Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have information of the order in which students were classified in regard to their scores in a SAT test. I know the distribution of scores for each student is uniform with support [a,b]. I also know that:

  • 17 students scored less than 3
  • 13 students scored between 3 and 5
  • 58 students scored more than 5

How can I form reasonable estimates for a and b?

Thanks

share|improve this question
1  
Please do not crosspost in the future. It is impolite and strongly discouraged. –  cardinal Feb 3 '13 at 3:48
add comment

1 Answer

up vote 1 down vote accepted

Assume that $u$ students scored in $(x-r,x)$, $v$ students scored in $(x,x+s)$ and $w$ students scored in $(x+s,x+s+t)$, where $(u,v,w,x,s)$ are known and $(r,t)$ are unknown. The likelihood is $$ L(r,t)={u+v+w\choose u,v,w}\left(\frac{r}{r+s+t}\right)^u\left(\frac{s}{r+s+t}\right)^v\left(\frac{t}{r+s+t}\right)^w. $$ Thus, $L(r,t)$ is maximal when the partial derivatives of $L$ with respect to $r$ and $t$ are zero, that is, when $$ \frac{u}{r}=\frac{u+v+w}{r+s+t}=\frac{v}{t}. $$ Solving these yields $$ r=s\frac{u}v,\qquad t=s\frac{w}v, $$ a result which common sense could suggest directly. In the notations of the problem, $a=x-r$ and $b=x+s+t$ hence $a=3-\frac{17}{13}\cdot2$ and $b=5+\frac{58}{13}\cdot2$.

share|improve this answer
    
Thanks, Did! But I don't understand the meaning of the first term you have in the lieklihood. I also don't get why you raise each term of the likelihood to the power of u, v, and w. Could you clarify that, please? –  user1172558 Feb 2 '13 at 23:51
    
The multinomial factor accounts for the number of ways of assigning tags less than 3 or between 3 and 5 or more than 5 to the students. Once every student is tagged, the remaining product is the probability that the score of each agrees with their tag. Here is an analogy: you throw a coin twice with probabilities $h$ and $t$ to get heads and tails. Then the probability to get one head and one tail is $2ht$ (that is, $ht$ for heads then tails + $th$ for tails then heads). –  Did Feb 3 '13 at 0:02
    
OK! But is there any easier way to solve this problem? I was trying to approach it just using the mean of a uniform distribution (a+b)/2. Any suggestion? –  user1172558 Feb 3 '13 at 0:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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