Linear Contrasts:



Linear Contrasts & Orthogonality Practice:

1) Control Placebo Drug α Drug β Drug δ

1 2 3 4 5

3 4 7 4 10

2 6 8 4 9

A) Conduct a simple ANOVA to determine if drug condition significantly affects error.

ANOVA

Errors

| |Sum of Squares |df |Mean Square |F |Sig. |

|Between Groups |62.400 |4 |15.600 |4.105 |.032 |

|Within Groups |38.000 |10 |3.800 | | |

|Total |100.400 |14 | | | |

B) Test the significance of the following contrasts:

Control & Placebo vs. Drug α , Drug β & Drug δ

Control vs. Placebo

Drug α vs. Drug β & Drug δ

Drug β vs. Drug δ

Contrast Coefficients

|Contrast |Drug |

| |Control |Placebo |Drug Alpha |Drug Beta |Drug Delta |

|1 |-3 |-3 |2 |2 |2 |

|2 |-1 |1 |0 |0 |0 |

|3 |0 |0 |-2 |1 |1 |

|4 |0 |0 |0 |-1 |1 |

Contrast Tests

| | |Contrast |Value of Contrast |Std. Error |t |

|Between Groups |512.400 |4 |128.100 |25.620 |.000 |

|Within Groups |50.000 |10 |5.000 | | |

|Total |562.400 |14 | | | |

A) Test the significance of the following contrasts:

Control vs. Bambi, Hostel, Gigli, & Bring It On!

Control vs. Bambi

Hostel vs. Gigli & Bring It On!

Gigli vs. Bring It On!

Contrast Coefficients

|Contrast |Movie |

| |Control (Nature|Bambi |Hostel |Gigli |Bring It On! |

| |Movie) | | | | |

|1 |3 |3 |-2 |-2 |-2 |

|2 |1 |-1 |0 |0 |0 |

|3 |0 |0 |2 |-1 |-1 |

|4 |0 |0 |0 |1 |-1 |

Contrast Tests

| |Contrast |Value of Contrast |Std. Error |t |df |Sig. (2-tailed) | |Depressionscores |Assume equal variances |1 |-32.0000 |7.07107 |-4.525 |10 |.001 | | | |2 |-4.0000 |1.82574 |-2.191 |10 |.053 | | | |3 |-25.0000 |3.16228 |-7.906 |10 |.000 | | | |4 |-7.0000 |1.82574 |-3.834 |10 |.003 | | |Does not assume equal variances |1 |-32.0000 |6.45497 |-4.957 |5.548 |.003 | | | |2 |-4.0000 |1.29099 |-3.098 |2.941 |.055 | | | |3 |-25.0000 |3.10913 |-8.041 |4.195 |.001 | | | |4 |-7.0000 |2.38048 |-2.941 |2.249 |.086 | |

C) Prove Orthogonality

Partition Proof

SScon1 (102.4) + SScon2(24) + SScon3(312.5) + SScon4(73.5) = SSbet (512.4)

Proof of Orthogonality:

Contrast 1 vs. Contrast 2:

3(1) + 3(-1) + -2(0) + -2(0) -2(0) = 0

Contrast 1 vs. Contrast 3:

3(0) + 3(0) + -2(2) + -2(-1) + -2(-1) = 0

Contrast 1 vs. Contrast 4:

3(0) + 3(0) + -2(0) + -2(1) + -2(-1) = 0

Contrast 2 vs. Contrast 3:

1(0) + -1(0) + 0(2) + 0(-1) + 0(-1) = 0

Contrast 2 vs. Contrast 4:

1(0) + -1(0) + 0(0) + 0(1) + 0(-1) = 0

Contrast 3 vs. Contrast 4:

0(0) + 0(0) + 2(0) + -1(1) + -1(-1) = 0

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