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 Hawaii, the Solomon 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 release to point of recovery in km/100. The following data are representative of one study using drift bottles to study ocean currents.

x days 75 76 35 91 203
y km/100 14.7 19.6 5.4 11.7 35.2
(a) Verify that ?x = 480, ?y = 86.6, ?x2 = 62,116, ?y2 = 2005.34, ?xy = 10991.4, and r ? 0.94059. ?x ?y ?x2 ?y2 ?xy r
(b) Use a 1% level of significance to test the claim ? > 0. (Use 2 decimal places.)
t _____
critical t ______
Conclusion
A) Reject the null hypothesis, there is sufficient evidence that ? > 0.
B) Reject the null hypothesis, there is insufficient evidence that ? > 0.
C) Fail to reject the null hypothesis, there is insufficient evidence that ? > 0.
D) Fail to reject the null hypothesis, there is sufficient evidence that ? > 0.
(c) Verify that Se ? 4.4072, a ? 1.2893, and b ? 0.1670.
Se____
a_____
b____
(d) Find the predicted distance (km/100) when a drift bottle has been floating for 60 days. (Use 2 decimal places.)
______km/100
(e) Find a 90% confidence interval for your prediction of part (d). (Use 1 decimal place.)
lower limit km/100 ______
upper limit km/100 ______
(f) Use a 1% level of significance to test the claim that ? > 0. (Use 2 decimal places.)
t____
critical t_____
Conclusion
A) Reject the null hypothesis, there is sufficient evidence that ? > 0.
B) Reject the null hypothesis, there is insufficient evidence that ? > 0.
C) Fail to reject the null hypothesis, there is insufficient evidence that ? > 0.
D) Fail to reject the null hypothesis, there is sufficient evidence that ? > 0.

Answer 1

Explanation:

Hello!

Given the variables

X: number of days it takes to recover a drift bottle after release.

Y: distance from point of release to point of recovery (km/100)

X: 75; 76; 35; 91; 203

Y: 14.7; 19.6; 5.4; 11.7; 35.2

a)

n=5

∑X= 75 + 76 + 35 + 91 + 203= 480

∑X²= 75² + 76² + 35² + 91² + 203²= 62116

∑Y= 14.7 + 19.6 + 5.4 + 11.7 + 35.2=86.60

∑Y²= 14.7² + 19.6² + 5.4² + 11.7² + 35.2²= 2005.34

∑XY= (75*14.7) + (76*19.6) + (35*5.4) + (91*11.7) + (203*35.2)= 10991.40

`r= \frac{sumXY-((sumX)(sumY))/(n) }{\sqrt{[sumX^2-((sumX)^2)/(n) ][sumY^2-((sumY)^2)/(n) ]} } = \frac{10991.40-(480*86.60)/(5) }{\sqrt{[62116-(480^2)/(5) ][2005.34-(86.60^2)/(5) ]} } = 0.94`

b)

The claim is that there is a positive association between these two variables, symbolically ρ > 0

H₀: ρ ≤ 0

H₁: ρ > 0

α: 0.01

The statistic is
`t= (r√((n-2)) )/(√((1-r^2)) ) ~~t_(n-2)`

`t_(H_0)= (0.94√(5-2) )/(√(1-0.94^2) )= 4.77`

This test is one-tailed to the right, meaning that you'll reject the null hypothesis to high values of t. There is only one critical value that defined the rejection region and it is:

`t_(n-2;1-\alpha )= t_(3;0.99)= 4.541`

The rejection region is t ≥ 4.541

The calculated value is greater than the critical value, so the decision is to reject the null hypothesis.

Conclusion

A) Reject the null hypothesis, there is sufficient evidence that ρ > 0.

c)

The estimated regression equation is

^Y= a + bX

a= Y[bar] -bX[bar]

`b= (sumXY-((sumX)(sumY))/(n) )/(sumX^2-((sumX)^2)/(n) )`

`b= (10991.40-(480*86.60)/(5) )/(62116-((480)^2)/(5) )= 0.17`

Y[bar]= ∑Y/n= 86.60/5= 17.32

X[bar]= ∑X/n= 480/5= 96

a= 17.32 - 0.17*96= 1.29

The estimated regression line is ^Y= 1.29 + 0.17X

And the estimated variance Se²

`S_e^2= (1)/(n-2)[(sumY^2-((sumY)^2)/(n) )-b^2(sumX^2-((sumX)^2)/(n) )]`

`S_e^2= (1)/(3)[(2005.34-((86.60)^2)/(5) )-(0.17)^2(62116-((480)^2)/(5) )] = 19.42`

Se= 4.406= 4.41

d)

You have to find the value of ^Y/X=60 days, simply replace the value of X on the estimated equation:

^Y= 1.29 + 0.17*60= 11.49

The distance the bottle will travel if it drifts for 60 days will be 11.49 km/100

e)

90% CI for ^Y/X₀= 60

(a+bX₀) ±
`t_(n-2;1-\alpha /2)` *
`\sqrt{S_e^2((1)/(n) + ((x_0-X[bar])^2)/(sumX^2-((sumX)^2)/(n) ) )}`

`t_(n-2;1-\alpha /2)= t_(3;0.95)= 2.353`

11.49 ± 2.353 *
`\sqrt{19.42((1)/(5) + ((60-96)^2)/(62116-((480)^2)/(5) ) )}`

[5.99; 16.98]km/100

f)

The hypotheses are:

H₀: β ≤ 0

H₁: β > 0

α: 0.01

`t= (b-\beta )/(Sb)~~t_(n-2)`

`t_(H_0)= (0.17-0)/(0.03)= 4.80`

This hypothesis test is one-tailed to the right, the critical value is:

`t_(n-2;1-\alpha )= t_(3; 0.90)= 1.638`

The rejection region is t ≥ 1.638, since the calculated statistic is greater than the critical value, the decision is to reject the null hypothesis. Then there is a positive linear regression between the distance from the release point and the number of days a bottle drifted.

Conclusion:

A) Reject the null hypothesis, there is sufficient evidence that β > 0.

I hope this helps!