ciss.midp | R Documentation |
Calculate sample size for a binomial proportion based on a mid-p confidence interval width specification.
ciss.midp(p0, d, alpha, nMax=1e6)
p0 |
hypothesized upper bound (if below 0.5, if above 0.5 then lower bound) on the parameter p in the binomial distribution |
alpha |
an |
d |
half width of the confidence interval |
nMax |
Largest n to check. Interrupt iterations when this value is reached |
Fosgate (2005) discusses the need for improved sample size
calculations in cases where the binomial proportion is close to 0 and
1. To improve on this, calculation on confidence intervals based on
the mid-p method are suggested where computation of the upper and
lower limit are combined into one formula. Given lower and upper
bounds p_l
and p_u
of the (1-alpha
)*100%
confidence interval, one finds the sample size n
as the solution
to
\frac{1}{2} f(x;n,p_l) + \frac{1}{2} f(x;n,p_u) + (1 -
F(x;n,p_l)) + F(x-1;m,p_u) = \alpha
where f(x;n,p)
denotes the probability mass function (pmf) and
F(x;n,p)
the (cumulative) distribution function of the binomial
distribution with size n
and proportion p
evaluated at
x
. The function then returns \lceil n \rceil
. Note that
in this approach (p_l,p_u) = p_0 \pm d
, which has to be a subset
of (0,1)
. Another option would be to choose the lower and upper
independent specifically.
In the above, x
is found as the integer value, such that
x/n
is as close as possible to the hypothesized value p0
as possible.
An alternative approach to determine sample sizes based on the mid-p
approach is to manually find the sample size n
such that the
interval obtained by binom.midp
has a length less than
2\cdot d
.
the necessary sample size n
M. Höhle
Fosage, G.T. (2005) Modified exact sample size for a binomial proportion with special emphasis on diagnostic test parameter estimation, Statistics in Medicine 24(18):2857-66.
binom.midp
, ciss.binom
#Fosgate approach
ciss.midp(p0=0.2,alpha=0.05,d=0.05)
#Iterative increase of n using the general purpose function
ciss.binom( p0=0.2, alpha=0.05, ci.fun=binom.midp, d=0.05)
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