Question

Ocean currents are important in studies of climate change, as well as ecology studies of dispersal of plankton. Drift bottles are used to study ocean currents in the Pacific near Kawal, the Solomon Islands, New Guines, and others x represent the number of days to recovery of a bottlewer release and represent the distance from point of release to point of recovery in km/100. The following data are representative of one wody w drody Currents days 204 195 (a) Verity that x 477, Ly 6.8.27 -62,323,7 - 2034.38, Exy - 1108 and 94568 1471 X 200 Ery 1100 0 4500 (15. level of cance to test the camp > 0.se? Decimale Conclusion Reject the types, there we evidence that Reject the others, there is evidence that Toto je trenutypothes, there is not evidence that Tutto reject the rules, there is an evidence that (c) verty that s, 432,1676 5.4 DO (0) Find the predicted to when tri bottle has been floating for a decimale) 121 Les Find a new for your ren of parte delle Tower km 100 km/100

Answer 1

Solos given that the date ý x days 7 79 بنا 192 204 y kn/100 14.3 19.5 5.8 113 35.9 y 7 14.3 79 19.5 XoY YY 1015.3 5041 204.49 [1540.56241 380.25 179.8 961 33.64 11039.6 8464 (27.69 1208.81 31 5.8 92 11.3 2014 35.9 17223.6 | 41616 = 477 Ey=86.8 Exp=1098.8 Ex=6923 / E72 20 3485 : Ex=477 ; &y=86.8; Exy=11098.8 ; &x2=62323;&y? 2034.8F n. Exy - EX. Ey V[n 8x = (5x)+). {n-sy? (Ey)"] 5x( 11098.8) – 477x8608 155462323 - (477)+][5x2034.8 - (86.932] 850945684 b) Ho :p=0 ta:p>O Sample size n= 5 de = -2 = 5-2 n-2 .94568 t= 1-0.945682 5.03836 critical value Sol Loool d=3 Critical = 4.541

Since t= 5.03836 > critical value t= 4.541, we reject the hull hypothesis 2. There å Enough Evidence t to clain that the population Comelation p is greater than o, at the 0.01 Sigripirence level c) Se= 4.3129; a = 1.3737; and 6= 0. 1676 Now Ss xx &x; ²- { x: ) 2 n 62323-5 (477)? = 16817.2. Soy PCE) 2034.88== *( 86.8)? - 528.032 { "Xey: - ħ (int) (5 %. Ssxy is ज (477) (868) 11098.8- X = 2818.08 slope and y. Intercept Coefficierly are computed using the following formula 2018.08 Ssxy Sixx 16817.2 0.16757 0.16757 A SX 477 -2)954 17.36. - 0176x(95.4) ja e PS-14:36 = 1.3738

a 1.3438 6= 0.1676 we know that the total dom of Squares ilg SS Total = SS yy = SS - 528.03. SS regression = Bi (Ssxy) = 0.1675-7X (2818.08) = 472.2256 we know that SS Since so total TSS regression Ewol Semer Ss Total is regression 528.03.2 – 472.9256 = 55.8063 the mean squared Erol i Hence, MSE SSEL n-2 55.8063 5-2 18.57h4 = NMSE = (18.6021 4.3128 4.3128 Se = 443128 written as d) Based on part () regression Equation can be y= 1.3738 +0.1676*X Now At x=to Y = 1.3738 +0.1676 x 7o y = 13.1058 Y- 13.11 (m/100) 1 e, f g h Not available