In a study of the relationship between angular stomatitis and occupati

Question

In a study of the relationship between angular stomatitis and occupation, the occupation of each individual was recorded as : P (Professional), S (Skilled) or U (Unskilled). There were 88 people with angular stomatitis and 100 without the disease. The
distribution of 188 people by occupational classification and angular stomatitis is as
follows :

28 Jun 2018

Answer :

Word Count : 934

You have a 3×2 table with occupation as exposure (P = Professional, S = Skilled, U = Unskilled) and angular stomatitis as outcome (present/absent). Let the six cell counts be: for Professionals, diseased = a, non-diseased = b; for Skilled, diseased = c, non-diseased = d; for Unskilled, diseased = e, non-diseased = f. Given totals: diseased (D+) = a + c + e = 88; non-diseased (D−) = b + d + f = 100; N = 188. Row totals are n₁ = a + b (P), n₂ = c + d (S), n₃ = e + f (U). 1) Descriptive layout and basic measures Report the table with row and column totals. Compute prevalence of angular stomatitis within each occupation: * P: p₁ = a/(a + b) * S: p₂ = c/(c + d) * U: p₃ = e/(e + f) Pick a reference category (commonly Professionals). Derive prevalence ratios (PRs) and odds ratios (ORs): * PR(S vs P) = \[c/(c + d)] ÷ \[a/(a + b)] * PR(U vs P) = \[e/(e + f)] ÷ \[a/(a + b)] * OR(S vs P) = (c/d) ÷ (a/b) = (c·b)/(a·d) * OR(U vs P) = (e/f) ÷ (a/b) = (e·b)/(a·f) For each PR or OR, include 95% CIs. For ORs, use the log method: SE{ln(OR)} = √(1/a + 1/b + 1/c + 1/d). CI: ln(OR) ± 1.96·SE, then exponentiate. For PRs, use _________ _________ _____ __________ __________ _________ _______ _________ ___ _____.
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