I assumed that U = 0.3 and x = 1

P = (e ^ -0.3) * (0.3 ^ 1) / 1! = 0.22

This is not the 95% I thought it should be..any ideas? Thank you so much!

- Thread starter radwa
- Start date

I assumed that U = 0.3 and x = 1

P = (e ^ -0.3) * (0.3 ^ 1) / 1! = 0.22

This is not the 95% I thought it should be..any ideas? Thank you so much!

Of course if you have 10 or 100 test tubes matters for the certainty/uncertainty.

Also please note that "at least one" means "one or more". So that P(Y>=1) = 1 - P(Y=0).

Correct, say 10 test tubes

How do you know that there are any DNA in the tube? Is it because they duplicates many times and you can e.g. se a change in colour? Then it is about "at least 1".

Maybe it can have some interest for others about a likelihood interval.

Code:

```
# This is an R program
# formulation of the problem
# if you have n=10 test tubes and get changed colour in 3 of 10, ie y=3
# and no changed colour in n-y = 10-3 = 7
# When there is a change in colour there can be 1 or more DNA molecules in the test tube
# compute the lambda from a Poisson distribution
# P(Y = y) = lamb^y*exp(-lamb)/factorial(y)
# P(Y = 0) = exp(-lamb)
# P(Y >=1) = 1- P(Y = 0) = 1 - exp(-lamb)
#
# probability for 7 no-change-of-colour, ie 7 with Y=0
# P(Y = 0)^7 = (exp(-lamb))^7
#
# probability for 3 change-of-colour,
# ie (P(Y >=1))^3 = (1 - exp(-lamb))^3
# probability for 3 change-of-colour AND 7 no-change-of-colour
# (1 - exp(-lamb))^3 * (exp(-lamb))^7
#
# give a sequence of lambda values, here called lamb_sekv:
lamb_sekv <- seq(from=0.0, to=2, by= 0.025)
lamb_sekv
# compute the Likelihood (called "like") for each of lambd_sekv:
like <- ((1 - exp(-lamb_sekv))^3) * ((exp(-lamb_sekv))^7)
# plot the liklihood values for the lambda values
plot(lamb_sekv, like, type ="l", xlim = c(0, 1.5))
abline(v=0.30, col="red")
abline(v=0.35, col="blue")
# also: draw red line at the intuitive value of 0.30 = 3 of 10
# also draw a blue line for what looks like the max likelihood
# look at it on narrow scale:
plot(lamb_sekv, like, type ="l", xlim = c(0.15, 0.50))
abline(v=0.30, col="red")
abline(v=0.35, col="blue")
# print out the different lambda values and the likelihood
cbind(lamb_sekv, like)
# compute the relative likelihood, so that the curve will have a max=1
# 2.222396e-03 is the mx value of the variable "like"
like_rel <- like/(2.222396e-03)
# Plot the relative likelihood
plot(lamb_sekv, like_rel, type ="l", xlim = c(0, 1.5))
abline(v=0.30, col="red")
abline(v=0.35, col="blue")
abline(h=0.15, col="black")
# the black line shows where the relative likelihood is 0.15
# the relative likelihood at 0.15 has got a value under 1 on lamba
# this is called a "likelihood interval", not a "confidence interval"
# but it can be interpreted as confidence interval
# explanations in Pawitan "In all likelihood"
# which values for lamb_sekv is larger than 0.15
cbind(lamb_sekv, like_rel)
# the 0.15 (or more exact 0.146607) comes from (the likelihood ratio test) :
# -2 log(like) = 3.84
# log(like) = -3.84/2, like=exp(-3.84/2)
# exp(-3.84/2)
# [1] 0.146607
# so the estimated Poisson intensity (ie the lambda)
# is somewhat less than 1
# the lambda is estimated 0.35 and between 0.08 to 0.93
```

It is interesting that this method is similar to what the microbiologist call the "most probabel number (MPN)" (an ancient title). Nowdays it would be called a maximum likelihood estimate from the Poisson distribution. This method was actually mentioned in R A Fishers "Foundations of Mathematical Statistics" in 1922.