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   40   48   51   44   51

(a) Find Σx, Σy, Σx2, Σy2, Σxy, and r. (Round r to three decimal places.)
Σx =   438
Σy =   278
Σx2 =   32264
Σy2 =   12978
Σxy =   20429
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
A- Reject the null hypothesis, there is sufficient evidence that ρ > 0.
B-Reject the null hypothesis, there is insufficient evidence that ρ > 0.
C-Fail to reject the null hypothesis, there is insufficient evidence that ρ > 0.
D-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 = 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
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

a)

X	Y	XY	X²	Y²
67	44	2948	4489	1936
64	40	2560	4096	1600
75	48	3600	5625	2304
86	51	4386	7396	2601
73	44	3212	5329	1936
73	51	3723	5329	2601

	X	Y	XY	X²	Y²
total sum	438.000	278.000	20429.00	32264.000	12978
mean	73.0000	46.3333

sample size ,   n =   6          
here, x̅ =Σx/n =   73.0000   ,   ȳ = Σy/n =   46.33333333  
                  
SSxx =    Σx² - (Σx)²/n =   290.000          
SSxy=   Σxy - (Σx*Σy)/n =   135.000          
SSyy =    Σy²-(Σy)²/n =   97.333          
estimated slope , ß1 = SSxy/SSxx =   135.000   /   290.000   =   0.4655
                  
intercept,   ß0 = y̅-ß1* x̄ =   12.3506          
                  
so, regression line is   Ŷ =   12.35   +   0.47   *x
                  
SSE=   (Sx*Sy - S²xy)/Sx =    34.4885          
                  
std error ,Se =    √(SSE/(n-2)) =    2.9363          
                  
correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.8035        

b)

Ho:   ρ = 0       tail=   1  
Ha:   ρ > 0              
n=   6              
alpha,α =    0.05              
correlation , r=   0.8035              
t-test statistic = r*√(n-2)/√(1-r²) =        2.700     

critical t-value =    2.1318        
DF=n-2 =   4          

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

----------------

std error ,Se =    √(SSE/(n-2)) =    2.9363

estimated slope , ß1 = SSxy/SSxx =   135.000   /   290.000   =   0.4655
                  
intercept,   ß0 = y̅-ß1* x̄ =   12.3506          

so, regression line is   Ŷ =   12.35   +   0.47   *x

---------------------------

Predicted Y at X=   80   is                  
Ŷ =   12.351   +   0.466   *   80   =   49.59

----------------------------

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    1.701              
margin of error,E=t*Std error=t* S(ŷ) =   2.1318   *   1.7011   =   3.6266
                  
Confidence Lower Limit=Ŷ +E =    49.592   -   3.6266   =   45.965
Confidence Upper Limit=Ŷ +E =   49.592   +   3.6266   =   53.219

-------------------------------

Ho:   ß1=   0          
H1:   ß1 >   0          
n=   6              
alpha=   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    2.936   /√   290   =   0.1724
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.4655   /   0.1724   =   2.700
                  
t-critical value=    2.132   [Excel function: =T.INV(α,df) ]          
Degree of freedom ,df = n-2=   4              
p-value =    0.0271              
  
A-Reject the null hypothesis, there is sufficient evidence that β > 0.

THANKS

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