How do I know if a Binomial model is appropriate?I have a question which is...





How do I know if a Binomial model is appropriate?
I have a question which is about the number of weeks out of 5 in which an event occurs. I have a frequency table with a sample of 40 - with x = 0 , 1 , 2 , 3 , 4 , 5 and freq, 2,7,11,12,6,2.
I have worked out the unbiased population mean and estimate - but then I'm not sure whether binomial what I need or not? I have to decide if a bionomial model is appropriate.
I can see that the data is discrete but its not binary like "event happens" or "event does not happen". It seems relatively symmetrical - and almost normally distributed? I'm not really sure how to work this out? Is a binomial model right or not?

Answer & Explanation

Henry Arellano

Henry Arellano


2022-11-25Added 12 answers

Step 1
If this is your first chi-squared test, the clues in the comments may be a bit too sparse. Without working the problem for you, I offer the following more complete outline: (Use it along with whatever examples your text or class notes may have to offer.)
It is appropriate to try a binomial model, and obviously n = 5. From the given data you can find the sample mean of the 40 observations.
By looking at the PDF of Binom(5,0.495). you can find the expected counts E i . (multiply the probabilities by 40.) Your observed counts are F = ( 2 , 7 , 11 , 12 , 6 , 2 ) .
Step 2
Next, you can find the chi-squared statistic Q = i = 0 5 ( F i E i ) 2 E i , which is approximately distributed as C h i s q ( ν = 4 ) . [Ordinarily, a chi-squared test with 6 categories would have ν = 6 1 = 5 , but you have used the data to estimate parameter p, so you 'lose' a degree of freedom for that and ν = 4. ]]
I got Q = 1.1815. The critical value for a chi-squared test with ν = 4 at the 5% level is the 95th percentile c = 9.487 of C h i s q ( ν = 4 ) . You can find this number in printed tables of the chi-squared distribution or using software (as with R below).
qchisq(.95, 4)
This means that you would reject the null hypothesis that the data are consistent with B i n o m ( n = 5 , p = 0.495 ) only if Q c = 9.487.
There is one remaining difficulty. The chi-squared test is usually deemed to be accurate only if all expected counts exceed 5. Your first and last about:blanks are too small. One cure for this is to combine 'categories' 0 and 1, and 'categories' 4 and 5. In each tail, combine categories by adding the two observed frequencies and adding the two expected frequencies.
You will now have four categories and ν = 4 1 1 = 2 degrees of freedom. Re-compute Q and find the new c (as below). [According to my computations, you will still not reject H 0 . ]]
qchisq(.95, 2)
[1] 5.991465

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