This preview shows pages 1–8. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 LargeSample 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 ztable 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, largesample 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 ,. ( ) 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 H : π= π ) ( 2 value : : ) ( value : : ) ( value : : ) 1 ( : Statistic Test 2 / ^ 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 Largesample test works well when n π and n (1 π ) ≥ 5 Ginkgo and Acetaz for AMS...
View
Full
Document
This note was uploaded on 06/04/2011 for the course STA 6166 taught by Professor Staff during the Fall '08 term at University of Florida.
 Fall '08
 Staff
 Binomial, Probability

Click to edit the document details