# 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}) = [m^{x } ]/[x! ]e^{-m }

where $$m$$ is the expected total number of correctly answered questions from the N questions. Since this equals N/k we can rewrite the above as:

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

one-tailed p-value = P(X \leq x) = (sum from 1 to 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( [x-(N/k)] divided by [Square Root{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 + [4.67

^{2 }]/[2] + [4.67^{3 }]/[6])e^{-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 accounts for *less likely* occurrences than the observed 3 correctly answered questions, given random responses. So the *exact* two-sided p-value is the sum of the Poisson probabilities P(X $$\leq$$ 3) and P(X $$\geq$$ 6) which equals 0.64.

P(X $$\leq$$ 3) is approximately equal to P(X $$\geq$$ 6) because the Poisson distribution is symmetric about its expected value, N/k, the two-tailed p-value can also be computed as approximately equal to twice the one-tailed p-value = 0.31 x 2= 0.62. This is close to the p-value of 0.64 using the sum of the poisson probabilities above.

The approximate or exact p-values from a Poisson distribution both conclude that there is no evidence to suggest a score of 3 on a three-choice task of 14 questions differs from chance responses.

We can also evaluate probabilities of observing 3 correct responses due to 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 the Normal distribution probabilities are, like those of the Poisson distribution, 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 of the number of correct responses!

As with the Poisson distribution we conclude there is no evidence to suggest getting 3 questions correct on a three-choice task of 14 questions differs from chance. The exact Poisson two-sided p-value and its Normal approximation may be computed using a spreadsheet.

In practice for over 30 questions (N) the Poisson and Normal approximations should closely agree. For less than 30 questions the Poisson is preferable as it is a discrete distribution assuming, as in this example, only integer values can occur (ie that numbers of correctly answered questions are whole numbers).