Question #110670
The average rate of emission radioactive particles from a source was measured over a long period, and found to be 10 particles per unit time. After an experimental treatment had been applied to the source, a further sample was examined and emitted 17 particles in unit time. Test at 5% the null hypothesis that the rate of emissions is unchanged.
1
Expert's answer
2020-04-27T18:55:39-0400

The number of emitted particles has Poisson distribution.

P{ξ=k}=λkk!eλ,k=0,1,P\{\xi=k\}=\frac{\lambda^k}{k!}e^{-\lambda}, k=0,1,\ldots

We have λ1=10,λ2=17.\lambda_1=10, \lambda_2=17.

We will take samples of sizes n1=20n_1=20 and n2=20n_2=20 from the first and the second populations.

P{ξ=20}=102020!e100.0019.P{ξ=20}=172020!e170.0692.P\{\xi=20\}=\frac{10^{20}}{20!}e^{-10}\approx 0.0019.\\ P\{\xi=20\}=\frac{17^{20}}{20!}e^{-17}\approx 0.0692.

We get relative frequencies:

ω1=0.0019ω2=0.0692\omega_1=0.0019\\ \omega_2=0.0692

H0:p1=p2;H1:p1p2where pi is probability that event A (the particle emits)will happen in the i-th population.H_0: p_1=p_2; H_1: p_1\neq p_2\\ \text{where } p_i \text{ is probability that event A (the particle emits)}\\ \text{will happen in the i-th population}.

We will use the following random variable:

u=ω1ω2(ω1+ω2)(1(ω1+ω2))(1n1+1n2)uobs0.8281Φ(ucr)=1α2=0.475ucr=1.96(,1.96)(1.96,) — critical region.uobs does not fall into the critical region. So we accept H0.u=\frac{\omega_1-\omega_2}{\sqrt{(\omega_1+\omega_2)(1-(\omega_1+\omega_2))(\frac{1}{n_1}+\frac{1}{n_2}})}\\ u_{obs}\approx -0.8281\\ \Phi(u_{cr})=\frac{1-\alpha}{2}=0.475\\ u_{cr}=1.96\\ (-\infty,-1.96)\cup (1.96,\infty)\text{ --- critical region}.\\ u_{obs}\text{ does not fall into the critical region. So we accept } H_0.

The rate of emission did not change after an experimental treatment.


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