Answer:
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
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-\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](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7BsumXY-%5Cfrac%7B%28sumX%29%28sumY%29%7D%7Bn%7D%20%7D%7B%5Csqrt%7B%5BsumX%5E2-%5Cfrac%7B%28sumX%29%5E2%7D%7Bn%7D%20%5D%5BsumY%5E2-%5Cfrac%7B%28sumY%29%5E2%7D%7Bn%7D%20%5D%7D%20%7D%20%3D%20%5Cfrac%7B10991.40-%5Cfrac%7B480%2A86.60%7D%7B5%7D%20%7D%7B%5Csqrt%7B%5B62116-%5Cfrac%7B480%5E2%7D%7B5%7D%20%5D%5B2005.34-%5Cfrac%7B86.60%5E2%7D%7B5%7D%20%5D%7D%20%7D%20%3D%200.94)
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= \frac{r\sqrt{(n-2)} }{\sqrt{(1-r^2)} } ~~t_{n-2}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7Br%5Csqrt%7B%28n-2%29%7D%20%7D%7B%5Csqrt%7B%281-r%5E2%29%7D%20%7D%20~~t_%7Bn-2%7D)
![t_{H_0}= \frac{0.94\sqrt{5-2} }{\sqrt{1-0.94^2} }= 4.77](https://tex.z-dn.net/?f=t_%7BH_0%7D%3D%20%5Cfrac%7B0.94%5Csqrt%7B5-2%7D%20%7D%7B%5Csqrt%7B1-0.94%5E2%7D%20%7D%3D%204.77)
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](https://tex.z-dn.net/?f=t_%7Bn-2%3B1-%5Calpha%20%7D%3D%20t_%7B3%3B0.99%7D%3D%204.541)
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= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }](https://tex.z-dn.net/?f=b%3D%20%5Cfrac%7BsumXY-%5Cfrac%7B%28sumX%29%28sumY%29%7D%7Bn%7D%20%7D%7BsumX%5E2-%5Cfrac%7B%28sumX%29%5E2%7D%7Bn%7D%20%7D)
![b= \frac{10991.40-\frac{480*86.60}{5} }{62116-\frac{(480)^2}{5} }= 0.17](https://tex.z-dn.net/?f=b%3D%20%5Cfrac%7B10991.40-%5Cfrac%7B480%2A86.60%7D%7B5%7D%20%7D%7B62116-%5Cfrac%7B%28480%29%5E2%7D%7B5%7D%20%7D%3D%20%200.17)
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} )]](https://tex.z-dn.net/?f=S_e%5E2%3D%20%5Cfrac%7B1%7D%7Bn-2%7D%5B%28sumY%5E2-%5Cfrac%7B%28sumY%29%5E2%7D%7Bn%7D%20%29-b%5E2%28sumX%5E2-%5Cfrac%7B%28sumX%29%5E2%7D%7Bn%7D%20%29%5D)
![S_e^2= \frac{1}{3}[(2005.34-\frac{(86.60)^2}{5} )-(0.17)^2(62116-\frac{(480)^2}{5} )] = 19.42](https://tex.z-dn.net/?f=S_e%5E2%3D%20%5Cfrac%7B1%7D%7B3%7D%5B%282005.34-%5Cfrac%7B%2886.60%29%5E2%7D%7B5%7D%20%29-%280.17%29%5E2%2862116-%5Cfrac%7B%28480%29%5E2%7D%7B5%7D%20%29%5D%20%3D%20%20%2019.42)
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) ±
* ![\sqrt{S_e^2(\frac{1}{n} + \frac{(x_0-X[bar])^2}{sumX^2-\frac{(sumX)^2}{n} } )}](https://tex.z-dn.net/?f=%5Csqrt%7BS_e%5E2%28%5Cfrac%7B1%7D%7Bn%7D%20%2B%20%5Cfrac%7B%28x_0-X%5Bbar%5D%29%5E2%7D%7BsumX%5E2-%5Cfrac%7B%28sumX%29%5E2%7D%7Bn%7D%20%7D%20%29%7D)
![t_{n-2;1-\alpha /2}= t_{3;0.95}= 2.353](https://tex.z-dn.net/?f=t_%7Bn-2%3B1-%5Calpha%20%2F2%7D%3D%20t_%7B3%3B0.95%7D%3D%202.353)
11.49 ± 2.353 * ![\sqrt{19.42(\frac{1}{5} + \frac{(60-96)^2}{62116-\frac{(480)^2}{5} } )}](https://tex.z-dn.net/?f=%5Csqrt%7B19.42%28%5Cfrac%7B1%7D%7B5%7D%20%2B%20%5Cfrac%7B%2860-96%29%5E2%7D%7B62116-%5Cfrac%7B%28480%29%5E2%7D%7B5%7D%20%7D%20%29%7D)
[5.99; 16.98]km/100
f)
The hypotheses are:
H₀: β ≤ 0
H₁: β > 0
α: 0.01
![t= \frac{b-\beta }{Sb}~~t_{n-2}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7Bb-%5Cbeta%20%7D%7BSb%7D~~t_%7Bn-2%7D)
![t_{H_0}= \frac{0.17-0}{0.03}= 4.80](https://tex.z-dn.net/?f=t_%7BH_0%7D%3D%20%5Cfrac%7B0.17-0%7D%7B0.03%7D%3D%204.80)
This hypothesis test is one-tailed to the right, the critical value is:
![t_{n-2;1-\alpha }= t_{3; 0.90}= 1.638](https://tex.z-dn.net/?f=t_%7Bn-2%3B1-%5Calpha%20%7D%3D%20t_%7B3%3B%200.90%7D%3D%201.638)
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!