3
$\begingroup$

I'm trying to understand how uncertain an estimation of a demand distribution of a good is. If I take a random sample of n people from the representative consumer population who give me their personal cutoff/choke prices (assuming they accurately report them), I can construct a demand curve with n points representing an estimation of the proportion of the population who would buy the good at that price.

Given that I used n independent pieces of information to calculate n points, should I be using a t-distribution with 1 degree of freedom to calculate the confidence interval for each point? And, if I want to reduce this interval by increasing the degrees of freedom, I either need to take more samples, or increase the granularity of the demand curve to get more d.f. at each point?

$\endgroup$
3
  • $\begingroup$ Could you clarify how you would construct a demand curve? Do your n points constitute the "curve", or would you having calculated the n points use regression to fit a smooth curve to them? $\endgroup$ Commented Jun 18, 2014 at 21:01
  • $\begingroup$ n "choke price" points can be aggregated into a bar chart (the "demand curve"), where each bar represents the proportion of people in the sample who would buy the good at that price. It is a cumulative distribution. You can, of course, smooth this discrete bar chart into a smooth continuous curve, but as it doesn't add new information (and the application I'm using this for doesn't require it) I don't bother. $\endgroup$
    – Escher
    Commented Jun 19, 2014 at 13:04
  • $\begingroup$ A regression doesn't add new information, but it could perhaps make more effective use of the information you have. However, I have answered your question on the basis that each estimated point simply reflects the sample proportion of people who would buy the good at that price. $\endgroup$ Commented Jun 20, 2014 at 10:03

1 Answer 1

0
$\begingroup$

You should use a $t$-distribution with $n-1$ degrees of freedom (or if $n$ is large the normal distribution which provides a good approximation to the $t$-distribution). A larger sample, increasing $n$, would narrow the confidence interval both because it increases the degrees of freedom (although once $n$ increases beyond 30 this effect is quite small) and because it reduces the standard error.

For a given price the sample contains, say, $m$ people who would buy the good at that price. Thus the proportion of the population who would buy the good at that price can be estimated as $m/n$. This applies to any price, not just to the $n$ cutoff prices given by the people in the sample. So you can calculate any number of points, taking a point to be identified by a price and its associated estimated proportion.

The number of points that are or could be calculated has no relevance to the degrees of freedom used in calculating a confidence interval for the proportion at any one price (the situation is quite different from a regression in which $n$ observations are used to jointly estimate $k$ parameters, implying $n-k$ degrees of freedom). The degrees of freedom here depend only on the size $n$ of the sample used to estimate that proportion and its standard error, from which we deduct 1 because use of the same sample to estimate both the proportion and its standard error implies a loss of 1 degree of freedom.

The value of $t_{n-1}$ at the desired confidence level times the estimated standard error gives the half-width of the confidence interval.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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