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In this paper, we are going to discuss the Repeated Measures ANOVA. We are going to consider an experiment done by students who wanted to check the consistency of marking of the lecturers. The students submitted the same essay to four different lecturers and the grade given by each of the four lecturers. The total number of essays assigned to each lecturer was eight. Here, the independent variable is the lecturer who is marking while the dependent variable is the marks given by each lecturer.

In this paper, we are going to look at the hypothesis of the ANOVA test, and then the descriptive statistics of the variables, the posthoc tests to test the assumptions and lastly the main ANOVA test (effects of the main variables). The assumption of this test is the sphericity assumption, which will be tested by Mauchly’s Test. All the data analysis is done using IBM SPSS.**Hypothesis**

Ho: All the tutors are consistent in marking

Ha: The tutors are not consistent in marking**Descriptive Statistics**

The mean and standard deviation of the marks given by each tutor are given in the table below.Descriptive Statistics | |||

Mean | Std. Deviation | N | |

Dr. Field | 68.88 | 5.643 | 8 |

Dr. Smith | 64.25 | 4.713 | 8 |

Dr. Scrote | 65.25 | 6.923 | 8 |

Dr. Death | 57.38 | 7.909 | 8 |

*Table 1*: Table of descriptive statistics

From the table above, we see that Dr. Field has the highest mean which is 68.88 with a standard deviation of 5.643. He is followed by Dr. Scrote who has a mean of 65.25 with a standard deviation of 6.923. Then we have Dr. Smith who has a mean of 64.25 and a standard deviation of 5.6 and lastly is Dr. Death with a mean of 57.38 and a standard deviation of 7.9.**Post-hoc tests**

We assess the assumption of sphericity. Here, the data should have significant diversion from spericity. Below is the output of Mauchly’s Test of Sphericity.

Mauchly’s Test of Sphericity^{b} | |||||||

Measure:MEASURE_1 | |||||||

Within Subjects Effect | Mauchly’s W | Approx. Chi-Square | df | Sig. | Epsilon^{a} | ||

Greenhouse-Geisser | Huynh-Feldt | Lower-bound | |||||

Tutor | .131 | 11.628 | 5 | .043 | .558 | .712 | .333 |

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

a. 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. | |||||||

b. Design: Intercept Within Subjects Design: Tutor |

*Table 2*: Post-Hoc test

From Table 2 above, we see that we have a p-value of 0.043. Since this p-value is less than 0.05, it means that the assumption of sphericity has been violated. Also, the table 2 above shows the level of sphericity. We have Greenhouse-Geisser value as 0.558 and Huynh-Feldt value as 0.712. It is known that if a data is perfectly spherical, then both of the estimates should be closer to 1 (Andey, 2013). For the Greenhouse-Geisser value, we always compute the lowest value as 1/(k-1), where k is the number of levels. Therefore, since we have four conditions, the lowest bound is 0.333. From the Greenhouse-Geisser value above, we see that its coefficient is closer to 0.333 (lower value) than it is to 1 which is the upper value. This implies that there is a substantial diversion from sphericity.**Main ANOVA**

The table below shows the ANOVA output, where we are comparing the means of the four tutors.Tests of Within-Subjects Effects | ||||||

Measure:MEASURE_1 | ||||||

Source | Type III Sum of Squares | df | Mean Square | F | Sig. | |

Tutor | Sphericity Assumed | 554.125 | 3 | 184.708 | 3.700 | .028 |

Greenhouse-Geisser | 554.125 | 1.673 | 331.245 | 3.700 | .063 | |

Huynh-Feldt | 554.125 | 2.137 | 259.329 | 3.700 | .047 | |

Lower-bound | 554.125 | 1.000 | 554.125 | 3.700 | .096 | |

Error(Tutor) | Sphericity Assumed | 1048.375 | 21 | 49.923 | ||

Greenhouse-Geisser | 1048.375 | 11.710 | 89.528 | |||

Huynh-Feldt | 1048.375 | 14.957 | 70.091 | |||

Lower-bound | 1048.375 | 7.000 | 149.768 |

*Table 3:* Main ANOVA test

From the table above, we see that the four means are significantly different since F(3,1)= 3.7, p-value=.0.28. Since the p-value is less than 0.05, we reject the null hypothesis and conclude that the tutors are not consistent in marking**Conclusion**

From the above tests, we have seen that the assumption of sphericity has been met. This gives us a go ahead to conduct the ANOVA test. The ANOVA test above shows that the lecturers are not consistent in marking. Different lecturers give different marks for the same test.

References

Andey Field, (2013). *Discovering Statistics Using IBM SPSS Statistics*. Washington: Sage Publication