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.

x	67	64	75	86	73	73
y	44	41	48	51	44	51

(A) Find a 90% confidence interval for y when x = 80. (Round your answers to one decimal place.)

lower limit	%
upper limit	%

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

t =
critical t =

C. Conclusion

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

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

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

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

Answer 1

	x	y	(x-xbar)^2	(y-ybar)^2	(x-xbar)*(y-ybar)
	67	44	36	6.25	15
	64	41	81	30.25	49.5
	75	48	4	2.25	3
	86	51	169	20.25	58.5
	73	44	0	6.25	0
	73	51	0	20.25	0
sum	438	279	290	85.5	126
mean	73	46.5	sxx	syy	sxy

slope=b1=sxy/sxx = 0.4345
intercept=b0=ybar-(slope*xbar) = 14.7828
SST = SYY = 85.5
SSR = sxy^2/sxx = 54.7448
SSE = syy-sxy^2/sxx = 30.7552
r^2 = SSR/SST = 0.6403
error variance s^2 = SSE/(n-2) = 7.6888
S^2b1 = s^2/sxx = 0.0265
standard error b1=se(b1) = sqrt(s^2b1) = 0.1628
test statistics = b1/se(b1) = 2.6683

so, regression line is   Ŷ =   14.7828   +   0.4345   *x

Predicted Y at X= 80   is                  
Ŷ =14.7828+0.4345*80 =   49.54

a)

90% confidence interval for y when x = 80.

std error ,Se =sqrt(SSE/(n-2) )= sqrt(7.6888)=2.7729

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) = 2.7729*sqrt((1/6)+(80-73)^2/290) =1.6064   
margin of error,E=t*Std error=t* S(ŷ) =   2.1318   *   1.6064 =   3.4246
                  
Confidence Lower Limit=Ŷ +E =    49.54   -   3.4246 =   46.1
Confidence Upper Limit=Ŷ +E =   49.54 +   3.4246 =   53.0

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b)

slope hypothesis test               tail=   1
Ho:   ß1=   0          
H1:   ß1 >   0          
n=   6              
alpha=   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    2.7729   /√   290   =   0.0265
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.4345/0.0265   =   2.67

t stat = 2.67
                  
t-critical value=    2.13   [Excel function: =T.INV(α,df) ]          
Degree of freedom ,df = n-2=   4              

decision : TEST STAT> CRITICAL , reject Ho         

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C. Conclusion :Correct option is A

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

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