Question

What is the optimal time for a scuba diver to be on the bottom of the ocean? That depends on the depth of the dive. The U.S. Navy has done a lot of research on this topic The Navy defines the optimal time to be the time at each for the best balance between length of work period and decompression time after surfacing. Let x = depth of dive in meters, and let y = optimal time in hours. A sandom sample of divers gave the following data 333513 20.5 227 1.78 1.03 0.75 () Find 2x, 2.2. Er? Dxy, and r. (Round r to three decimal places) 2012 2-2713 Exy 14.532 0.961 C) a level of significance to test the chain that po. (Round your answers to two decimal places) - 3143 Conclusion Reynect the hypothesis. There is not evidence that Folltore the full hypothesis. There is what even that Rect the others. There were that Fol toe te pas. There is under evidence that d) Find the predicted optimal maher for a dive depth of 3 meters (Round your answer to the decimal (w) Pindan concentral for when x 33 meters (Round you were to two decimal) tower per im

Answer 1

S.No X XY 1 2 3 4 5 6 7. Total 13.1 23.3 29.2 38.3 51.3 20.5 22.7 198.4 2.68 2.18 1.78 1.03 0.75 2.38 2.2 13 x, 171.6100 542.8900 852.6400 1466.8900 2631.6900 420.2500 515.2900 6601.2600 ΣΧ Y, 7.1824 4.7524 3.1684 1.0609 0.5625 5.6644 4.8400 27.2310 ΣΥ 35.11 50.79 51.98 39.45 38.48 48.79 49.94 314.5320 Mean 28.343 1.8571 ΣΧΥ

	n=	7.0000
	X̅=ΣX/n	28.34
	Y̅=ΣY/n	1.86
	sx=(√(Σx2-(Σx)2/n)/(n-1))=	12.77
	sy=(√(Σy2-(Σy)2/n)/(n-1))=	0.7174
	Cov=sxy=(ΣXY-(ΣXΣY)/n)/(n-1)=	-8.9875
	r=Cov/(Sx*Sy)=	-0.9812
	slope= β̂1 =r*Sy/Sx=	-0.0551
	intercept= β̂0 ='y̅-β1x̅=	3.4199

a)

ΣX =	198.400
ΣY=	13.000
ΣX2 =	6601.260
ΣY2 =	27.231
ΣXY =	314.532
r =	-0.981

b)

t=r*(√(n-2)/(1-r2))=	-11.37
critical t =	-3.36

c)

Se =√(SSE/(n-2))=	0.15161
a=	3.41986
b=	-0.05514

d)

predicted val=3.42+33*-0.055=

1.60

e)

standard error of PI=s*√(1+1/n+(x0-x̅)2/Sxx)=		0.1636
for 80 % CI value of t =	1.476
margin of error E=t*std error   =		0.242
lower confidence bound=xo-E =		1.36
Upper confidence bound=xo+E=		1.84

f_)

t=r*(√(n-2)/(1-r2))=	-11.37
critical t =	-3.36

g)

std error of slope =se(β1) =s/√Sxx=		0.0048
for 90 % CI value of t =	2.015
margin of error E=t*std error   =		0.010
lower confidence bound=xo-E =		-0.065
Upper confidence bound=xo+E=		-0.045