How do I compare observed numbers correct to those expected by chance in a multi-choice task?

Suppose a questionnaire has a set of k possible responses to N questions of which one of the k choices is the correct response. Assuming by chance each of the k responses to each question is equally likely the number expected by chance equals N/k.

Suppose we observe x $$\leq$$ N responses for a patient. We wish to see how likely x responses are given the patient responds at random to k choices on each of N questions.

To do this we can assume the number of correct responses, x, follows a Poisson distribution, Po(m) of general form:

$$P(X=x|\mbox{random responses}) = \frac{m^{text{x}}{x!}e}text{-m}$$

where $$m$$ is the expected total number of correct responses for N questions. Since this equals N/k we can rewrite the above as:

$$P(X=x|\mbox{random responses}) = \frac{(N/k)^{text{x}}{x!}e}text{-(N/k)}$$

one-tailed p-value = $$P(X \leq x) = \sum_text{1}^text{x}P(X=x|\mbox{random responses})$$ = 0.5(two-tailed p-value)

For large N the Poisson distribution is approximately Normally distributed with mean and variance both equal to N/k so we can analogously obtain a one-tailed p-value as:

$$P(X \leq x) = Probit(\frac{x-(N/k)}{\sqrt{N/k}})$$.

Example

Suppose we have 14 questions each with 3 possible responses of which only one is correct and a patient gets a total score of 3 correct responses. We wish to determine how likely it is we would observe 3 or fewer responses given the patient has responded at random (1 sided p-value).

The expected number of correct responses assuming the patient is responding at random is 14/3=4.67.

Using the Poisson distribution

P(X $$\leq$$ 3)

= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

$$ = (1 + 4.67 + \frac{4.67^{text{2}}{2} + \frac{4.67}text{3}}{6})e^text{-4.67}$$

= 0.31.

We can think of the p-value as the sum of numbers correct which are less likely assuming random responses than the observed one. It turns out that P(X $$\geq$$ 6) also account for less likely occurrences given random responses than the observed 3 correct.

Since P(X $$\leq$$ 3) is approximately equal to P(X $$\geq$$ 6) because the Poisson distribution is symmetric the two-tailed p-value = 0.31 x 2= 0.62.

We conclude there is no evidence to suggest a score of 3 on a three-choice task of 14 questions differs from chance.

Using the Normal approximation to the Poisson distribution:

$$P(X \leq 3) = Probit (\frac{3-4.67}{\sqrt{4.67}}) = Probit(-0.772) = 0.22$$ with a two-sided p-value of 0.44. The two-tailed probability equals $$P(X \leq 3) + P(X \geq 6.33)$$ since ths Normal distribution probabilities are symmetric about the mean of 4.67. Of course we can't observed 6.33 correct responses but this is a continuous approximation to the discrete Poisson distribution - like joining lines between frequency bars on a histogram!

We again conclude there is no evidence to suggest a score of 3 on a three-choice task of 14 questions differs from chance. These calculations may be computed using a [attachment:multichoice.xls spreadsheet.]

[WORK IN PROGRESS: Exact two-tailed p-value for the Poisson distribution giving the chance of numbers correct occurring which are no more likely than the m=number correct observed assuming random responses.]

For N > 30 the Poisson and Normal approximations should closely agree. For N < 30 the Poisson is preferable as it is a discrete distribution assuming, as in this example, only integer values can occur.