# Data Analysis

The data collected yield the following counts:

Total number of Cases: 112
Total number of Controls: 224
Number of Cases who ingested Endurobrick: 28
Number of Controls who ingested Endurobrick: 56
Number of Cases who consumed Quench-it: 50
Number of Controls who consumed Quench-it: 56

8. How would you set up the classic 2x2 table using the above information to test the hypothesis that cases are more likely to have ingested EnduroBrick than controls?

none:
&nbsp Case Control Total
Exposed (Endurobrick) 28 56 84
Unexposed (No Endurobrick) 84 168 252
Total 112 224 336
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Odds of exposure among cases (# of Cases exposed) / (# of Cases unexposed)

28 / 84 = 0.333

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Odds of exposure among controls (# of Controls exposed) / (# of Controls unexposed)

56 / 168 = 0.333

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OR = (Odds of Exposure among Cases) / (Odds of Exposure among Controls)

(28/84) / (56/168) = 1.0 or 0.33 / 0.33 = 1.0

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Odds of disease among exposed (# of Cases Exposed) / (# of Controls Exposed)

28 / 56 = 0.50

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Odds of disease among unexposed (# of Cases Unexposed) / (# of Controls Unexposed)

84 / 168 = 0.50

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OR = Odds of disease among Exposed / Odds of disease among Unexposed

(28/56) / (84/168) = 1.0 or 0.50 / 0.50 = 1.0

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Individuals with Susser Syndrome (cases) have the same odds of having ingested EnduroBrick as those without Susser Syndrome (controls). Conversely, individuals who ate EnduroBrick have the same odds of developing Susser Syndrome as those who did not eat EnduroBrick. An Odds Ratio = 1.0 suggests that there is no association between Susser Syndrome and EnduroBrick ingestion.

9. How would you set up the classic 2x2 table using the above information to test the hypothesis that cases are more likely to have consumed Quench-It than controls?

none:
Case Control Total
Exposed (Quench-it) 50 56 106
Unexposed (No Quench-it) 62 168 230
Total 112 224 336
none:

Odds of exposure among cases (# of Cases exposed) / (# of Cases unexposed)

50 / 62 = 0.806

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Odds of exposure among controls (# of Controls exposed) / (# of Controls unexposed)

56 / 168 = 0.333

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OR = (Odds of Exposure among Cases) / (Odds of Exposure among Controls)

OR = (50/62) / (56/168) = 2.4 or 0.806 / 0.333 = 2.4

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Odds of disease among exposed (# of Cases Exposed) / (# of Controls Exposed)

50 / 56 = 0.893

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Odds of disease among unexposed (# of Cases Unexposed) / (# of Controls Unexposed)

62 / 168 = 0.369

none:

OR = Odds of disease among Exposed / Odds of disease among Unexposed

OR = (50/56) / (62/168) = 2.4 or 0.893/0.369 = 2.4