"Let $X_i$ where $i = 1,...,5$ be independent Poisson-distributed random variables with expected value be $\mu$. We get $\bar{x}=16$. Find the lower limit of the confidence interval where the confidence level is 95%."

What I've done: since $V(X)$ for Poisson-distribution is $\mu$ I believe $\sigma = \sqrt{\mu} = 4$. To calculate D I do $\frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{5}}$. I then find the lower limit by $\mu -D \cdot \lambda_{\frac{\alpha}{2}} = 16-\frac{4}{\sqrt{5}}\cdot1.96 = 12.49$ which is the wrong answer. What am I doing wrong/misunderstanding here?

  • $\begingroup$ What is D and what exactly is the method you are following? You appear to be using a normal approximation which is not justified for a sample size of 5. $\endgroup$ – StubbornAtom May 23 '19 at 16:05
  • $\begingroup$ Do you have to apply a "continuity correction" of sorts? $\endgroup$ – Matthias May 23 '19 at 16:05
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    $\begingroup$ I think you can adopt the answer here to your question. Also see en.m.wikipedia.org/wiki/…. $\endgroup$ – StubbornAtom May 23 '19 at 16:22
  • $\begingroup$ Yes, I just realized that I followed a formula which requires it to be a normal distribution, which this obviously isn't. $\endgroup$ – gbgult May 23 '19 at 16:23

The total number of events counted is $T \sim \mathsf{Pois}(\Lambda = 5\lambda),$ where $X_i \sim \mathsf{Pois}(\lambda).$

Using the normal approximation to a Poisson distribution we have $$Z = \frac{T-\Lambda}{\sqrt{\Lambda}} \stackrel {aprx}{\sim} \mathsf{Norm}(0,1).$$

For the asymptotic Wald interval, one estimates the denominator as $\sqrt{80},$ to obtain $$P\left(-1.96 <\frac{T-\Lambda}{\sqrt{80}} < 1.96\right) \approx 0.95.$$

So that an approximate 95% confidence interval for $\Lambda$ is of the form $T \pm 1.96\sqrt{80},$ which computes to $(71.06, 88.95).$ Upon division by $5$ we have the approximate 95% CI $(14.21, 17.79)$ for $\lambda.$ This is the traditional CI for the Poisson mean, and probably what you are expected to compute.

However, notice that two approximations were involved above: (i) normal approximation to Poisson and (ii) $\sqrt{\Lambda} \approx \sqrt{T}$ $= \sqrt{80}.$ Thus the Wald CI may not be accurate for small values of $T.$

Consider a test of $H_0: \Lambda = \Lambda_0$ against $H_a: \Lambda \ne \Lambda_0,$ rejecting when $|Z_0| \ge 1.96,$ where $Z_0 = (T - \Lambda)/\sqrt{\Lambda}.$ Let us 'invert this test' to get a CI, consisting of 'non-rejectable' values of $\Lambda_0.$

Then upon solving a quadratic inequality, one gets the improved CI for $\Lambda:$ $$T + 2 \pm 2\sqrt{T+1}.$$

This method gets rid of the second assumption. [In solving the inequality, we have used the slight simplification of equating 1.96 to 2 in several places.]

For our data this improved 95% CI computes to $(64, 100)$ for $\Lambda$ and hence to $(12.8, 20.0)$ for $\lambda.$

Note: For more on improved Poisson CIs, perhaps see this Q&A or this one.

Addendum: Mainly in response to question in @StubblornAtom's comment. The simulation in R below shows that for $\Lambda = 80,$ the "95%" Wald interval has a little less than 95% coverage. By contrast, the slightly longer adjusted CI has very nearly 95% coverage.

lam = 80
m = 10^6;  lcl.w = ucl.w = lcl.a = ucl.a = numeric(m)
for(i in 1:m) {
   t = rpois(1, lam)
   me = 1.96*sqrt(t);  lcl.w[i] = t-me; ucl.w[i] = t+me 
   me = 1.96*sqrt(t+1);  lcl.a[i] = t+2-me; ucl.a[i] = t+2+me
mean(lcl.w < lam & ucl.w > lam)
[1] 0.944699          # Wald int: aprx 94.4% coverage
mean(ucl.w - lcl.w)
[1] 35.00591          # Avg len Wald: 35.0
mean(lcl.a < lam & ucl.a > lam)
[1] 0.9499            # Adj int: aprx 95.0%
mean(ucl.a - lcl.a)
[1] 35.2254           # Avg len Adj: 35.2
  • $\begingroup$ And while estimating $\Lambda$ by $T=80$, isn't $\Lambda\approx 80$ already large enough for normal approx so that the Wald CI is justified? $\endgroup$ – StubbornAtom May 24 '19 at 4:41
  • $\begingroup$ Thanks. Fixed my typo and (I hope) made it clear I was changing from $\mu$ to $\lambda.$ // $\Lambda \approx 80$ is pretty much OK for (i) the normal approximation, but not quite so good for (ii) estimating the standard error. $\endgroup$ – BruceET May 24 '19 at 4:44
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    $\begingroup$ Thanks. Fixed my typo and (I hope) made it clear I was changing from $μ$ to $λ.$ // $Λ≈80$ is pretty much OK for (i) the normal approximation, but not quite so good for (ii) estimating the standard error. // See Addendum with simulation showing approx coverage of Wald and adjusted intervals // Also perhaps see graphs at end of first link. // At least a dozen kinds of CIs have been seriously recommended for Poisson mean $\lambda.$ $\endgroup$ – BruceET May 24 '19 at 5:36

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