A true-false test consists of 20 questions, each of which has one correct answer: true, or false. One point is awarded for every correct answer, but one point is taken off for each wrong answer. Suppose a student answers every question by guessing at random, independently of other questions. Let S be the student’s score on the test. Find Standard Error of test score SE(S)?



SE(S)= $\sqrt{\frac{P*(1-P)}{n}}= \sqrt{\frac{0.5*(1-0.5)}{20}}= 0.1118$? yet is wrong

  • $\begingroup$ You are using the wrong formula for SE. $\endgroup$ – André Nicolas May 9 '13 at 5:39

Let $X_i=1$ if the student is right on the $i$-th question, and $0$ if she is not right. Let $M_i$ be the student's mark on the $i$-th question. Then $M_i=2X_i-1$. The test score $S$ is then given by $$S=M_1+M_2+\cdots+M_{20}.$$

It is standard that the standard error of $X_i$ is $\sqrt{(1/2)(1-1/2)}=1/2$. So the standard error of $2X_i-1$ is $(2)(1/2)=1$.

The standard error of the sum $M_1+M_2+\cdots+M_{20}$ is therefore $(\sqrt{20})(1)$.

Remark: when you divided the (not quite right) standard error per question by $\sqrt{20}$, you were finding the standard error of the average mark per question, not the standard error of the total mark.

We could also find the SE of each $M_i$ directly. The variance of $M_i$ is $E(M_i^2)-(\mu_i)^2$, where $\mu_i$ is the mean of $M_i$. But the mean of $M_i$ is $0$. And $M_i^2=1$ always, so $E(M_i^2)=1$. It follows that the variance of $M_i$ is $1$, and therefore the SE of $M_i$ is $1$.

The $M_i$ are assumed independent. The total mark is the sum of the $M_i$, so has variance $(20)(1)$, and therefore standard deviation $\sqrt{20}$.

  • $\begingroup$ SE of the S the standard error of S _ (SD of the Test) =SQRT(n)*SD this is correct formula? $\endgroup$ – statistics-student13 May 9 '13 at 5:56
  • $\begingroup$ The standard error (standard deviation) of $S$ is $\sqrt{20}$ times the standard deviation on any particular question. That turns out to be $1$. So the SE of $S$ is $\sqrt{20}$. $\endgroup$ – André Nicolas May 9 '13 at 5:58
  • $\begingroup$ Andre Nicolas Thanks, understood! $\endgroup$ – statistics-student13 May 9 '13 at 6:00

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