3
$\begingroup$

In a multiple choice test with 4 answer options, the probability of guessing correct is 25% (assuming you have no knowledge about the materials at all). Thus, regardless of sample size, my estimated mean would be 25 (dependent variable = percentage correct).

What would be the standard deviation here?

I have found this formula:

sqrt(N * p * (1-p))

So for N = 40, the standard deviation would be:

sqrt(40 * 0,25 * (1-0,25)) = 0,74.

Is this correct? If so, why is the standard deviation dependent on the sample size?

$\endgroup$
3
$\begingroup$

The formula you have is the standard deviation of the number correct (under binomial sampling), not the proportion correct.

However, because the proportion correct is the number correct multiplied by $\frac{1}{N}$, we can use basic properties of variance to figure out that the standard deviation of the sample proportion is $\sqrt{p(1-p)/N}$ (also called the standard error of a proportion).

[Note that here $p$ is the population proportion not the sample proportion]

On the "why" question: it's the usual situation - standard deviations of means of independent, identically distributed random variables relate to sample size.

Again from basic properties of the variance,

$\text{Var}(X_1+X_2+...+X_N)=\text{Var}(X_1)+\text{Var}(X_2)+...+\text{Var}(X_N)$

$\qquad=N\text{Var}(X_1)$

Hence $\text{Var}(\bar{X})=\frac{1}{N^2}.N\text{Var}(X_1)=\frac{1}{N}\text{Var}(X_1)$

So the standard deviation of the mean is the standard deviation (square root of variance) of an individual observation, divided by $\sqrt{N}$, . (I tend to call it the "sigma on root n effect".)

This is the standard error of the mean, and in the case of a binomial, the individual Bernoulli-trial standard deviation is $\sigma=\sqrt{p(1-p)}$

--

In addition, you probably don't want to calculate with the percentage correct (25); better to stick with the proportion correct, or the number correct. (The calculations can be converted to work with percentages, but I recommend you stick with the proportion.)

$\endgroup$
  • $\begingroup$ Isn't your first formula the variance of the sample proportion rather than the standard deviation? And is the standard deviation of the proportion equivalent to the standard error of the sample? $\endgroup$ – Robert Kubrick Dec 14 '16 at 15:54
  • $\begingroup$ @Robert Thanks, edits in converting from variances to standard deviations ended up without the square root being put in. Hopefully I fixed that everywhere needed. On the second thing, strictly speaking a sample doesn't have a standard error, a statistic does. When we say "the standard error" without any other qualifier, we generally intend the standard error of the mean, which is the sample proportion in this case. (If you're suggesting an edit on that basis you'll have to be more specific, sorry.) $\endgroup$ – Glen_b Dec 15 '16 at 0:13
  • $\begingroup$ About my second question, I was having some confusion with the terminology. Yes, the standard deviation of the number correct is the standard error of the proportion (or mean), as you pointed out in the answer. $\endgroup$ – Robert Kubrick Dec 15 '16 at 15:31
  • $\begingroup$ No, the standard deviation of the number correct is not the standard error of the proportion. The two are related (each one is the other scaled by a constant which depends on $N$). $\endgroup$ – Glen_b Dec 15 '16 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.