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I have a Binomial distribution: $X$~$B(100,0.15)$. I used: $X$~$Po(15)$ as I considered n large and p small, so felt it to be a good approximation. However, the solution in the book used a normal approximation with:

$X$~$B(100,0.15)$ being approximately $X$~$N(np, npq)$

Question is, is my method valid in using the poisson approximation, and if not, why?

Edit: photo of orig question. It’s part d) the above question pertains to. From an old alevel maths textbook (S2) orig question from S2 exam paper

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The approx is correct, but using the Gaussian approx (with an opportune correction factor) you surely will reach the same result in a faster way (and perhaps a better result)

Using a poisson the calculations can be long and noisy...in this situation you could calculate the exact value with the binomial...

Just for the sake of completeness, will you add the request of the exercise? In this way we can look out at the 3 scenarios (exact binomial Vs Poisson Vs Gaussian approx)

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    $\begingroup$ Hi, thanks for your reply. Added the original question. As for gaussian approximation - i’ve never learnt it so the only choices I had were between poisson and normal. $\endgroup$
    – Noobcoder
    Jul 31 '20 at 10:25
  • $\begingroup$ @Noobcoder : this is very useful en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem $\endgroup$
    – tommik
    Jul 31 '20 at 10:29
  • $\begingroup$ @Noobcoder "Gaussian approximation" and "normal approximation" are the same thing. $\endgroup$
    – Ian
    Jul 31 '20 at 16:27
  • $\begingroup$ Reasonable start (+1). // @Noobcoder: Ease of computation for exact binomial and for Poisson and normal approximations depends on whether you're evaluating discrete binomial and Poisson probabilities 'by hand' from the PDF formula, or whether you use software. In my answer I try to show how good the approximations are for parts (a)-(c). $\endgroup$
    – BruceET
    Jul 31 '20 at 20:40
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(a) Test $H_0: p = .15$ against $H_a: p < .15,$ Under $H_0,$ the number of purple pins in the box is $X\sim\mathsf{Binom}(n=20,p=.15).$ Having seen one pin, the P-value of the test is $$P(X \le 1) = P(X=0)+P(X=1)=0.1756$$ (to four places).

Thus, even though we have seen less than the expected $E(X) = np = 20(.15) = 3$ purple pins, we do not have evidence from this one bit of data to reject $H_0$ at the 5% level of significance. (We could have rejected with a P-values smaller than $0.05 = 5\%.)$

The P-value is not difficult to compute exactly, using the PDF formula for the null binomial distribution. The exact answer from R (where pbinom is a binomial CDF and dbinom is a binomial PDF) can be obtained in either of two ways, the first of which is a little simpler:

pbinom(1, 20, .15)
[1] 0.1755579
dbinom(0:1, 20, .15)
[1] 0.03875953 0.13679835
sum(dbinom(0:1, 20, .15))
[1] 0.1755579

So there is no need for approximate solutions using Poisson or normal approximations. But, since you asked about approximations, let's look to see how good they are.

Poisson: Because $E(X) = 3,$ the best Poisson distribution to use for approximation is $X \stackrel{aprx}{\sim}\mathsf{Pois}(3).$ Again here, it is easy to compute the approximate P-value using the Poisson PDF. Using R, $P(X \le 1)\approx 0.1991,$ as shown below.

ppois(1, 3)
[1] 0.1991483

Normal: Because $\mu =E(X) = 3$ and $\sigma = \sqrt{np(1-p)} = \sqrt{2.55} = 1.5969,$ we use the approximating distribution $X \stackrel{aprx}{\sim}\mathsf{Norm}(\mu=3,\sigma=1.5989)$ to get $P(X \le 1) = P(X < 1.5) \approx 0.1738.$ This normal approximation is found in R below. [You could get about the same result from printed normal tables by standardizing.]

n=20; p=.15; mu=n*p; sg=sqrt(n*p*(1-p))
mu; sg
[1] 3
[1] 1.596872
pnorm(1.5, mu, sg)
[1] 0.173779

In this problem the normal approximation gives a slightly better approximation than Poisson. If you believe results from normal and Poisson approximations are nearly correct, either would be OK for testing $H_0$ at the 5% level because both give P-values substantially larger than $0.05.$

(b) This part is asking for the critical value $c$ such that $P(X \le c) \le 0.05).$

Using R, for the discrete binomial distribution, we can get a rough idea using the quantile function to see what value is just big enough to cut at least 5% of the probability from the lower tail of $\mathsf{Binom}(20, 0.15),$ which is $1.$ then use the CDF to see whether $c = 1$ or $0.$ The answer is that for this discrete distribution we can't cut exactly 5% from the lower tail. To cut just less than 5%, we need to use $c =0,$ which has $P(X \le c) = 0.0388 < 0.05.$

qbinom(.05, 20, .15)
[1] 1
pbinom(0:1, 20, .15)
[1] 0.03875953 0.17555788

If we're using the discrete Poisson distribution, we get the same value $c=0.$

qpois(.05, 3)
[1] 1
ppois(0:1, 3)
[1] 0.04978707 0.19914827

For the normal distribution, we can cut exactly 5% of the probability from the lower tail with the value 0.3734, which is not an integer, so we use the next smaller integer $0.$ [People who standardize may think they're testing at exactly the 5% level instead of the 3% level if they use $c^\prime$ that cuts 5% from the lower tail of standard normal, without checking what that means in terms of numbers of purple pins.]

qnorm(.05, mu, sg)
[1] 0.3733794
pnorm(0, mu, sg)
[1] 0.03014459

Finally, for parts (a) and (b), here is a plot of the PDF of $\mathsf{Binom}(20, .15)$ (bars), the PDF of $\mathsf{Pois}(3)$ (centers of open circles), and the density function of $\mathsf{Norm}(3, 1.5969).$

enter image description here

x = 0:20; PDF = dbinom(x, 20, .15) 
pdf.p = dpois(x, 3)
plot(x, PDF, type="h", lwd=2)
 abline(h=0, col="green2")
 points(x, pdf.p, col="red")
 curve(dnorm(x,mu,sg), -1, 21, add=T, col="blue", lwd=2)

(c) With a total of $Y=8$ purple pins in $n=100,$ the null distribution is $\mathsf{Binom}(100, .15),$ with $E(Y) = 100(.15) = 15.$ Thus we have seen what seems many fewer than the expected number of purple pins. The P-value is $P(Y\le 8) = 0.0275 < 0.05 = 5\%,$ and we reject $H_0.$ [Both approximations work better here on account of the larger $n:$ The Poisson approximation is $P(y\le 8) \approx 0.0374;$ The normal approximation gives $P(Y \le 8) \approx 0.0344.$]

pbinom(8, 100, .15)
[1] 0.02747569
n=100; p=0.15;  mu = n*p; sg=sqrt(n*p*(1-p));  mu; sg
[1] 15
[1] 3.570714
ppois(8, 15)
[1] 0.03744649
pnorm(8.5, mu, sg)
[1] 0.03435179

For values up to 40, here is a plot that show binomial probabilities along with Poisson and normal approximations for part (c).

enter image description here

Notes: (1) You really shouldn't use the normal approximation in (a) and (b). The usual rule-of-thumb for using a normal approximation to a binomial distribution is to have both $np$ and $n(1-p)$ greater than 5; here $np=3.$ Then, if $p$ is near $1/2,$ you can expect about two decimal places of accuracy from a normal approximation. Some authors say Poisson approximations work best for large $n$ and small $p,$ with a mean of moderate size, but few give specific numbers. [Some users on this site are eager to give 'counterexamples', with good Poisson fits to binomial distributions that break the rules for various rules-of-thumb.]

(2) From the level of your question, I'm guessing you know (a) how to use standardization with printed normal tables to get normal approximations and (b) how to use the binomial PDF to find individual binomial probabilities. If you need help with either, please leave a Comment and I'll try to return to edit my answer to include what you need.

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