Gerald van Belle “Statistical Rules of Thumb”
Andrew Gelman and Jennifer Hill’s “Data Analysis Using Regression and Multilevel/Hierarchical Models” (Chapter 20)
Brad Carlin
JoAnn Alvarez
Recognize denominator as the square of the standardized difference (\(\Delta ^2\)) \[ n_{group}=\frac{2(1.96 + 0.84)^2}{(\Delta)^2} \\ n_{group}=\frac{16}{(\Delta)^2} \]
10point difference in IQ between two groups (mean population IQ of 100, and a standard deviation of 20): \[ n_{group}=\frac{16}{((10090)/20)^2} \\ n_{group}=\frac{16}{(.5)^2} \]
total sample size of 128.
function in R
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Rearrange Lehr’s equation to solve for the detectible standardized difference between two populations: \[ n_{group}=\frac{16}{(\Delta)^2} \\ (\Delta)^2 = \frac{16}{n_{group}} \\ \Delta = \frac{4}{\sqrt{n_{group}}} \\ \]
unstandardized difference \[ \delta = \frac{4 \sigma}{\sqrt{n_{group}}} \]
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 \[ n_{group}=\frac{16(c.v.)^2}{(ln(\mu_0)ln(\mu_1))^2} \\ n_{group}=\frac{16(c.v.)^2}{(ln(r.m.))^2} \]
R function
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 \[ y_i \sim Pois (\lambda) \\ \sqrt{y_i} \sim Nl(\sqrt{\lambda}, 0.25) \]
apply Lehr’s equation by taking the square root \[ n_{group}=\frac{4}{(\sqrt{\lambda_1}\sqrt{\lambda_2})^2} \]
\(n_{group}=\frac{4}{(\sqrt{30}\sqrt{36})^2} = 15\) units of observations in each group.
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Given no events or outcomes in a series of \(n\) trials, the 95% upper bound of the rate of occurrence \[ \frac{3}{n} \]
Set probability to statistical significance *\(10.95 = 0.05 = Pr[\Sigma y_1 = 0] = e^{n \lambda}\) *\(n \lambda = ln(0.05) = 2.996\) *\(\lambda \approx 3/n\)
for RR *\(n_{group} = \frac{8(RR+1)/RR}{\frac{c}{c+d} (ln(RR))^2}\)
E.g. \(n_{group}=\frac{4}{(.3  .1)^2} = 100\)
conservative approximation only valid where sample size each group between 10 and 100
For rare, discrete outcomes, you need at least 50 events in a control group, and an equal number in a treatment group, to detect a halving of risk.
incorporating uncertainty is inherently Bayesian
redefine basic Lehr’s formula R function in terms of difference and standard deviation.
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can achieve same power under the same conditions with a few as 30 or as many as 160 observations in each group.
Power = \(\phi (\sqrt{\frac{n \theta ^2}{2 \sigma ^2}}1.96)\)
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use this info to define a Beta distribution to use as a prior in a binomial simulation
define parameters for Beta prior from mean and variance
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if prior is \[ \theta \sim Beta(a,b) \]
and the likelihood is \[ X \theta \sim Bin (N, \theta) \]
then the posterior is \[ \theta  X \sim Beta (X+a, NX+b) \]
combines Binomial \({20 \choose 15} p^{15}q^{2015}\) likelihood of external data with the \(Beta(9.2, 13.8)\),
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now don’t hit 80% power until a sample size of about 54.
E.g. take the \(Beta(14+24.2,2014+18.8)\) prior and shift 7 of the successes to failures as \(Beta(7+24.2,207+18.8)\)
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