Question

E1 Following the Fukushima disaster, an agency is measuring the concentration of the ra- dioactive isotope Cesium-137 in two regions in the Pacific. Region A is 100 miles from Fukushima and Region B is 200 miles away from Fukushima. Let ja and MB be the true concentration levels in Regions A and B, respectively. The agency is interested in the dif- ference between the concentration levels. It has measurements A1,..., An from Region A and measurements B1, ..., Bn from Region B (the n is the same). For each region, the measurements are measured in “microSieverts.” Assume that A1,..., An 19 N(ua, 400), B1, ..., Bn Hd N(MB, 400) and that the A's and B's are independent. Suppose that we want the probability that A - B is within 5 microSieverts of the difference MA - MB to be .98. What should n be?

Answer 1

Consider the mean of sample from Region A is să and the mean of sumple from Region B is to of A,, A2,..., An no Nl MA , 400) S then X = mean of sample no l MA, 400) also B, B2, ..., Bn NNC MB, 400) = X = mean of sample from B ~ N(Mg, you) Consider the variable y = x - X2 - then e[y]= € {x, J- E [X2] = up - MB and V (Y) = V(F) + vet) - 2 Cor CX7,82) But I, and to are independent. So v (4): v(X) + VC) = 40 + 400 Now 9 5 » p{ -BI-IMA-Moll 25} = 0.98 tỷ 15 - 4, 4sj = 0. P { ! You my paskets)} = 0.18 P { izi < 5n] -0.48 where Z= Yally ~ N(0) Standard Nasmal olm and a goo/n > P { 1z) < Sumica ļ = 0.98 3 px izle 5m | >0.98 Using Standard normal table p { iz) <2:33 } -0.98 - 800 in 800

Therefore, using 0 and ① ; - n= 13,18 ~ 14 sample size is 14. Hence

amble hom Regon A (ongide the mean is mean fsumpre fAom Region B the ond A,, A ,An N AD then of sampli mean alio B,62,,Bn NLAHo) X mean of sampe hom B Consi der the Vaniabl then E J-E L and 2 Cov C But and ase ndupendent v (i ) v Los Nouw 0.9P P -y D.8 where Standand No mal and 0.98 UsingStandrd momal table

These fore, using 5n 2.33 Hence Sam