Central limit theorem, when do we know if n isn't large enough

Question

I'm working on a statistics question, and I'm stumped on how to answer it.

Here is the question

According to a survey conducted by the American Bar Association, 1 in every 410
Americans is a lawyer, but 1 in every 64 residents of Washington, D.C., is a lawyer.


(a) Use the Central Limit Theorem to approximate the probability that there is at
least one lawyer in a random sample of 1500 Americans. Is n = 1500 large enough
for the approximation to work well?

Using the CLT, I found that the answer was approximately .95, but I don't know how to determine if the sample size is large enough for the approximation? What do you look for in this case?

The usual rule of thumb is $np\ge 10$ (or sometimes $5$) so no, the normal approximation should not be used here. — A.S., Feb 18 '16 at 04:04
You should use a Poisson approximation instead if you want to avoid exact calculations. — A.S., Feb 18 '16 at 04:24
The usual rule of thumb is that both $np > 5$ and $n(1-p) > 5$. Some authors use values other than 5. There are better rules of thumb, but this one is easiest to remember and works pretty well. It works best for $p$ near 1/2 and worst for probabilities far out in the tails. — BruceET, Feb 18 '16 at 05:56

score 1 · Answer 1 · answered Feb 18 '16 at 00:06

1

You can calculate the exact probability and compare it to the CLT approximation. The number of lawyers in a sample of $1500$ has binomial distribution with $n=1500$ and $p=1/410$, so the probability that there are no lawyers in your sample is $$(1-p)^n=\left(1-\frac1{410}\right)^{1500}.$$ Subtract this number from $1$ to get the exact prob of least one lawyer in your sample.

Aside: When you did the CLT approximation, did you apply the continuity correction? (Do you know about the continuity correction?)

answered Feb 18 '16 at 00:06

grand_chat

38,951

no what is the continuity correction? – user270494 Feb 18 '16 at 05:37
It is an attempt to 'coordinate' the discrete binomial distribution with the continuous normal. Example: for the binomial $P(3 < X < 7) = P(4 \le X \le 6) = P(3.5 < X < 6.5).$ Of these three forms of the same expression for the binomial, the last of them leads to the best normal approximation. – BruceET Feb 18 '16 at 06:02

BruceET · Accepted Answer · 2016-02-18T07:21:53.487

$Exact\; Binomial$: As per the Answer by @grand_chat. Let the number of lawyers in a random sample of 1500 Americans be $X \sim Binom(1500, 1/410)$. We seek $P(X \ge 1) = 1 - P(X = 0).$ From R, we have:

 1 - (409/410)^1500
 ## 0.9743447
 1 - dbinom(0, 1500, 1/410)
 ## 0.9743447

$Poisson\; approximation\; to\; binomial$. We must have $\lambda = E(X) = 1500/410 = 3.658537,$ so that $Y \sim Pois(\lambda).$ Again, we seek $P(Y \ge 1) = 1 - P(Y=0).$

 1 - exp(-lam)
 ## 0.9742298
 1 - dpois(0, lam)
 ## 0.9742298

$Normal\; approximation$. For the binomial distribution, $\mu = np = 1500/410,$ $\sigma^2 = np(1-p) = \mu(409/410).$ Now let $W \sim Norm(\mu, \sigma).$ We seek $P(W \ge 1) = P(W > .5) = 1 - P(W < .5).$

 mu = 1500*(1/410);  var = mu*(409/410); sg = sqrt(var)
 1 - pnorm(.5, mu, sg)
 ## 0.9508693

Of course $P(W > .5)$ can be found from printed normal tables by standardizing: $P(W > .5) = P[Z = (W - \mu)/\sigma > (.5 - \mu)/\sigma)].$ This seems to be the result you obtained. [Note: Application of the 'continuity correction' leads us to seek $P(W \ge .5)$ instead of $P(W \ge 1).$]

$Summary\; comments.$ The Poisson approximation is very good (see figure below). Although the 'usual rule' for using the the normal approximation as given in the comment does not hold, the normal approximation is not very good here for two additional reasons: (a) We are dealing with a probability in the tail of the binomial distribution, (b) we are dealing with a skewed binomial distribution ($p$ far from 1/2). In statistical practice, normal approximations are less frequently used nowadays than before software was available to do exact computations.

The figure below shows the binomial distribution (black bars), its Poisson approximation (purple dots), and the normal curve that matches the binomial mean and variance.

Larginess of $n$ has nothing to do with the quality of Poisson approximation. In fact, if $n$ was smaller, the approximation would be even better (in relative terms). What matters is that $p$ is small and we are interested in probabilities of events not far from the mean. — A.S., Feb 18 '16 at 05:56
@A.S. Best just to refer to figure in this case. Edited summary accordingly. — BruceET, Feb 18 '16 at 07:16

Central limit theorem, when do we know if n isn't large enough

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