Mathematics, 18.06.2020 02:57 gooby1324

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.

Answers: 2

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Answer from: leslie4759

Step-by-step 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

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

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

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 $t= \frac{r\sqrt{(n-2)} }{\sqrt{(1-r^2)} } ~~t_{n-2}$

$t_{H_0}= \frac{0.94\sqrt{5-2} }{\sqrt{1-0.94^2} }= 4.77$

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.

The estimated regression equation is

^Y= a + bX

a= Y[bar] -bX[bar]

$b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }$

$b= \frac{10991.40-\frac{480*86.60}{5} }{62116-\frac{(480)^2}{5} }= 0.17$

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

$S_e^2= \frac{1}{n-2}[(sumY^2-\frac{(sumY)^2}{n} )-b^2(sumX^2-\frac{(sumX)^2}{n} )]$

$S_e^2= \frac{1}{3}[(2005.34-\frac{(86.60)^2}{5} )-(0.17)^2(62116-\frac{(480)^2}{5} )] = 19.42$

Se= 4.406= 4.41

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

90% CI for ^Y/X₀= 60

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

$t_{n-2;1-\alpha /2}= t_{3;0.95}= 2.353$

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

[5.99; 16.98]km/100

The hypotheses are:

H₀: β ≤ 0

H₁: β > 0

α: 0.01

$t= \frac{b-\beta }{Sb}~~t_{n-2}$

$t_{H_0}= \frac{0.17-0}{0.03}= 4.80$

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!

Answer from: Quest

Teri ma ki chhut‌‌‌‌‌‌

Answer from: Quest

Ineed with this same question!
In a circle with a radius of 36.9 m, an arc is intercepted by a central angle of 8π5 radians.<

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