Poisson distribution: sample size vs rate

Question

I was looking at some exercises on Poisson distribution and was confused by the answers. Here are the questions:

• Sam's Party Supplies has had trouble with balloons. Experience has shown that the ones he buys from his present supplier, Rooty Balloons, are 2% defective. Al, representing Tooty Balloons, promises him 1% defectives if he will switch his order to them. Al gives Sam a sample package of 10 Tooty balloons. Sam blows them all up, and one of them turns out to be no good. "Looks to me like we have 10% defectives here," he tells Al. What does Al say in response?

• Sam tries nine more packages of ten balloons each, and all of them are free of defects, except one package which has 1 bad balloon in it. Does Sam sign the Tooty contract?

The questions are from this page. (Click the blue box for answers).

What Al and Sam say both make sense, so why is Al's answer wrong? Is it because the r=0.2 simply comes from larger samples? Would that mean if we try 100 packages and we observe a different r, it would be even more accurate?

Answer 1

Al is claiming that the table means you should expect $9$ good balloons and one bad balloon in a single package, whereas the table actually means that in a sample of $10$ packages, you should expect $9$ to have no bad balloons, and one to have a single bad balloon. Al's interpretation is manifestly ridiculous, as it is consistent with a $10\%$ failure rate, not a $1\%$ rate.

The actual experience, $8$ good packages and $2$ with a single defect each, conforms exactly to the $r=.2$ column, that is a rate $.2$ defects per $10-$item sample.

Answer 2

No wonder you are perplexed. In my view the question in the link is not very well written and the answer is misleading. So I will put the 'answer key' aside and try to say something correct.

The evidence on hand, after presumably independently sampling and inflating 100 Tooty balloons, is that we found 2 defectives in 100.

(1) An 'Agresti-Coull' 95% confidence interval for the true proportion of defectives is $(0.0015, 0.0754),$ which includes both $.01$ and $.02.$ So, with such a small sample size, there is not sufficient evidence to distinguish between 1% and 2% failure rates.

(2) Testing $H_0: p \le .01$ vs. $H_0: p > .01,$ the P-value is $0.264.$ There no evidence that 2 defectives in 100 is significantly more than 1%.

sum(dbinom(2:100, 100, .01))
[1] 0.264238

I see no reason to bring the Poisson distribution into this discussion.