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Question: Over time, with instruction in between, each student completed a quiz

ID: 3291297 • Letter: Q

Question

Question: Over time, with instruction in between, each student completed a quiz five times. The quiz scores or the measurement of student achievement over time would serve as the dependent variable while instruction would serve as the independent variable/treatment/factor. Provide a narrative discussion of the tables, including a discussion of SPHERICITY. DATA BELOW:

Pairwise Comparisons

Measure:   MEASURE_1

(I) instruction

(J) instruction

Mean Difference (I-J)

Std. Error

Sig.b

95% Confidence Interval for Differenceb

Lower Bound

Upper Bound

1

2

-.514*

.179

.049

-1.028

-.001

3

-.514*

.126

.001

-.874

-.154

4

-.333

.137

.168

-.727

.060

5

-.400

.215

.658

-1.017

.217

2

1

.514*

.179

.049

.001

1.028

3

.000

.164

1.000

-.469

.469

4

.181

.173

1.000

-.316

.678

5

.114

.129

1.000

-.255

.483

3

1

.514*

.126

.001

.154

.874

2

.000

.164

1.000

-.469

.469

4

.181

.143

1.000

-.229

.591

5

.114

.205

1.000

-.475

.703

4

1

.333

.137

.168

-.060

.727

2

-.181

.173

1.000

-.678

.316

3

-.181

.143

1.000

-.591

.229

5

-.067

.212

1.000

-.676

.542

5

1

.400

.215

.658

-.217

1.017

2

-.114

.129

1.000

-.483

.255

3

-.114

.205

1.000

-.703

.475

4

.067

.212

1.000

-.542

.676

Based on estimated marginal means

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

b. Adjustment for multiple comparisons: Bonferroni.

Descriptive Statistics

Mean

Std. Deviation

N

quiz1

7.47

2.481

105

quiz2

7.98

1.623

105

quiz3

7.98

2.308

105

quiz4

7.80

2.280

105

quiz5

7.87

1.765

105

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

instruction

.400

93.851

9

.000

.640

.657

.250

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

a. Design: Intercept

Within Subjects Design: instruction

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.

Pairwise Comparisons

Measure:   MEASURE_1

(I) instruction

(J) instruction

Mean Difference (I-J)

Std. Error

Sig.b

95% Confidence Interval for Differenceb

Lower Bound

Upper Bound

1

2

-.514*

.179

.049

-1.028

-.001

3

-.514*

.126

.001

-.874

-.154

4

-.333

.137

.168

-.727

.060

5

-.400

.215

.658

-1.017

.217

2

1

.514*

.179

.049

.001

1.028

3

.000

.164

1.000

-.469

.469

4

.181

.173

1.000

-.316

.678

5

.114

.129

1.000

-.255

.483

3

1

.514*

.126

.001

.154

.874

2

.000

.164

1.000

-.469

.469

4

.181

.143

1.000

-.229

.591

5

.114

.205

1.000

-.475

.703

4

1

.333

.137

.168

-.060

.727

2

-.181

.173

1.000

-.678

.316

3

-.181

.143

1.000

-.591

.229

5

-.067

.212

1.000

-.676

.542

5

1

.400

.215

.658

-.217

1.017

2

-.114

.129

1.000

-.483

.255

3

-.114

.205

1.000

-.703

.475

4

.067

.212

1.000

-.542

.676

Based on estimated marginal means

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

b. Adjustment for multiple comparisons: Bonferroni.

Explanation / Answer

Over time, with instruction in between, each student completed a quiz five times.  The quiz scores or the measurement of student achievement over time would serve as the dependent variable while instruction would serve as the independent variable/treatment/factor.

Here we have given pairwise comparison.

We can test the hypothesis that,

H0 : mu1 = mu2 Vs H1 : mu1 not= mu2

where mu1 and mu2 are two population means.

Assume alpha = level of significance = 0.05

If P-value < alpha then reject H0 and we get significant result about that pair.

Here we see that pair (1,3), (2,1) are significant while remaining are insignificant pairs.

Also we have given the output for Mauchly's Test of Sphericitya

Here the test of hypothesis is,

H0 : The error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix.

H1 : The error covariance matrix of the orthonormalized transformed dependent variables is not proportional to an identity matrix.

Assume alpha = level of significance = 0.05

Here we see that p-value = 0.000

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion :  The error covariance matrix of the orthonormalized transformed dependent variables is not proportional to an identity matrix.

We get significant result about this test.

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