Hi,
Some suggestion about the arguments of the function defined below. Since
theta is calculated with the value of lambda1 and lambda2, there is no need
to include theta in the argument. Or, your function can be defined as
function(lambda1, lambda2, significance.level)
cover <- function(theta
Hi,
Given the function cover, it's very likely that you will get 0 for both s1
and s1 with small value of lambda1 and lambda2. In that case the sum s will
be 0. With s being 0, you will have issue with the expression in pi <-
s2/s and root <- ((s2/s)*(1-s2/s)+k/(4*s))^(1/2). You need to take
Hi,
In your function cover, lambda1 and lambda2 are used but not in the
argument of the function. I suppose that you need to have lambda1 and
lambda2 in the argument of the function cover, like function(lambda1,
lambda2, n, significance.level).
Give it a try.
cover <- function(lambda, n,
Hi,
After some thought, I found the treatment of sample mean equal 0 was not
appropriate. I modified the function likelihood.ratio.test.Poisson.
resulting.matrix now has 0.0512 as the average of type I error.
function(lambda, sample.size, significance.level)
{
reject <- 0
sample.mean <- m
Hi,
My first question is what is your n when you say fixed n. I assume the
lambda is the mean of the poisson distribution that you want to take sample
from.
Another question is about the sample size. It does not make too much
sense to make a sample of size 1.
Let's assume that you want t
Hi,
Roughly reading the code, I find this statement "phat <- x / m" is
probably incorrect since this will give you the set of 100 observed x
values /100. I redefine the function cover with three inputs: lambda
for the parameter of the poisson distribution, sample.size and
significance.
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