Question

Let x be a random variable that represents the percentage of successful free throws a professional basketball player makes in a season. Let y be a random variable that represents the basketball player makes in a season. A random sample of n = 6 professional basketball players gave the following information. 75 86 73 73 44 3948 51 (a) Verify that Ex - 438, 2y = 277, 2x2 = 32264, y = 12899, Exy = 20365, and r * 0.803. 67 54 44 51 Xy 277 32 32264 3y2 12800 Exy 20365 800 (b) Use a 5% level of significance to test the claim that p > 0. (Round your answers to two decimal places) t critical Conclusion Reject the null hypothesis, there is sufficient evidence that p > 0. Reject the null hypothesis, there is insufficient evidence that p > 0. Fall to reject the null hypothesis, there is insufficient evidence that p > 0. Fail to reject the null hypothesis, there is sufficient evidence that p > 0. (c) Verify that S - 3.1357, 3 * 9.918, 00.4966, and x = 73.000. S, 3.1357 a 9918 b 4000 13 (d) Find the predicted percentage y of successful field goals for a player with x 8% successful free throws. (Round your answer to two decimal places) (e) Find 90% confidence interval for y when x + 1). (Round your answers to one decimal place.) lower limit upper limit (1) Use a level of significance to test the claim that > 0. (Round your answers to two decimal places)

Answer 1

S.No X XY Ε 1 2 3 4 67 64 75 86 73 73 438 Υ 44 39 48 51 X;? 4489.0000 4096.0000 5625.0000 7396.0000 5329.0000 5329.0000 32264.0000 ΣΧ Y, 1936.0000 1521.0000 2304.0000 2601.0000 1936.0000 2601.0000 12899.0000 ΣΥ 2948.00 2496.00 3600.00 4386.00 3212.00 3723.00 20365.0000 5 6 Total 44 51 277 Mean 73.000 46.1667 ΣΧΥ

	n=	6.0000
	X̅=ΣX/n	73.00
	Y̅=ΣY/n	46.17
	sx=(√(Σx2-(Σx)2/n)/(n-1))=	7.62
	sy=(√(Σy2-(Σy)2/n)/(n-1))=	4.7081
	Cov=sxy=(ΣXY-(ΣXΣY)/n)/(n-1)=	28.8000
	r=Cov/(Sx*Sy)=	0.8032
	slope= β̂1 =r*Sy/Sx=	0.4966
	intercept= β̂0 ='y̅-β1x̅=	9.9184

b)

t=r*(√(n-2)/(1-r2))=	2.70
critical t =	2.13

e)

standard error of PI=s*√(1+1/n+(x0-x̅)2/Sxx)=		3.8551
for 90 % CI value of t =	2.132
margin of error E=t*std error   =		8.218
lower confidence bound=xo-E =		42.9
Upper confidence bound=xo+E=		59.4

f)

t=r*(√(n-2)/(1-r2))=	2.70
critical t =	2.13

g)

std error of slope =se(β1) =s/√Sxx=		0.1841
for 90 % CI value of t =	2.132
margin of error E=t*std error   =		0.393
lower confidence bound=xo-E =		0.104
Upper confidence bound=xo+E=		0.889