Session 4 - How to Analyze Rates and Proportions.docx

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Cell and Systems Biology
William Navarre

HMB325H © Lisa| Page 113 S E S S I O N 4 : H O W T O A N A LY Z E R AT E S A N D P R O P O R T I O N S ( C H A P T E R 5 ) LEARNING OBJECTIVES 1. recognize & describe the characteristics of the types of data (i.e. levels of measurement) associated w proportions 2. describe & calculate descriptive statistics used for proportions, including the standard deviation, standard error of a proportion, & confidence interval 3. describe & complete a difference of proportions analysis, including a confidence interval 4. describe Yates correction for continuity & its rationale 5. describe & carry out contingency table analysis 6. describe & determine the chi-square (χ ) test statistic 7. describe & carry out the Fischer exact test RATES & PROPORTIONS 1. information not measured quantitatively 2. measured categorically (ex. nominal or dichotomous scales) ex. male/female, alive/dead 3. no arithmetic relationship bw categories 4. need ways to describe/summarize & analyze this info DESCRIPTIVE STATISTICS 1. summarize by counts or proportions (percentages) ¿withfactor m ‘average’()= ¿∈wholesample = n 2. measure data’s variability 1. average ‘distance’ from the ‘mean’: −¿ mean ' 1¿ ¿ ' ('mean )¿ √¿ 5. ‘variability’: 'S D = (proportion)(1−proportion) √ 6. proportions range from 0% (no one has it) to 100% (everyone has it) 1. at what proportion is variability lowest? when is it highest? 1. lowest – everyone in the sample has the exact same results (no variability), at the extremes 1. 0% – no one has the outcome of interest 2. 100% – everyone has the outcome of interest 3. SD = 0 2. highest – in the middle, 50% STANDARD ERROR OF PROPORTION 7. measures how well a sample proportion estimates the ‘true’ (population) proportion (the variability) among diff sample proportions (aka ‘percentages’) SD SE= 8. recall, √ n 9. so, HMB325H © L| Page 2 2013 √(percent) 1−percent) (percent 1−percent) SEP= = n √ n percent = proportion 10. SEP enables calculation of confidence interval of a proportion (using normal distribution) HMB325H © Lis| Page 3013 11. ex. CI of a proportion: A survey of 106 patients indicates that 32 (30.2%) are still experiencing pain a month after their surgery 1. 95% confidence interval: SEP: (p 1−p ) (.302 .698 ) = =.045(4.5) √ n √ 106 CI: 95 CI=p±1.96(SEP)=.302±1.96(.045)=.215¿.389 = 21.5% to 38.9% 1. just over 30% are still experiencing pain, & the CI tells us that the true percentage of patients experiencing pain will range from as low as 21.5% to as high as 38.9% w an accuracy of 19/20 (95%) DIFFERENCE OF PROPORTIONS 2. ex. death after halothane vs morphine: does the risk of post-operative death differ? 1. is this difference consistent w a real clinical effect or simply random variation 2. need a statistical test enabling us to measure the probability of a difference this extreme or more extreme arising by chance t= differenceof means 3. recall the t test assessed the difference bw 2 groups: SEof difference 4. in the case of proportions, this would be: (p −p ) 1 2 (SE p1+SEp )2 5. plus 2 ‘refinements’ & 1 assumption: 1. estimating & using a ‘pooled proportion’ pooled– rather than using the 2 proportions separately 2. Yates correction for continuity – dealing w continuous data in t tests & ANOVA, dealing w simple counts (step-wise increments) in proportions 3. ‘large sample’ assumption ESTIMATING & USING p pooled 6. if the hypothesis that the 2 samples are drawn from the same population, then each sample provides an independent estimate of that pop proportion 7. as a result, we can estimate a single, pooled estimate of the proportion as: (m1+m 2 p pooled (n1+n 2 ‘LARGE SAMPLE’ASSUMPTION (p −p ) 1 2 8. because the ratio (SE p 1SEp ) 2 is approx normally distributed only for sample¿¿>5 ppooledample¿5 sufficiently large sample sizes, and (1−p pooled¿ ppooledroportion HMB325H © Li| Page 42013 9. the Fishe
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