Appropriate use of the binomial test to compare expected and actual selection frequencies?

119 Views Asked by At

We have experimental data of food selection frequencies among a population of primates across a series of 15 consecutive trials. Individually in each trial, each subject was presented with two food items simultaneously (Food A and Food B) and their selection was noted. Only one of the two foods could be selected, although rejections (immediate drops without consumption) and null selections (nothing selected) were recorded. Therefore, we considered four outcomes: Food A, Food B, rejection, null.

Our null hypothesis is that the each food type would each be selected half the time. I was thinking of using a binomial test to compare actual frequency of Food A to the expected frequency of Food A, and doing the same for Food B, in order to see if there is a significant difference between actual and expected frequency for each. In doing so, I would treat the selection of the food of interest as a success (e.g., Food A), and any other behavior (reject, null, or selection of other food) as a failure (e.g., not Food A). In this way, we have a dichotomous outcome and the use of the binomial test seems to make sense.

Question 1: Does this seem like an acceptable approach? Or is this an inappropriate use of the binomial test because I would be treating null, rejects, and the "other" food selection as a single outcome? If not appropriate, is there a different statistical test I could use to compare expected selection frequency of each food to its actual frequency?

Question 2: My sample contains 10 primates. Is it appropriate to aggregate selection frequency across subjects when conducting the binomial test?

1

There are 1 best solutions below

3
On

Unless 'null' and 'reject' occur frequently, it seems better to drop those responses and make the question about a binary choice between A and B.

Say you have 12 subjects who gave unequivocal A or B answers. Let $X$ be the number of subjects out of 12 that selected A. Then the only responses that reflect a significant difference are $X = 0, 1, 2$ indicating a preference for B and $X = 10, 11, 12$ indicating a preference for A.

Under the null hypothesis that A and B are equally likely to be chosen $P(3 \le X \le 9) \approx 0.96.$ So the significance level of the procedure I describe is about 4%.

The computation in R statistical software is:

sum(dbinom(3:9, 12, .5))
## 0.9614258

enter image description here

If you have more than about a dozen subjects or if you had something else in mind, please edit your question or leave a Comment, and maybe one of us can be of further help.

Note: For your second question, with $n = 10$ definitive selections of A or B, only results $X = 0,1$ and $X = 9,10$ lead to rejection. The logic is similar. The significance level is about 2%. [Including 3 or 8 as significant results would lead to a significance level above 10%, which does not seem persuasive.]