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Published byMelvyn Fisher Modified over 9 years ago
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Hypothesis Testing Field Epidemiology
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Hypothesis Hypothesis testing is conducted in etiologic study designs such as the case-control or cohort as well as the experimental study designs. An hypothesis is a a statement of association between exposure (predictor) and an outcome (disease or health event). Hypotheses are one-tailed or two tailed. The null hypothesis states that there is no association.
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Examples Smoking is not associated with lung cancer (Null hypothesis) Smoking is associated with a higher incidence of lung cancer (One-tailed hypothesis) Smoking is associated with a lower incidence of lung cancer (or it is protective) (One-tailed hypothesis) Smoking has some association with lung cancer (uncertain of how it influences lung cancer) (Two-tailed hypothesis)
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Rules of Thumb Usually there is one main hypothesis and a couple of secondary hypotheses The more specific you are in your statement of hypothesis, the easier it will be to answer your question Usually stated in the paper as “The purpose of the study is to….”
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Epidemiologic Decision Making
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Relative Risk R.R.= a/(a+b) ------------ c/(c+d) RR = the likelihood of developing the disease in the exposed group compared to the unexposed group
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Relative Risk for a disease exposure RR = 75/100 = 3.00 25/100 C.I. (2.10 - 4.29)
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Relative Risk for preventive intervention RR = 25/100 =.50 50/100 C.I. (.39-.79)
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Relative Risk Calculation RR = =
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Attributable Risk AR = Ie - Io the difference between incidence rates in the exposed and nonexposed groups
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Odds Ratio a/c b/d or the odds of exposure in disease compared to odds of exposure in non diseased a*d b*c mathematically equivalent to the simpler formula
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Odds Ratio O.R. = 60 * 70 = 3.50 40 * 30
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T-test - Continuous data Formula t-test = mean A - mean B - diff Null variance for the entire study pop Which group is more immuno-suppressed?
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C.I. For Mean CD4
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T-test If the sample sized are different - first must pool the variances pooled var = (4015-1)71.0 + (955-1)84.9 =74 (4015+955-2) t-test = 34.8-29.9 - 0 =4.9 =16 ________________ 41(1/4015+1/955).23
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Normally Distributed Data
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Non-normally distributed data
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2 Test of statistical association Used to determine statistical association for categorical data 2 = (O - E) 2 E
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2 Test - Categorical data
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2 Test of statistical association Used to determine statistical association for categorical data 2 = (O - E) 2 E
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2 Calculation (100-200) 2 + (400-300) 2 + (300-200) 2 + (200-300) 2 200 300 2 = 166.7, 1 D.F. (look up in table)
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Multivariable techniques Continuous OutcomeCategorical Outcome Linear regressionLogistic Regression Generalized estimating equations (GEE) Cox Regression ANOVAGEE Polychotomous
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