ইতিমধ্যেই এই প্রশ্নের কিছু চমৎকার উত্তর আছে, কিন্তু আমি উত্তর চাই কেন , মান ত্রুটি কি এটা কেন আমরা ব্যবহার সবচেয়ে খারাপ ক্ষেত্রে যেমন, এবং কিভাবে মান ত্রুটি সঙ্গে পরিবর্তিত হয় এন ।p=0.5n
ধরা যাক আমরা মাত্র একটি ভোটারের সমীক্ষা নিই, আসুন তাকে বা তার ভোটারকে ফোন করুন এবং জিজ্ঞাসা করুন "আপনি কি বেগুনি দলের পক্ষে ভোট দিবেন?" আমরা উত্তরটি "হ্যাঁ" এর জন্য 1 এবং "" না "হিসাবে কোড করতে পারি। ধরা যাক যে "হ্যাঁ" এর সম্ভাবনা । আমাদের কাছে এখন বাইনারি র্যান্ডম ভেরিয়েবল এক্স 1 যা সম্ভাব্যতা পি সহ 1 এবং সম্ভাব্যতা 1 - পি সহ 0 হয় । আমরা বলি যে এক্স 1 হল সাফল্যের সম্ভাবনার সাথে একটি বার্নুইলি পরিবর্তনশীল পি , যা আমরা এক্স 1 ∼ বি ই r এন ও ইউ আই এল এল আই ( পি ) লিখতে পারিpX1p1−pX1pX1∼Bernouilli(p)। প্রত্যাশিত, অথবা গড়, মান দেওয়া হয় ই ( এক্স 1 ) = Σ এক্স পি ( এক্স 1 = এক্স ) যেখানে আমরা উপর যোগফল সব সম্ভব ফলাফলের x এর এক্স 1 । তবে কেবল দুটি ফলাফল রয়েছে, 0 সম্ভাব্যতার সাথে 1 - পি এবং 1 সম্ভাব্যতা পি সহ 1 , সুতরাং যোগফলটি কেবল ই ( এক্স 1 ) = 0 ( 1 - পি ) + 1 ( পি )X1E(X1)=∑xP(X1=x)xX11−pp । থাম এবং চিন্তা কর. এটি আসলে পুরোপুরি যুক্তিসঙ্গত বলে মনে হচ্ছে - যদি বেগুনি পার্টির সমর্থনকারী ভোটার 1 এর 30% সম্ভাবনা থাকে এবং আমরা "হ্যাঁ" বলি এবং "না" বললে 0 বলে ভেরিয়েবল 1 হয়ে যাব, তবে আমরা চাই এক্স 1 গড়ে 0.3 হওয়ারআশা করে expectE(X1)=0(1−p)+1(p)=pX1
এর মনে কি আমরা বর্গ ঘটে যাক । যদি এক্স 1 = 0 হয় তবে এক্স 2 1 = 0 এবং যদি এক্স 1 = 1 হয় তবে এক্স 2 1 = 1 । তাই আসলে এক্স 2 1 = এক্স 1 উভয় ক্ষেত্রে। যেহেতু তারা একই, তাই তাদের অবশ্যই একই প্রত্যাশিত মান থাকতে হবে, তাই ই ( এক্স 2 1 ) = পি । এটি আমার একটি Bernouilli পরিবর্তনশীল ভ্যারিয়েন্স গণক এর একটি সহজ উপায় দেয়: আমি ব্যবহার ভী একটিX1X1=0X21=0X1=1X21=1X21=X1E(X21)=pVar(X1)=E(X21)−E(X1)2=p−p2=p(1−p) and so the standard deviation is σX1=p(1−p)−−−−−−−√.
Obviously I want to talk to other voters - lets call them voter 2, voter 3, through to voter n. Let's assume they all have the same probability p of supporting the Purple Party. Now we have n Bernouilli variables, X1, X2 through to Xn, with each Xi∼Bernoulli(p) for i from 1 to n. They all have the same mean, p, and variance, p(1−p).
XiX=∑ni=1Xi. I can calculate the mean or expected value of X by using the rule that E(X+Y)=E(X)+E(Y) if those expectations exist, and extending that to E(X1+X2+…+Xn)=E(X1)+E(X2)+…+E(Xn). But I am adding up n of those expectations, and each is p, so I get in total that E(X)=np. Stop and think. If I poll 200 people and each has a 30% chance of saying they support the Purple Party, of course I'd expect 0.3 x 200 = 60 people to say "yes". So the np formula looks right. Less "obvious" is how to handle the variance.
There is a rule that says
Var(X1+X2+…+Xn)=Var(X1)+Var(X2)+…+Var(Xn)
but I can only use it
if my random variables are independent of each other. So fine, let's make that assumption, and by a similar logic to before I can see that
Var(X)=np(1−p). If a variable
X is the sum of
n independent Bernoulli trials, with identical probability of success
p, then we say that
X has a binomial distribution,
X∼Binomial(n,p). We have just shown that the mean of such a binomial distribution is
np and the variance is
np(1−p).
Our original problem was how to estimate p from the sample. The sensible way to define our estimator is p^=X/n. For instance of 64 out of our sample of 200 people said "yes", we'd estimate that 64/200 = 0.32 = 32% of people say they support the Purple Party. You can see that p^ is a "scaled-down" version of our total number of yes-voters, X. That means it is still a random variable, but no longer follows the binomial distribution. We can find its mean and variance, because when we scale a random variable by a constant factor k then it obeys the following rules: E(kX)=kE(X) (so the mean scales by the same factor k) and Var(kX)=k2Var(X). Note how variance scales by k2. That makes sense when you know that in general, the variance is measured in the square of whatever units the variable is measured in: not so applicable here, but if our random variable had been a height in cm then the variance would be in cm2 which scale differently - if you double lengths, you quadruple area.
Here our scale factor is 1n. This gives us E(p^)=1nE(X)=npn=p. This is great! On average, our estimator p^ is exactly what it "should" be, the true (or population) probability that a random voter says that they will vote for the Purple Party. We say that our estimator is unbiased. But while it is correct on average, sometimes it will be too small, and sometimes too high. We can see just how wrong it is likely to be by looking at its variance. Var(p^)=1n2Var(X)=np(1−p)n2=p(1−p)n. The standard deviation is the square root, p(1−p)n−−−−−√, and because it gives us a grasp of how badly our estimator will be off (it is effectively a root mean square error, a way of calculating the average error that treats positive and negative errors as equally bad, by squaring them before averaging out), it is usually called the standard error. A good rule of thumb, which works well for large samples and which can be dealt with more rigorously using the famous Central Limit Theorem, is that most of the time (about 95%) the estimate will be wrong by less than two standard errors.
Since it appears in the denominator of the fraction, higher values of n - bigger samples - make the standard error smaller. That is great news, as if I want a small standard error I just make the sample size big enough. The bad news is that n is inside a square root, so if I quadruple the sample size, I will only halve the standard error. Very small standard errors are going to involve very very large, hence expensive, samples. There's another problem: if I want to target a particular standard error, say 1%, then I need to know what value of p to use in my calculation. I might use historic values if I have past polling data, but I would like to prepare for the worst possible case. Which value of p is most problematic? A graph is instructive.
The worst-case (highest) standard error will occur when p=0.5. To prove that I could use calculus, but some high school algebra will do the trick, so long as I know how to "complete the square".
p(1−p)−−−−−−−√=p−p2−−−−−√=14−(p2−p+14)−−−−−−−−−−−−−−√=14−(p−12)2−−−−−−−−−−−√
The expression is the brackets is squared, so will always return a zero or positive answer, which then gets taken away from a quarter. In the worst case (large standard error) as little as possible gets taken away. I know the least that can be subtracted is zero, and that will occur when p−12=0, so when p=12. The upshot of this is that I get bigger standard errors when trying to estimate support for e.g. political parties near 50% of the vote, and lower standard errors for estimating support for propositions which are substantially more or substantially less popular than that. In fact the symmetry of my graph and equation show me that I would get the same standard error for my estimates of support of the Purple Party, whether they had 30% popular support or 70%.
So how many people do I need to poll to keep the standard error below 1%? This would mean that, the vast majority of the time, my estimate will be within 2% of the correct proportion. I now know that the worst case standard error is 0.25n−−−√=0.5n√<0.01 which gives me n−−√>50 and so n>2500. That would explain why you see polling figures in the thousands.
In reality low standard error is not a guarantee of a good estimate. Many problems in polling are of a practical rather than theoretical nature. For instance, I assumed that the sample was of random voters each with same probability p, but taking a "random" sample in real life is fraught with difficulty. You might try telephone or online polling - but not only has not everybody got a phone or internet access, but those who don't may have very different demographics (and voting intentions) to those who do. To avoid introducing bias to their results, polling firms actually do all kinds of complicated weighting of their samples, not the simple average ∑Xinthat I took. Also, people lie to pollsters! The different ways that pollsters have compensated for this possibility is, obviously, controversial. You can see a variety of approaches in how polling firms have dealt with the so-called Shy Tory Factor in the UK. One method of correction involved looking at how people voted in the past to judge how plausible their claimed voting intention is, but it turns out that even when they're not lying, many voters simply fail to remember their electoral history. When you've got this stuff going on, there's frankly very little point getting the "standard error" down to 0.00001%.
To finish, here are some graphs showing how the required sample size - according to my simplistic analysis - is influenced by the desired standard error, and how bad the "worst case" value of p=0.5 is compared to the more amenable proportions. Remember that the curve for p=0.7 would be identical to the one for p=0.3 due to the symmetry of the earlier graph of p(1−p)−−−−−−−√
