# How many people bought wine?

Rephrased a problem trying to solve for work in terms of people buying wine, also included progress made so far.

Set-up: Customers enter a winery with the option of buying bottles of wine. Those who chose not to purchase can leave through one exit while those who wish to purchase go to the cash register. After exiting the register they are asked a survey "How many bottles of wine did you just purchase?". All have to answer. They can give any number between zero and their true value, but not more. If they answer 'zero' nothing is recorded. Any other answer is added to the survey total along with incrementing the number of positive respondents.

Knowing:

• $T$ : Actual number of bottles sold (known, from inventory)
• $R$ : Total bottles recorded sold from survey
• $A$ : Total respondents that gave non-zero answer
• Find $X$, total number of actual customers.
• Also given $U$ : Theoretical maximum for answer.

U can be explained that the maximum number of customers can be either the total number of bottles of wine on the shelf (one bottle per customer), or the capacity of the store in terms of customers.

Small illustrative example: • $T = 12$ bottles actually sold (all squares)
• $R = 7$ bottles recorded being sold (black squares)
• $A = 3$ people gave non-zero answers (brackets with at least one black square)
• $X = 4$ actual number of customers (all brackets; unknown)

Progress so far:

At first tried simple proportion $\hat{X}=\frac{A T}{R}$ but found it over-estimated the true population. This led to exploring what properties the estimate should have. Found at least six:

1. The minimum $X$ can be is $A$. (At least $A$ people bought wine.)
2. The maximum $X$ can be is the minimum of $U$ and $T-R+A$. Define this to be $V$. (Reason: If we know that $A$ people bought $R$ bottles then the remaining $T-R$ bottles could in theory be bought by $T-R$ individual people. Take the minimum as $U$ could be lower due to store capacity)
3. $X$ can only be $A$ when $R=T$ (All bottles were bought, nothing to estimate)
4. $X$ can only be the maximum when $R=A$ (Every respondent said they bought one bottle)
5. If they actually purchased a bottle, they purchased at least one.
6. If they are recorded as purchasing a bottle, they purchased at least one.

 and  imply it could be a weighted average between $A$ and $V$ for some value $w$

$$\hat{X}=Aw + V(1-w)$$

 and  imply the weight could be linear

$$w=\frac{R-A}{T-A}$$

when ($R=A$ then $\hat{X}=V$) and ($R=T$ then $\hat{X}=A$).

 and  imply the weight should be modified. Each customer who bought wine, bought at least one; and each customer who gave a non-zero answer in the survey also bought at least one.Out of $R$ recorded only $R-A$ are 'free', the first $A$ were needed to assign 1-1 to each customer. Same argument suggests the denominator should be $T-X$. With $X$ unknown substitute with $\hat{X}$

Summary:

Define

• $V=Min[U,T-R+A]$
• $w=\frac{R-A}{T-\hat{X}}$
• Solve for $\hat{X}$ in the equation: $\hat{X}=Aw + V(1-w)$

The new estimate gave significantly better answers than simple proportion estimate.

Questions

1. Is this the best estimate that can be made with given information? (Am I missing an assumption or property which would make it better?)
2. Is there a way to estimate the variance?
• "Best estimate" in what sense - what is being optimized? Is anything known about the distribution of values customers report? – Glen_b Mar 18 '14 at 19:49
• @Glen_b Good point. I guess I should have phrased "better estimate" but that gives the same question - what is being optimized? Ideally would like MVUE, but realize that may not be possible. As for the distribution, we only have the values T,R,A,U. Individual customer responses are not known, only their totals (R and A) – sheppa28 Mar 19 '14 at 15:31
• Then on what basis can you make assumptions about their behavior? They might all do whatever it takes to be at one or the other of the extremes (as far as is consistent with the constraints), or they might be more-or-less concentrated anywhere in between, or they might be in two groups, or... who knows? – Glen_b Mar 19 '14 at 23:13