Dumb math question: Get min sample size based on result?

[ComputerGuy2006]

If there is a survey with 10.21% people picked option A and 89.79% picked option B, how can I figure out the minimum amount of respondents that are mathematically possible?

Now obviously there might be some rounding going on, but is there a good mathematical way to figure this out without writing a script to brute force it?

 

[Ken g6]

Rounding makes a hash of things. For instance, if the rounding was to the nearest percent, you'd say there were ten people.

 

[RichUK]

It's 10, lol

 

[UncleWai]

A/(A+B)= 0.1021

B/(A+B)= 0.8979

Work it out

 

[ComputerGuy2006]

A/(A+B)= 0.1021

B/(A+B)= 0.8979

Work it out

I can't hence my post. You just listed 100% of what I already know about this

 

[ComputerGuy2006]

Rounding makes a hash of things. For instance, if the rounding was to the nearest percent, you'd say there were ten people.

It cannot be 10, because if there was 10 people that would be 10%, you would never 'round' to 10.21. Im thinking more along the lines of the number might be 10.2028 and they rounded to 10.21.

 

[SOFTengCOMPelec]

141, 1381 = 10.2099927...
Very close with a tiny bit of rounding

1,381 People

Since the exact rounding method/system has NOT been specified. The question is probably ambiguous.
E.g. If NO rounding is permitted. Then 1,021 , 10,000 (people) is probably the correct answer.

 

[ComputerGuy2006]

Tnx, but how did you get those numbers (the 141)? Is there a math way to figure this out or did you just write a simple php/python script or something?

This questioned bugged me because 10k is also what I figured, yet that number is impossible (should be closer to 500 total people included). This is also the only report item that specifies decimals. Thinking the HR person who did it messed up on that report or did something weird.

 

[SOFTengCOMPelec]

You are right, I used a simple script/program to do it.
There probably is a quicker mathematical solution, but off-hand, I don't know what it is.

As it stands, I don't think the question is quite right, because it is ambiguous as regards rounding/precision.

 

[compcons]

I can't hence my post. You just listed 100% of what I already know about this

Solve the first equation for a. It will have some numbers and b as a variable. Then substitute that new a in for all a in the second equation. The will give b as an actual number. The go back and substitute that for b in the first equation and solve for a.

 

[frozentundra123456]

I dont know how to do it personally, but there is something called a "power" calculation. It tells you the sample size needed to get a statistically significant result based on the uncertainty you are willing to accept. We use this every time we submit a grant for an animal or human study to estimate the number of subjects necessary. Like I said though, our statisticians do the calculations, so I dont know how to do it.

 

[pete6032]

Is there a way to do this in Excel using goal seek?

 

[Paperdoc]

SOFTengCOMPelec has the right answer, and it's based on logic more than algebra. To start with we have only ONE variable: the fraction of the sample that picked Option A is 0.1021. Then there's the logical statement that the ONLY other choice was Option B, and hence the fraction that made that choice is already known.

The point about rounding is crucial. In attempting to answer OP's question we can make an assumption (since the info is not provided) that the results quoted are absolutely precise with NO rounding. The precision of the the result is 4 significant digits, so the possible variability in it is no more than than ± 1 in 10,000. Thus the MINIMUM sample size is 10,000. Now, a larger sample size could also generate exactly that result of 0.1021 (with an infinite number of zeroes after that to indicate absolute precision) ONLY if the sample size is exactly an integer multiple of 10,000. That is, the sample size also could be 20,000, 30,000, 100,000 or 1,000,000, etc. But it cannot be 19,287, for example, because that could not generate that exact result without any rounding. IF we assume that some rounding has been done already, then all bets are off and you can NOT determine a unique minimum sample size.

In terms of mathematics, this type of statistic falls into the realm of a Binary Distribution (that is, there are only two possible choices for an individual observation), an area widely used in studies of the occurence of defects and of reliability or failure of equipment.

 

[JTsyo]

A/(A+B)= 0.1021

B/(A+B)= 0.8979

Work it out

That won't work since there isn't an unique solution.

 

[Paperdoc]

That won't work since there isn't an unique solution.

In fact, because there are ONLY TWO possible answers ( option A or option B), then by definition we know that, for the fractions who chose those options, A + B =1, with NO rounding errors at all! So the original equations proposed amount to:

A / (A +B) = A / 1 = A
B / (A + B) = B / 1 = B

That is NOT two equations in two independent unknowns; they cannot be "solved" using algebraic methods. As SOFTengCOMPelec and I said, this is a logic problem, and it can ONLY yield a unique solution if you ADD the ASSUMPTION that the fractions provided have NO rounding errors involved. Otherwise there is NO unique solution.