# Is my answer to this Poisson distribution story problem correct?

Andreia's secretary makes random errors in his work at an average rate if 1.7 errors every 100 words.

(a) Andreia offers the secretary a choice of one of two bonus schemes, based on a sample of 40 pieces of work each consisting of 100 words. In scheme A, the secretary will receive the bonus if more than 10 of the 40 pieces of work contain no errors. In scheme B, the bonus is awarded if the total number of errors in all 40 pieces of work is fewer than 56

Explain which bonus scheme you would advise the secretary to choose.

(b) Following the bonus scheme, Andreia randomly selects a single 500-word piece of work from the secretary to test if there is any evidence that the secretary's error rate has reduced. Stating your hypotheses clearly, and using a 5% level of significance, find the critical region for this test.

My approach to solving these is as follows.

(a) Naturally one would choose the most likely of these two criteria.

For scheme A:

Let $$X$$ be the number of errors made with respect to a given 100-word document. Then $$X \sim Poisson(1.7).$$

$$P(X=0) = \frac{1.7^{0}e^{-1.7}}{0!} = e^{-1.7}.$$

My impression is that if $$Y$$ is the number of pieces of work for which the secretary makes no errors, then $$Y \sim B(40,e^{-1.7}).$$

Consequently, the secretary's chance of receiving the bonus under scheme A is given by $$P(Y >10) = 1-P(Y \leq 10)$$ $$= 1 - \Bigg[{40\choose{0}} (e^{-1.7})^{0}(1-e^{-1.7})^{40} + {40\choose{1}} (e^{-1.7})(1-e^{-1.7})^{39} + \dots + {40\choose{10}} (e^{-1.7})^{10}(1-e^{-1.7})^{30} \Bigg].$$

I think scheme B can be solved by simply scaling the the Poisson parameter $$\lambda = 1.7$$ by a factor of $$40.$$ This produces a new error rate of 68 errors per 4000 words, so that if $$Z$$ is the number of errors made over the course of the 40 100-page documents, then $$P(Z < 56) = \sum_{i=0}^{55} \frac{68^{i}e^{-68}}{i!}.$$

I'm not necessarily interested in the final answer. I'd just like to know whether my reasoning is correct.

(b) I would suggest that $$H_{0}$$ state that the secretary's error rate has not changed, and that $$H_{1}$$ state that the secretary's error rate has reduced.

Say $$X_{1}$$ represents the number of errors that the secretary makes for each 500 word passage.

If we scale $$\lambda$$ by a factor of $$5,$$ we get a hypothetical error rate of 8.5 errors for every 500-words (assuming H_{0}). I think the way to obtain the critical value would be to read a Poisson CDF table and find the value of $$x$$ (where x represents a given value of the random variable $$X_{1}$$) that best corresponds to $$\lambda = 8.5$$ and $$P(X_{1} \leq x) = 0.05).$$

Am I mistaken?

• For (b) it's not a scaling but a sum instead. Use that the sum of independent Poissons is Poisson again. (The result remains the same.) Or use CLT. Commented May 16, 2022 at 12:52
• @MichaelHoppe I'm sorry but I don't really understand what you mean. Could you please elaborate? Commented May 16, 2022 at 14:02

The hypothesis test in $$(b)$$ is $$H_0:\lambda= 1.7$$ against $$H_a:\lambda<1.7$$. If you take $$X$$ as the number of errors identified in your $$500$$ page document, then $$X|H_0\sim \text{Poisson}(8.5)$$. Take $$\hat{c}=\max\{x\in \mathbb{N}\cup \{0\}:\mathbb{P}(X\leq x|H_0)<0.05\}$$ With a calculator we find that $$\hat{c}=3$$ and our critical region is $$X\in\{0,1,2,3\}$$. This means we will reject $$H_0$$ if we observe $$3$$ or less errors in a five hundred page document and we don't reject $$H_0$$ otherwise.