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W.R. Wilcox, Clarkson University
Explanation of menus in Piface for sample size determinations
(Java Applets for Power and Sample Size by R.V. Lenth, 2006)
Before beginning, read hypothesis testing. Use Help for each dialog. Note that these are to be used
BEFORE doing the experiment, although preliminary (“pilot”) experiments may be needed to
estimate standard deviations. Experiment runs should be done in random order.
CI for one proportion
For a binomial distribution, each member of which can have only one of two
values (say 0 or 1). This applet finds the sample size n required to determine a
specified confidence interval for the proportion pi of members of a population
that have value 1. If p is the proportion with value 1 in the sample, then the
confidence limits for pi are p  ME, where ME is the “margin of error.” The
Confidence (= 1 - ) is the probability that pi lies within this interval, with the
default being 95%. Either enter the population size N or click off “Finite
Population” for an infinitely large population. The worst case, i.e. requiring the
largest sample size n, is for pi = 0.5. Move the slider for Margin of Error to the
desired value, and read off the value of n required to achieve this.
(See also Binomial histogram applet.)
Test of one proportion
For an infinite population with a binomial distribution this applet gives the
sample size required for there to be probability Size (significance level) of
concluding that pp0 when p = p0 (null hypothesis) is actually true, and
Power chance of rejecting the null hypothesis in favor of the Alternative
hypothesis for the specified values of p0 and p. Typically you would set
p0 = 0.5, |p – p0| to be the minimum difference you want to detect,
Alternative to p != p0 (i.e., p  p0), Alpha = 0.05 (maximum Size), and
Power = 0.8 or slightly larger. The Applet then gives you the required
sample size; because the binomial distribution is discrete the Size will be
probably be less than 0.05 and the Power will differ from 0.8. The
example shown here indicates that we need 206 samples for a 4.307%
chance of concluding that p  p0 when actually p = p0, and a 80.75%
chance of detecting a difference in proportion of 0.1. (Note that !=
signifies , not equal to.) (See also Binomial histogram applet.)
Test comparing two proportions
This is for two infinite populations with binomial distributions with
proportions p1 and p2. Alpha is the probability of concluding that p1  p2
when they are actually the same (null hypothesis), Power  1 - , and  is
the chance of concluding p1 = p2 when p1 and p2 are actually as selected. In
practice, one sets p = 0.5, |p1 – p2| to be the minimum difference you want
to detect, and then move the n sliders until the Power  0.8. The example
shown here indicates that we need samples of 411 in order for there to be a
5% chance of concluding p1  p2 when actually p1 = p2, and 80.39%
chance of detecting a difference of 0.1 between p1 and p1.
CI for one mean
This applet is for a normal distribution with estimated standard deviation Sigma
(). It finds the sample size n required to determine a specified confidence
interval with Confidence that the true population mean  = y  ME, where y is
the sample mean and ME is the Margin of Error. Thus, for the example shown
here, there is a 95% probability that  = y  0.6428 for a sample size of 40
taken from an infinite population with  = 2.01. Normally, one would have an
estimate of Sigma from prior experiments, set the desired ME, and read off the
number of samples required. See the “Pilot study” applet for a general method
to estimate .
One-sample t test (or paired t)
This applet is for a normal distribution from an infinitely large population with
estimated standard deviation sigma (). It gives the sample size n required so
that there is probability power of detecting a specified difference (effect size)
|mu – mu_0| between the mean of the population () and a given value 0, and
chance alpha of concluding that the null hypothesis is false when it is actually
true (i.e. that   0 when actually  = 0). For the example shown here, if  =
2 it takes 128 samples for a 5% chance of concluding that   0 when actually
 = 0, with an 80.15% probability of detecting a difference of 0.5 between 
and 0. Note that this applet can also be used for a paired-sample test to
compare two populations by taking the same number of samples from each and
treating the differences between paired samples as being taken from a single
population. The null hypothesis in this case would be that the means of the two
populations are the same, and the power would give the probability of detecting
a specified difference. See the “Pilot study” applet for a general method to
estimate .
Two-sample t test (pooled or Satterthwaite)
Before beginning, read the excellent Help page
with this applet, the purpose of which is to
determine the sample sizes required to compare the
means of two infinitely large populations with
normal distributions and estimated standard
deviations sigma1 (1) and sigma2 (2). In this
example, with 1 = 1.2, 2 = 0.59, the optimal
sample sizes are n1 = 69 and n2 = 34 in order for
there to be a  = 5% chance of concluding that
12 when actually 1=2, and an 80.17% chance
of detecting |1 - 2| = 0.5. See “Some Practical
Guidelines
for
Effective
Sample
Size
Determination,” Russell V. Lenth, The American
Statistician, Vol. 55, No. 3. (Aug. 2001), pp. 187193. See the “Pilot study” applet for a general
method to estimate .
Linear regression
Read the excellent Help. In the example given here, we
are considering 3 factors (independent variables;
“predictors” here). Based on prior experience, we can
expect the standard deviation of the actual values of the
result (dependent variable) from its predicted values
(“ErrorSD”) to be 0.2. For the values of the jth predictor
we plan to use the standard deviation will be 0.1 (SD of
x[j] = 0.1). We want a 5% chance of concluding that the
coefficient for the j’th variable, beta[j], is not 0 if actually
j = 0, and a 90.01% chance of detecting j = ±0.8. The
applet tells us that 68 runs are required. If this is too
many runs, increase SD of x[j] (by increasing the range
over which x[j] is to be varied), increase the value of beta
you want to detect, or decrease the Power.
Balanced ANOVA (any model)
For design of experiments. Read Help. This requires knowledge of
factorial analysis using Analysis of Variance (ANOVA). Example: We
have two factors, A and B, with 3 levels of each. We want to be able to
detect a change of 40 in the result with probability greater than 90%,
given a standard deviation of 25 and a probability of type-I error of 5%.
Thus, for the t-test method 4 replicates gives a Power = 0.9653. The other methods give somewhat different
results, and are explained, for example, in Chapter 3 of Montgomery’s “Design and Analysis of
Experiments.”
R-square (multiple correlation)
Read Help.
Generic chi-square test
Read Help
Generic Poisson test
This Applet is for the Poisson distribution. Read Help.
Pilot study
Generic method for estimation of required sample size assuming normal distribution(s). Read Help.
Last revised January 20, 2011. Please contact W.R. Wilcox with comments, corrections and suggestions for
improvement.