# I am measuring airborne bacteria in a new high-risk room; how do I determine sample size?

I am measuring airborne bacteria in a new high-risk room. The bacteria are measured with Petri plates that have area 19,5 cm$$^2$$. I have 30 old measurements results, that give average bacteria counts 2,8 CFU/plate from other High-Risk rooms. The probability of the bacteria count is somewhat Poisson distributed. I want to use the old measurements to determine how many samples(plates) I need to use to show with 95 – 97% confidence that the new room conforms to the old ones. Max bacteria allowed is 50 CFU/plate for a High-Risk room.

When I started looking into your answer I noticed that you used 3.8 instead of 2.8 from the original question. So, I recalculated your answer with 2.8 instead and tried to figure the whole thing out. I wanted to ask you if my recalculations are in the vicinity of correct and if not, where I might have faltered?

Test. Suppose you decide to use n=8 plates. Assuming a Poisson distribution, you currently have λ=2.8 CFU per plate, which would be λT=8(2.8)=22.4 CFU per 8 plates. You could test H0: λT=22.4 against Ha: λT>22.4 at the 5% level of significance. To do so, you would reject H0 in favor of Ha, if the number T of CFU per dozen plates has T≥30. That is, P(T≥30|H0)=1−P(T≤30|H0)=0.043<0.05, assuming the null distribution T|H0∼Pois(λT=8(2.8)=22.4). [Computations in R.]

ppois(30, 2.8*8)
[1] 0.951017
1 - ppois(30, 2.8*8)
[1] 0.04898297


Then the power against the particular alternative λa=8(3.75)=30 is P(X≥31|λa=)=0.99. So, if the true rate per plate is λ=6 CFU, our test has a 99% chance of detecting that much increase.

1 - ppois(30, 6*8)
[1] 0.9963514


In the plot below, the null Poisson distribution (blue bars) is sufficiently different from the alternative distribution, for a satisfactory test using critical value 34 (dotted vertical line).

R code for plot:

t = 20:100;  PDF = dpois(t, 22.4); PDF.a = dpois(t, 48)
plot(t-.1, PDF, type="h", lwd=2,col="blue",
main="Null and Alternative Distn's")
lines(t+.1, PDF.a, type="h", lwd=2, col="brown")
abline(h=0, col="green2")
abline(v = 34, lwd=2, lty="dotted")


Confidence interval. A reasonably accurate 95% confidence interval for a Poisson rate based on T observed counts is of the form T+2 ±1.96T+1−−−−−√. If you observe T=48, then a 95% CI for λT is (22.4, 48) and a 95% CI for the rate λ per plate is (2.8, 6).

38 + qnorm(c(.025,.975))*sqrt(37)
[1] 26.0 - 49.9
(38 + qnorm(c(.025,.975))*sqrt(37))/12
[1] 2.17 - 4.16

• What matters most is the pattern of air circulation in the room. There is literature (in the domain of industrial hygiene) on how to model those patterns and where best to locate measurements. Since you mention "conforms," it also sounds like you are trying to meet some kind of standards or regulatory requirements. It is likely they specify something about how to sample and how many samples to take. – whuber Mar 1 at 17:45

There is not sufficient information in your question for a responsible answer. Indeed, in view of @whuber's Comment you may not have sufficient information without further research.

Therefore the following explores a tentative scenario with speculative Poisson rates, and is not intended as an answer to your question. Perhaps it will provide useful background information about Poisson tests and confidence intervals.

Test. Suppose you decide to use $$n = 12$$ plates. Assuming a Poisson distribution, you currently have $$\lambda = 3.8$$ CFU per plate, which would be $$\lambda_T = 12(3.8) = 45.6$$ CFU per $$12$$ plates.

You could test $$H_0: \lambda_T = 45.6$$ against $$H_a: \lambda_T > 45.6$$ at the 5% level of significance. To do so, you would reject $$H_0$$ in favor of $$H_a,$$ if the the number $$T$$ of CFU per dozen plates has $$T \ge 58.$$ That is, $$P(T \ge 58|H_0) = 1 - P(T \le 57|H_0) = 0.043 < 0.5,$$ assuming the null distribution $$T|H_0 \sim \mathsf{Pois}(\lambda_T = 12(3.8) = 45.6).$$
[Computations in R.]

ppois(57, 3.8*12)
[1] 0.9569788
1 - ppois(57, 3.8*12)
[1] 0.04302121


Then the power against the particular alternative $$\lambda_a = 12(6) = 72$$ is $$P(X \ge 58|\lambda_a = 72) = 0.96.$$ So if the true rate per plate is $$\lambda = 6.0$$ CFU, our test has a 96% chance of detecting that much increase.

1 - ppois(57, 6*12)
[1] 0.9600458


In the plot below, the null Poisson distribution (blue bars) is sufficiently different from the alternative distribution, for a satisfactory test using critical value $$58$$ (dotted vertical line).

R code for plot:

t = 20:100;  PDF = dpois(t, 45.6); PDF.a = dpois(t, 72)
plot(t-.1, PDF, type="h", lwd=2,col="blue",
main="Null and Alternative Distn's")
lines(t+.1, PDF.a, type="h", lwd=2, col="brown")
abline(h=0, col="green2")
abline(v = 57.5, lwd=2, lty="dotted")


Confidence interval. A reasonably accurate 95% confidence interval for a Poisson rate based on $$T$$ observed counts is of the form $$T + 2\pm 1.96\sqrt{T+1}.$$ If you observe $$T = 60,$$ then a 95% CI for $$\lambda_T$$ is $$(46.7,\, 77.3)$$ and a 95% CI for the rate $$\lambda$$ per plate is $$(3.89,\, 6.44).$$

62 + qnorm(c(.025,.975))*sqrt(61)
[1] 46.69219 77.30781
(62 + qnorm(c(.025,.975))*sqrt(61))/12
[1] 3.891016 6.442317

• Thank you for this detailed answer. I will need some time to review it for a deeper understanding, Thank you so much for your quick response :) – Arni Mar 2 at 14:30