2. Suppose you are a veterinarian who is interested in the distribution of fleas

Question

2. Suppose you are a veterinarian who is interested in the distribution of fleas on dogs. Below is a sample of dogs and each flea on each dog is represented by a dot on that dog's body. Based on this sample, test whether the number of fleas per dog follows a Poisson distribution. Enter the data in the gray section of the answer spreadsheet. Calculate the expected frequencies in the green section In this example, we are using the Poisson distribution as our hypothesis. To test the hypothesis that the number of fleas per dog follows a Poisson distribution, we will use a test statistic that measures the degree to which the observed values match the expected values under our null hypothesis. Our test statistic is (Observed - Expected)2 Expected and is summed over all categories. If the observed values are close to expected, the test statistic values will be near zero. If expected values tend to be different than expected, then our test statistic will be large in value.

Explanation / Answer

Goodness-of-fit Test:

H0: The number of fleas per dog follows a poisson probability distribution.

Ha: The number of fleas per dog does not follow a poisson probability distribution.

The data provided is of diagram of dogs with fleas shown on their bodies and the initial frequency distribution is as shown below:

The expected frequencies are calculated by using Poisson distribution function given by,

F(x) = Meanxe-Mean/x!

To calculate the Mean fleas on dog, we need to find the total fleas by normal mean function, i.e.

0 flea on 7 dogs = 0 fleas

1 flea on 7 dogs = 7 fleas

2 fleas on 2 dogs = 4 fleas

3 fleas on 4 dogs = 12 fleas

4 fleas on 2 dogs = 8 fleas

5 fleas on 1 dog = 5 fleas

6 fleas on 1 dog = 6 fleas

Total fleas = 42 fleas

So the Mean = 42 / 24 = 1.75 fleas

So the poisson probabilitu calculation would be

f(0) = (1.750)(e-1.75) / (0!) = 0.1738

f(1) = (1.751) (e-1.75) / (1!) = 0.3041

f(2) = (1.752) (e-1.75) / (2!) = 0.2661

f(3) = (1.753) (e-1.75) / (3!) = 0.1552

f(4) = (1.754) (e-1.75) / (4!) = 0.068

f(5) = (1.755) (e-1.75) / (5!) = 0.0238

f(6) = (1.756) (e-1.75) / (6!) = 0.007

Expected Frequency calculation is

0 flea = 24 * 0.1738 = 4.17

1 flea = 24 * 0.3041 = 7.3

2 fleas = 24 * 0.2661 = 6.4

3 fleas = 24 * 0.1552 = 3.73

4 fleas = 24 * 0.0679 = 1.63

5 fleas = 24 * 0.0237 = 0.57

6 fleas = 24 * 0.0069 = 0.17

Since the expected frequency for 5 fleas and 6 fleas is less than 1, we can add these frequencies to 4 fleas as the sum of 0.57 and 0.17 would still be less than 1.

So >4 fleas would be 1.63 + 0.57 + 0.17 = 2.37

From the table above we can see that Chi-square value = 6.098 at df = k-2 where k is number of categories and from table we know that there are 5 categories, So df = 3.

From the question, it is known that the critical value at df = 3 is 7.81 (assuming that it is at 95% confidence level), then in comparing the test statistic which is 6.098 which is less than critical value 7.81, then it must be very well from the similar mean. So, the null hypothesis must not be rejected.

This means that the number of fleas on dog follows Poisson distribution.

No. of fleas on dog Observed Frequency(O) Expected Frequency(E) Test Statistic = (O-E)2 / E 0 7 4.17 1.9206 1 7 7.3 0.0123 2 2 6.4 3.025 3 4 3.73 0.02 >= 4 4 2.37 1.121 Total 24 Chi-Square 6.098

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