chapter10 - Chapter 10 Categorical Data Analysis Inference for a Single Proportion Goal Estimate proportion of individuals in a population with a

# chapter10 - Chapter 10 Categorical Data Analysis Inference...

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Chapter 10 Categorical Data Analysis
Inference for a Single Proportion ( π ) Goal: Estimate proportion of individuals in a population with a certain characteristic ( π ). This is equivalent to estimating a binomial probability Sample: Take a SRS of n individuals from the population and observe y that have the characteristic. The sample proportion is y / n and has the following sampling properties: ) 5 ) 1 ( , : thumb of (Rule samples large for normal ely approximat : Shape 1 : Error Standard Estimated ) 1 ( : on distributi sampling of Dev. Std. and Mean : proportion Sample ^ ^ ^ ^ ^ ^ - - = - = = = π π π π π π σ π μ π π π π n n n SE n n y
Large-Sample Confidence Interval for π Take SRS of size n from population where π is true (unknown) proportion of successes. Observe y successes Set confidence level (1- α ) and obtain z α /2 from z -table m C z m n n y ± = - = = ^ 2 / ^ ^ ^ : for interval confidence % SE : error of Margin 1 SE : Error Standard Estimated : Estimate Point ^ ^ π π π π π π α π
Example - Ginkgo and Azet for AMS Study Goal: Measure effect of Ginkgo and Acetazolamide on occurrence of Acute Mountain Sickness (AMS) in Himalayan Trackers Parameter: π = True proportion of all trekkers receiving Ginkgo&Acetaz who would suffer from AMS. Sample Data: n= 126 trekkers received G&A, y =18 suffered from AMS ) 204 ,. 082 (. 061 . 143 . : for CI % 95 061 . ) 031 (. 96 . 1 : %) 95 % 100 ) 1 (( error of Margin 031 . 126 ) 86 )(. 14 (. SE 143 . 126 18 ^ ^ ± = = = - = = = = π α π π m
Wilson’s “Plus 4” Method For moderate to small sample sizes, large-sample methods may not work well wrt coverage probabilities Simple approach that works well in practice ( n 10) : Pretend you have 4 extra individuals, 2 successes, 2 failures Compute the estimated sample proportion in light of new “data” as well as standard error: m z m n n y ± - = + - = + + = ~ 2 / ~ ~ ~ : for interval confidence % 100 ) 1 ( SE : error of Margin 4 1 SE : Error Standard Estimated 4 2 : Estimate Point ~ ~ π π α π π π π α π
Example: Lister’s Tests with Antiseptic Experiments with antiseptic in patients with upper limb amputations (John Lister, circa 1870) n =12 patients received antiseptic y =1 died ) 40 ,. 0 ( ) 3988 ,. 0038 . ( 1913 . 1875 . : for CI % 95 1913 . ) 0976 (. 96 . 1 : %) 95 )100% - 1 ( error( of Margin 0976 . 16 ) 8125 (. 1875 . SE 1875 . 16 3 4 12 2 1 ~ ~ 2245 - ± = = = = = = + + = π α π π
Significance Test for a Proportion Goal test whether a proportion ( π ) equals some null value π 0 H 0 : π = π 0 ) ( 2 value - : : ) ( value - : : ) ( value - : : ) 1 ( : Statistic Test 2 / 0 0 0 0 0 ^ obs obs a obs obs a obs obs a o obs z Z P P z z RR H z Z P P z z RR H z Z P P z z RR H n z = = - < = - - = α α α π π π π π π π π π π Large-sample test works well when n π 0 and n (1- π 0 ) 5
Ginkgo and Acetaz for AMS Can we claim that the incidence rate of AMS is less than 25% for trekkers receiving G&A? H 0 : π = 0.25 H a : π < 0.25 0030 . ) 75 . 2 ( value - 645 . 1 : ) 05 . ( 75 . 2 039 . 107 . 126 ) 75 (. 25 . 25 . 143 . : Statistic Test 25 . 0 143 . 0 126 18 18 126 05 . 0 ^ = - = - = - = - = - = - = = = = = = Z P P z z RR z y n obs obs α π π Strong evidence that incidence rate is below 25% ( π < 0.25)
Comparing Two Population Proportions Goal: Compare two populations/treatments wrt a nominal (binary) outcome Sampling Design: Independent vs Dependent Samples Methods based on large vs small samples Contingency tables used to summarize data Measures of Association: Absolute Risk, Relative Risk, Odds Ratio
Contingency Tables Tables representing all combinations of
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