# Its deciding which Stats method to use that the problem

1. Apr 1, 2005

### Prester John

Its deciding which Stats method to use that the problem......

Hello, i hope you don't mind a biologist interloping on your boards :)

I have a statistics problem that i could do with some help with if anyone can (please!).

I'm taking 2 8ml samples from a volume of 200mls. Distributed evenly within the 200mls may be bacteria. If a sample gets 1+ bacteria in then the sample is flagged as positive, if it get 0 bacteria it is flagged as negative. I don't know how many bacteria are in the 200mls or in the samples, only positive or negative.

To make it more interesting each 200ml volume may or may not contain bacteria (but if it does the number of bacteria is a constant value).

Now i have tested 20,000 lots of 200mls. Of these 10 have flagged positive. 5 of these were positive in both samples, 5 only in 1 sample. What i really want to know is how many i have missed.

I was thinking that you should be able to use the data showing 5 in 2 samples, 5 in 1 sample only to estimate the number of bacteria in a 200ml volume (? binomial dist or similar) and then use that to determine a probability that both samples will have no bacteria in them from a 200ml volume containing bacteria.

Thankyou for reading and i'll throw in a little irrelevant factoid for your amusal: Your body has on and in it 100 Trillion bacteria, thats 10x the number of cells that make you up :yuck:

2. Apr 1, 2005

### honestrosewater

Does that include mitochondria?

3. Apr 1, 2005

### Prester John

Not for the last billion years or so!

4. Apr 1, 2005

### BicycleTree

How many what you have missed? 200 ml lots? Are you assuming that the test always gives the correct result for each 8 ml sample?

Without working this through, the number of bacteria in a volume of liquid should have a Poisson distribution. The samples of 8 ml would be described by a Bernoulli distribution. You need to find the parameter of the Poisson distribution, so you would find what parameter (for the Poisson distribution) would lead to the probability of at least 1 bacterium in 8 ml being equal to the sample proportion (15 out of 40,000). From this you could get a point estimate of the average # of bacteria in 200 ml, though I have no idea how accurate it would be.

Last edited: Apr 1, 2005
5. Apr 1, 2005

### BicycleTree

Also, if you do it that way you would have to assume that the numbers of bacteria in each of the lots of 200 ml are identically and independently distributed. You'd have to assume that you're not taking half the lots from sterile distilled water and the other half from a pond.

6. Apr 4, 2005

### Prester John

I was thinking it through over the weekend. If 50% of the time 1 out of 2 samples is positive, and 50% of the time 2/2 are positive does this imply that there are 3 "bacteria" per 16 ml sample ?
If there was 1 in 16 ml either or samples would be positive (never both)
If there was 2 per 16 ml the dist would be 2-0, 1-1, 0-2, leading to a distribution were 66%of the positives would have 1 sample positive, 33% in both samples
If there were 3 per 16ml he dist would be 3-0,2-1,1-2,0-3, ie 50% with 1 out of 2 samples positive, 50% with both samples positive?

(Assuming 100% detection of a "bacteria" in a sample)

Does that make sense?

Then assuming 3 "bacteria" per 16 ml (in a positive 200ml pool)its relativly easy to calculate the chance of not getting any bacteria in a 16ml sample??

I'll look up about the Bernoulli distribution and try to work it through that way. I know there are a lot of assumptions, some i know to be false but it is only the first attempt to assess the data. There are other data sets with different sample sizes etc that i can use this modelling to make predictions about.

7. Apr 4, 2005

### BicycleTree

You don't really need to use the Bernoulli distribution, the main one for the way I suggested is the Poisson distribution. The Bernoulli distribution is just the binomial distribution with size = 1.