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Have been given some data and the question says to determine if the data follows any distribution. It says to compare the observed data vs expected graphically and to test further. The distributions we've studied so far are normal, binomial and poisson so I assume it will be one of these. We've used these techniques so far: 1 and 2 sample t-tests, chi-squared tests of association and goodness of fit, one way and two way anova, confidence intervals, estimating abundance techniques, linear regression amongst a few others.

The data is as follows, 100 cows hooves are swabbed and checked for a bacteria. The results are as follows. The bacteria survives on the hooves for weeks, and on the grass for days.

No. of hooves that test positive per animal - Frequency
[0 - 25]
[1 - 5]
[2 - 15]
[3 - 30]
[4 - 25]

What I have done so far, I've calculated the normal, poisson and binomial distribution probabilities and the expected frequencies from them. But, I believe that the Binomial distribtuon is the right distribution as the number is limited (maximum of 4 hooves). I places the observed v expected in a bar chart to graphically compare and they do not appear to match.

I think my next steps should be carrying out a chi squared test of fitness but after that, I don't know what more to do or if I need to. Do I need to calculate a Confidence intervals, and what do I use it for? Is there anything else you would recommend?

Thanks in advance!

A Mixture Model for Exposed and Unexposed Animals

Suppose 1/4 of the cows are unexposed to bacteria,
and among the 3/4 of cows that are exposed the number of hooves with
bacteria is Binomial with n = 4 and p = 3/4.
This model gives the probability table of hooves
with bacteria shown in the table below.

This binomial distribution was deduced from
the fact that there are 225 hooves with bacteria out of 75
exposed animals for an average of 3 hooves per animal.
So the binomial mean
must be $mu = 3 = 4p$, whence $p = 3/4.$

Each of the probabilities for 1 through 4 hooves is
3/4 of the probabilities assigned by $Bin(4, 3/4).$
The probability for 0 is .25 plus the the 3/4 of the binomial
probability. (Probabilities are rounded to four places
and slightly 'fudged' in the fourth place so probabilities
add to 1. This method works without complication only because the binomial part of the
model contributes extremely little probability for 0 hooves.)

Expected counts are probabilities multiplied by 100 cows.
Observed counts are the data reported in the problem.

 Hooves       0      1      2      3      4

          ---------------------------------

 Prob     .2528  .0351  .1586  .3163  .2372  

 Exp      25.28   3.51  15.86  31.63  23.72

 Obs      25      5     15     30     25

The standard chi-squared goodness-of-fit test (as implemented
in R) gives the output shown below.

 prob=c(.2528, .0351, .1586, .3163, .2372)

 obs = c(25, 5, 15, 30, 25) 

 chisq.test(obs, p=prob)



 ##   Chi-squared test for given probabilities

 ##

 ## data:  obs 

 ## X-squared = 0.8353, df = 4, p-value = 0.9337

There is a warning message because the expected count in cell 1
is less than 5, putting the approximation of the chi-squared
statistic to the chi-squared distribution in some doubt.
However, an exact test (simulated permutation test) gave
a P-value of 0.9371.

So there is no question that that the observed counts are
consistent with the proposed probability model. (Other distributions
might fit as well, but the question implied we should look
for an answer based on a binomial or Poisson distribution.
The data fit the model almost 'too well', suggesting that
the data might have been contrived to make the solution to the problem easier to find.)