Question

Env. 6.6 The Department of Natural Resources (DNR) received a complaint from recreational fish- ermen that a community was releasing sewage into the river where they fished. These types of releases lower the level of dissolved oxygen in the river and hence cause damage to the fish resid ing in the river. An inspector from the DNR designs a study to investigate the fishermen's claim. Fifteen water samples are selected at locations on the river upstream from the community and fifteen samples are selected downstream from the community. The dissolved oxygen readings in parts per million (ppm) are given in the following table Upstream Downstrea 3.2 3.4 3.7 3.9 3.63.8 3.9 3.6 .2 4.8 5.1 5.0 494.85.0 4.7 4.7 5.0 4.65.2 5.0 4.94.7 3.3 4.53.7 39 3.8 3.7

Answer 1

Solution Given let x = Upstream X2 - Downstaeum (X-X) 2 (X2 72 72 3.2 5.2 4.8 5. 0.086 0.011 3.4 3.7 0.037 3.9 0.009 4.9 4.8 3.6 3.8 3.9 3.6 tal 3.3 I.5 3.7 3.9 3.8 3.7 0.000 0.01) 0.009 0.043 0.093 0.00g 0.094 0.086 0.009 0.292 0-116 0.002 0.026 0-020 0.004 0-026 0-020 0-130 0.194 0.578 0.002 0.026 0.004 0.002 1.436 14.7 0.000 0.043 Total 73.6 1 56.) 6.489 Jet Ex= 736 X = 4.907 EX-X2 = 0.489

ŽX2=561 X = 3.74 {(x2-x2)2 - 1.436 Sny El X-X2 E(X2=X2,2 n- = V 0.489 0.489 = V 1.436 14 14 Si=10.0349 0.0349 0.1025 5,= 0.1868 52=0.3202 Construct 951 confidence interval for the buc mean difference (4-12). C:s=(1-12 + Sptomtoztat hz let Sp=√(1-1) 53²+ M2-1) 52² nith₂-2 14(0.1868)2+1460.3202)2 - 28 0.0687 = 0.0687

sp= 0.2621 trezni+N2-2) = t10.025.28) = 2:048 by table CI=(4.907-3.74 € (0.262))(2.098) A SIS = ( 1.167 + (0.2621) (2.048) (0.3651)) = (1.167 + 0.1959) = (.).167 -0.19592 1.167 +0.1959) = (0.970) , 1.3629) = (0.97 11:36 ) confidence internal is the bue meon difference between the dissolved oxygen Opsbeam and downstream Clupsbeam-eldownsbeam) is (0.97_136) je (0.97 € 41-42 (136)