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	39	48	51	44	51

(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 S_e, 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 = 80% successful free throws. (Round your answer to two decimal places.)
%

(e) Find a 90% confidence interval for y when x = 80. (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)

	X	Y	XY	X²	Y²
total sum	438	277	20365	32264	12899

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.803
      

b)

Ho:   ρ = 0  
Ha:   ρ > 0  
n=   6  
alpha,α =    0.05  
correlation , r=   0.8032  
t-test statistic = r*√(n-2)/√(1-r²) =        2.697
DF=n-2 =   4  

critical t-value =    2.132

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

c)

sample size ,   n =   6          
here, x̅ =Σx/n =   73.000   ,   ȳ = Σy/n =   46.167  
                  
SSxx =    Σx² - (Σx)²/n =   290.00          
SSxy=   Σxy - (Σx*Σy)/n =   144.00          
SSyy =    Σy²-(Σy)²/n =   110.83          
estimated slope , ß1 = SSxy/SSxx =   144.000   /   290.000   =   0.49655
                  
intercept,   ß0 = y̅-ß1* x̄ =   9.91839          
                  
so, regression line is   Ŷ =   9.91839   +   0.49655   *x
                  
SSE=   (Sx*Sy - S²xy)/Sx =    39.3299          
                  
std error ,Se =    √(SSE/(n-2)) =    3.1357         
a=9.9184

b= 0.4966

d)

Predicted Y at X=   80   is                  
Ŷ =   9.918   +   0.497   *   80   =   49.64

e)

X Value=   80                      
Confidence Level=   90%                      
                          
                          
Sample Size , n=   6                      
Degrees of Freedom,df=n-2 =   4                      
critical t Value=tα/2 =   2.132   [excel function: =t.inv.2t(α/2,df) ]                  
                          
X̅ =    73.00                      
Σ(x-x̅)² =Sxx   290.000000                      
Standard Error of the Estimate,Se=   3.14                      
                          

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    1.817                      
margin of error,E=t*Std error=t* S(ŷ) =   2.1318   *   1.8166   =   3.8727      
                          
Confidence Lower Limit=Ŷ +E =    49.643   -   3.8727   =   45.8   
Confidence Upper Limit=Ŷ +E =   49.643   +   3.8727   =   53.5

f)

t stat = estimated slope/std error =ß1 /Se(ß1) =    0.4966   /   0.1841   =   2.697
                  
t-critical value=    2.132

  

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