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Cell and Systems Biology

CSB345H1

William Navarre

Fall

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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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