# Difference of two sample means from the same binomial distribution

Here is the question I am trying to answer: There are 4000 birds chirping in a forest and a bird either sings with a 30% probability or doesn't with an 70% probability. Each one is independent of the other and overall noise is proportional to the number of birds. The question then asks if you listen to the birds chirping in the forest at two different times, what's the probability that the noise at the second time is more than 3% larger than the noise at the first time. It is clear to me that this is a binomial distribution with n=4000 and p=0.3, and these noise levels are sample means but I don't know where to do from there. I don't know how to determine the probability that the second sample mean will be 3% higher. Would Bayes theorem help? Any help would be great!

• This is not a question about data or estimation: it is a pure probability calculation based on stated distributions. You are probably meant to use the Normal approximations to the distributions (they will be excellent for the parameters you are given).
– whuber
May 11, 2021 at 17:27
• Ok, sorry I put this in the wrong section of stack exchange! If I were to use a normal approximation how would i determine the probability that the second mean is 3% greater than the first? I only know how to determine the probability that a normal distribution will be greater than or less than a specific value May 11, 2021 at 17:47
• I did not intend to imply you posted this in the wrong place--I just wanted to keep you from going down the wrong path in your search for a solution. I do wonder, though, what "either sings with a 30% probability or doesn't with a 70% probability" means. Would that mean all birds have a 30% chance of singing or all birds have a 70% chance? If the chances could differ among birds, what are you assuming about how those chances are determined? The question doesn't appear to have enough information for a definite answer.
– whuber
May 11, 2021 at 18:22
• I believe it means that each individual bird has a 30% chance of singing and thus you can consider each bird to be a binomial variable with p=0.3 May 11, 2021 at 18:29
• Thanks--I misread the question. Although assuming the bird songs are independent is unrealistic, that evidently is one assumption you will need to make progress.
– whuber
May 11, 2021 at 18:31

As you note in a comment, the situation can be modeled by viewing each bird as a Bernoulli$$(0.3)$$ variable $$X_i,$$ $$i=1,2,\ldots, 4000,$$ and the total noise level is proportional to the sum of all the $$X_i.$$

The question cannot be answered without making several strong assumptions, of which the most important is that all the $$X_i$$ are independent. In that case the sum of the $$X_i$$ (let's call it $$Y$$) has a Binomial distribution with parameters $$n=4000$$ and $$p=0.3.$$ (Otherwise we would have little idea what distribution $$Y$$ might have.)

In such cases (characterized by both $$n\,p$$ and $$n(1-p)$$ being "large," with "large" depending on taste but typically exceeding $$5$$), the distribution of $$Y$$ is well approximated by a Normal distribution of mean $$\mu=p\,n = 0.3(4000)$$ and variance $$p(1-p)n = 0.3(0.7)(4000).$$ Let's not do the arithmetic here, and just leave the expressions general.

Another assumption you need to make is that the two observations of $$Y$$ are independent. Let's denote them $$Y_0$$ and $$Y_1.$$ The question asks,

What is the chance $$Y_1$$ is at least $$3\%$$ greater than $$Y_0$$?

The "at least" interpretation of "3% higher" is practically forced on us, because the chance the difference is exactly 3% higher will be astronomically small: it's not even an interesting question to ask. So that's a third assumption.

It helps to phrase this event mathematically as

$$Z = Y_1 - 1.03\, Y_0 \ge 0.\tag{*}$$

Why? Because (a) it reduces an apparent two-variable problem into a one-variable problem; (b) a linear combination of independent Normal variables is Normal, (greatly) simplifying the calculations; and (c) the constant of proportionality (between noise level and the $$Y_i$$) doesn't matter anymore: it just multiplies everything, leaving $$0$$ at $$0.$$

Probability calculations with Normal variables require knowledge of their parameters, tantamount to the mean and variance. Let's go get these.

The mean of $$Z$$ evidently is

$$E[Z] = E[Y_1] - 1.03E[Y_0] = (1-1.03)p(1-p)n = (-0.03)(0.3)(0.7)(4000).$$

Ordinarily, computing the variance requires knowing the covariance of the $$Y_i.$$ Our independence assumption sets that covariance to $$0,$$ giving

$$\operatorname{Var}(Z)=\operatorname{Var}(Y_1) + (-1.03)^2\operatorname{Var}(Y_0) + 2(-1.03)\operatorname{Cov}(Y_0,Y_1) = (1 + (-1.03)^2)p(1-p)(n),$$

etc. You can do the calculations and, since you now know the parameters of $$Z,$$ easily compute the answer $$(*),$$ $$\Pr(Z \ge 0).$$

• This is a very good approximation. It seems to give $0.1934547$ compared to an exact figure of $0.1934043$ as calculated on the same question posted at math.se May 11, 2021 at 19:24
• +1 for your answer on math, @Henry: thanks for doing the comparison. The approximation tends to improve when applied to differences of comparably-skewed Binomials, as here: the errors nearly cancel.
– whuber
May 11, 2021 at 20:45