339

188.733

.000

Gender

290.720

1

290.720

.480

.504

Error

6058.804

10

605.880

 

 

a. Is the assumption of sphericity violated? How can you tell? What does this mean in the context of interpreting the results?

Mauchly's Test of Sphericitya

Measure:   MEASURE_1 

Within Subjects Effect

Mauchly's W

Approx. Chi-Square

df

Sig.

Epsilonb

Greenhouse-Geisser

Huynh-Feldt

Lower-bound

Score

.001

56.876

20

.000

.441

.674

.167

Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix.

a. Design: Intercept + Gender

 Within Subjects Design: Score

b. May be used to adjust the degrees of freedom for the averaged tests of significance. Corrected tests are displayed in the Tests of Within-Subjects Effects table.

The above table depicts the results of Mauchly's Test of Spheriicty which tests for one of the assumptions of the ANOVA with repeated measures, namely, sphericity (homogeneity of covariance). This particular table is important for viewing as this assumption is commonly violated. In this case, since p-value is less than .05, I conclude that there are significant differences between the variance of difference. Therefore, the condition of sphericity has not been met.

b. Is there a main effect of gender? Is so, explain the effect. Use post hoc tests when necessary (or explain why they are not required in this specific case).

In this case, there is no main effect of gender since gender has a p-value of .504 which means that this is not significant at the 5% level. Also, in this case, the effect is not significant so there is no need for a post hoc test. Moreover, if the effect was significant, then we would not be able to perform the post hoc test since we only have two categories. Post hoc can be run if there are more than two classifications.

c. Is there a main effect tie (i.e. an increase in scores from Week 0 to Week 12)? If so, explain the effect. Use post hoc tests when necessary (or explain why they are not required in this specific case). Examine the output carefully and give as much detail as possible in your findings.

Tests of Within-Subjects Effects

Measure:   MEASURE_1 

Source

Type III Sum of Squares

df

Mean Square

F

Sig.

Score

Sphericity Assumed

3246.536

6

541.089

20.609

.000

Greenhouse-Geisser

3246.536

2.646

1227.164

20.609

.000

Huynh-Feldt

3246.536

4.045

802.659

20.609

.000

Lower-bound

3246.536

1.000

3246.536

20.609

.001

Score * Gender

Sphericity Assumed

182.155

6

30.359

1.156

.342

Greenhouse-Geisser

182.155

2.646

68.853

1.156

.341

Huynh-Feldt

182.155

4.045

45.035

1.156

.344

Lower-bound

182.155

1.000

182.155

1.156

.307

Error(Score)

Sphericity Assumed

1575.321

60

26.255

 

 

Greenhouse-Geisser

1575.321

26.456

59.546

 

 

Huynh-Feldt

1575.321

40.447

38.948

 

 

Lower-bound

1575.321

10.000

157.532

 

 

 

Pairwise Comparisons

Measure:MEASURE_1

(I) SCORE

(J) SCORE

Mean Difference (I-J)

Std. Error

Sig.a

95% Confidence Interval for Differencea

Lower Bound

Upper Bound

 

1

 

2

-4.500

3.383

.213

-12.039

3.039

3

-6.062*

2.412

.031

-11.437

-.688

4

-5.312*

2.117

.031

-10.029

-.596

5

-9.937*

2.693

.004

-15.938

-3.937

6

-15.813*

2.683

.000

-21.791

-9.834

7

-20.563*

3.524

.000

-28.415

-12.710

2

 

1

4.500

3.383

.213

-3.039

12.039

3

-1.562

1.542

.335

-4.998

1.873

4

-.812

2.324

.734

-5.990

4.365

5

-5.437*

1.914

.018

-9.701

-1.174

6

-11.313*

2.063

.000

-15.909

-6.716

7

-16.063*

3.154

.000

-23.089

-9.036

3

 

1

6.062*

2.412

.031

.688

11.437

2

1.562

1.542

.335

-1.873

4.998

4

.750

1.097

.510

-1.693

3.193

5

-3.875*

1.058

.004

-6.233

-1.517

6

-9.750*

1.336

.000

-12.726

-6.774

7

-14.500*

2.739

.000

-20.603

-8.397

4

 

1

5.312*

2.117

.031

.596

10.029

2

.812

2.324

.734

-4.365

5.990

3

-.750

1.097

.510

-3.193

1.693

5

-4.625*

1.019

.001

-6.895

-2.355

6

-10.500*

1.202

.000

-13.177

-7.823

7

-15.250*

2.711

.000

-21.291

-9.209

5

 

1

9.937*

2.693

.004

3.937

15.938

2

5.437*

1.914

.018

1.174

9.701

3

3.875*

1.058

.004

1.517

6.233

4

4.625*

1.019

.001

2.355

6.895

6

-5.875*

.716

.000

-7.471

-4.279

7

-10.625*

2.065

.000

-15.226

-6.024

6

 

1

15.813*

2.683

.000

9.834

21.791

2

11.313*

2.063

.000

6.716

15.909

3

9.750*

1.336

.000

6.774

12.726

4

10.500*

1.202

.000

7.823

13.177

5

5.875*

.716

.000

4.279

7.471

7

-4.750*

1.705

.019

-8.548

-.952

7

 

1

20.563*

3.524

.000

12.710

28.415

2

16.063*

3.154

.000

9.036

23.089

3

14.500*

2.739

.000

8.397

20.603

4

15.250*

2.711

.000

9.209

21.291

5

10.625*

2.065

.000

6.024

15.226

6

4.750*

1.705

.019

.952

8.548

Based on estimated marginal means

a. Adjustment for multiple comparisons: Least Significant Difference (equivalent to no adjustments).

*. The mean difference is significant at the .05 level.

The mean effect of score is significant at 5% level of significance. From the table, I am able to ascertain the F-value for the score factor, its associated significance level, and the effect size (Partial Eta Squared). Because my data violated the assumption of sphericity, I examine the values in the Greenhouse-Geisser row (if sphericity had not been violated, I would have looked under the Sphericity Assumed row). Thus, I can report that when using an ANOVA with repeated measures with a Greenhouse-Geiseer correction, the mean scores for weeks were statistically significantly different (F(2.646,60) = 20.609, p < 0.0005.

In addition, in looking at the above Paired Comparisons Table, I recognize the labels associated with score in the experiment from the Within-Subject Factors Table. This is a table which gives the significance level of differences between the individual time points. It can be seen that there was a significant difference in scores in training from pre to week 12. The p-values indicate the significant differences between the groups.

Statistics and Probability, Statistics

  • Category:- Statistics and Probability
  • Reference No.:- M91615574
  • Price:- $50

Priced at Now at $50, Verified Solution

Have any Question?


Related Questions in Statistics and Probability

Introduction to epidemiology assignment -assignment should

Introduction to Epidemiology Assignment - Assignment should be typed, with adequate space left between questions. Read the following paper, and answer the questions below: Sundquist K., Qvist J. Johansson SE., Sundquist ...

Question 1 many high school students take the ap tests in

Question 1. Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology exam 84,199 of them were female. In that same year,of the 211,693 students who too ...

Basic statisticsactivity 1define the following terms1

BASIC STATISTICS Activity 1 Define the following terms: 1. Statistics 2. Descriptive Statistics 3. Inferential Statistics 4. Population 5. Sample 6. Quantitative Data 7. Discrete Variable 8. Continuous Variable 9. Qualit ...

Question 1below you are given the examination scores of 20

Question 1 Below you are given the examination scores of 20 students (data set also provided in accompanying MS Excel file). 52 99 92 86 84 63 72 76 95 88 92 58 65 79 80 90 75 74 56 99 a. Construct a frequency distributi ...

Question 1 assume you have noted the following prices for

Question: 1. Assume you have noted the following prices for paperback books and the number of pages that each book contains. Develop a least-squares estimated regression line. i. Compute the coefficient of determination ...

Question 1 a sample of 81 account balances of a credit

Question 1: A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126. 1. Formulate the hypotheses that can be used to determine whether the mean of all acc ...

5 of females smoke cigarettes what is the probability that

5% of females smoke cigarettes. What is the probability that the proportion of smokers in a sample of 865 females would be greater than 3%

Armstrong faber produces a standard number-two pencil

Armstrong Faber produces a standard number-two pencil called Ultra-Lite. The demand for Ultra-Lite has been fairly stable over the past ten years. On average, Armstrong Faber has sold 457,000 pencils each year. Furthermo ...

Sppose a and b are collectively exhaustive in addition pa

Suppose A and B are collectively exhaustive. In addition, P(A) = 0.2 and P(B) = 0.8. Suppose C and D are both mutually exclusive and collectively exhaustive. Further, P(C|A) = 0.7 and P(D|B) = 0.5. What are P(C) and P(D) ...

The time to complete 1 construction project for company a

The time to complete 1 construction project for company A is exponentially distributed with a mean of 1 year. Therefore: (a) What is the probability that a project will be finished in one and half years? (b) What is the ...

  • 4,153,160 Questions Asked
  • 13,132 Experts
  • 2,558,936 Questions Answered

Ask Experts for help!!

Looking for Assignment Help?

Start excelling in your Courses, Get help with Assignment

Write us your full requirement for evaluation and you will receive response within 20 minutes turnaround time.

Ask Now Help with Problems, Get a Best Answer

Why might a bank avoid the use of interest rate swaps even

Why might a bank avoid the use of interest rate swaps, even when the institution is exposed to significant interest rate

Describe the difference between zero coupon bonds and

Describe the difference between zero coupon bonds and coupon bonds. Under what conditions will a coupon bond sell at a p

Compute the present value of an annuity of 880 per year

Compute the present value of an annuity of $ 880 per year for 16 years, given a discount rate of 6 percent per annum. As

Compute the present value of an 1150 payment made in ten

Compute the present value of an $1,150 payment made in ten years when the discount rate is 12 percent. (Do not round int

Compute the present value of an annuity of 699 per year

Compute the present value of an annuity of $ 699 per year for 19 years, given a discount rate of 6 percent per annum. As