Please help! Ocean currents are important in studies of climate change, as well
ID: 3309219 • Letter: P
Question
Please help!
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 Islands, New Guinea, and other islands. Let x represent the number of days to recovery of a drift bottle after release and y represent the distance from point of The following data are representative of one study using drift bottles to study ocean currents. My Notes Ask Your Teach olomon Pacific near Hawail, the S release to point of recovery in km/100. x days 72 76 32 9 208 m/100 14.6 19.3 5.1 11.6 35.2 thatEx=479, = 85.8, Ex2= 63,529, 2= 1985.26, 2xy= 11058.4, andra 0.94371. (a) Verify 479 x2 63529 2 1985.26 Exy 11058.4 r.94371 (b) Use a 1% level of significance to test the claim 0, (Use 2 decimal places.) t 4.94 critical t 4.50 Conclusion o Reject the null hypothesis, there is sufficient evidence that > 0 Reject the null hypothesis, there is insufficient evidence that > 0. Fail to reject the null hypothesis, there is insufficient evidence that > 0. Fail to reject the null hypothesis, there is sufficient evidence that > 0 (c) Verify that Se_ 4.3251, a~ 1.7439, and b~ 0.1609. Se 4.3251 a 1.7439 b .1609 (d) Find the predicted distance (km/100) when a drift bottle has been floating for 70 days. (Use 2 decimal places.) km/100Explanation / Answer
(a) Part a is correct.
(b) here dF = 5 -2 = 3
alpha = 0.01 (one tailed test)
tcritical = t0.01,3 = 4.5407
here t > tcritical
(c) Here the linear line equation
y = a + bx
y^ = 1.7439 + 0.1609 x
(d) x = 70 days
y^ = 1.7439 + 0.1609 * 70 = 13.01 km/100
(e) 90% confidenceinterval for the predicted value = y^ +- tcr se0
where tcr = t0.10,3 = 2.3534
se0 = se * sqrt [1/n + (x0 - x)2 / SSXX]
x = 479/5 = 95.8
SSXX = x2 -(x)2/n = 63529 - 4792/5 = 17640.8
90% confidenceinterval for the predicted value = y^ +- tcr se0
= 13.0068 +- 2.3534 * 4.3251 * sqrt [1/5 + (70 - 95.8)2 /17640.8]
= 13.0068 +- 2.3534 * 4.3251 * 0.4876
= (8.04, 17.97)
(f) Here for B > 0
t = b^1 / seb1
here seb1 = se/ sqrt(SSXX) = 4.3251/sqrt(17640.8) = 0.032564
so t = 0.1609/0.032564 = 4.942
critical t = 4.541
Option A is correct. There is sufficient evidence B > 0 and that null hypothesis is rejected.
(g) here 95% condidence interval = b^1 +- t0.05,3 seb1 = 0.1609 +- 3.1824 * 0.03256
= (0.0572, 0.2645)
= ( 0.06, 0.26) in two decimal places
Option D is correct. For every day of drift, the distance drift increases by an amount that falls under confidence interval.
(h) so here x = 50 days
so y^(50) = 1.7439 + 0.1609 * 50 = 9.79 KM /100