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 percentage of successful field goals a professional basketball player makes in a season. A random sample of n = 6 professional basketball players gave the following information.

x676575867373

y424048514451

(a) Find Σx, Σy, Σx2, Σy2, Σxy, and r. (Round r to three decimal places.)

Σx =

Σy =

Σx2 =

Σy2 =

Σxy =

r =

(b) Use a 5% level of significance to test the claim that ρ > 0. (Round your answers to two decimal places.)

t =

critical t =

Conclusion

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

Reject the null hypothesis, there is insufficient evidence that ρ > 0.    

Fail to reject the null hypothesis, there is insufficient evidence that ρ > 0.

Fail to reject the null hypothesis, there is sufficient evidence that ρ > 0.

(c) Find Se, a, b, and x. (Round your answers to four decimal places.)

Se =

a =

b =

x =

(d) Find the predicted percentage ŷ of successful field goals for a player with x = 85% successful free throws. (Round your answer to two decimal places.)

%

(e) Find a 90% confidence interval for y when x = 85. (Round your answers to one decimal place.)

lower limit    

%

upper limit    

%

(f) Use a 5% level of significance to test the claim that β > 0. (Round your answers to two decimal places.)

t =

critical t =

Conclusion

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

Reject the null hypothesis, there is insufficient evidence that β > 0.    

Fail to reject the null hypothesis, there is insufficient evidence that β > 0.

Fail to reject the null hypothesis, there is sufficient evidence that β > 0.

Answer 1

a)

ΣX =	439.000
ΣY=	276.000
ΣX2 =	32393.000
ΣY2 =	12806.000
ΣXY =	20335.000
r =	0.814

b)

test statistic t =		r*(√(n-2)/(1-r2))=		2.802
t crit =				2.13

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

c)

Se =√(SSE/(n-2))=				3.04680
a=				8.1875
b=				0.5168
X̅=				73.1667

d)

predicted value =

52.12

e)

std error of confidence interval =				s*√(1+1/n+(x0-x̅)2/Sxx)=			3.9490
for 90 % confidence and 4degree of freedom critical t=							2.132
lower limit =		43.70
uppr limit =		60.54

f)

test statistic t =		r*(√(n-2)/(1-r2))=		2.802
t crit =				2.13

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