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 65 75 86 73 73

y 44 42 48 51 44 51

(a) Verify that Σ x = 439, Σ y = 280, Σ x² = 32,393, Σ y² = 13,142, Σ x y = 20,599, and r < 0.784. (b) Use a 5% level of significance to test the claim that r 7 0.

(c) Verify that Se < 2.6964, a < 16.542, b < 0.4117, and x < 73.167.

(d) Find the predicted percentage yˆ of successful field goals for a player with x = 70% successful free throws.

(e) Find a 90% confidence interval for y when x = 70.

(f) Use a 5% level of significance to test the claim that b 7 0.

(g) Find a 90% confidence interval for b and interpret its meaning.

Answer 1

X	Y	XY	X²	Y²
67	44	2948	4489	1936
65	42	2730	4225	1764
75	48	3600	5625	2304
86	51	4386	7396	2601
73	44	3212	5329	1936
73	51	3723	5329	2601

a)

	X	Y	XY	X²	Y²
total sum	439	280	20599	32393	13142

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.784

b)

Ho:   ρ = 0  
Ha:   ρ > 0  
n=   6  
alpha,α =    0.05  
correlation , r=   0.7835  
t-test statistic = r*√(n-2)/√(1-r²) =        2.522
DF=n-2 =   4  
p-value =    0.0326  
Decison:   p value < α , So, Reject Ho  

c)

SSE=   (Sx*Sy - S²xy)/Sx =    29.0825
      
std error ,Se =    √(SSE/(n-2)) =    2.6964
estimated slope , ß1 = SSxy/SSxx =   112.333   /   272.833   =   0.4117
                  
intercept,   ß0 = y̅-ß1* x̄ =   16.542

here, x̅ =Σx/n =   73.167

d)

Predicted Y at X=   70   is                  
Ŷ =   16.542   +   0.412   *   70   =   45.363

e)

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) ]  

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    1.216              
margin of error,E=t*Std error=t* S(ŷ) =   2.1318   *   1.2161   =   2.5926
                  
Confidence Lower Limit=Ŷ +E =    45.363   -   2.5926   =   42.770
Confidence Upper Limit=Ŷ +E =   45.363   +   2.5926   =   47.955
  

f)

Ho:   ß1=   0          
H1:   ß1 >   0          
n=   6              
alpha=   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    2.696   /√   273   =   0.1632
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.4117   /   0.1632   =   2.522
                  
  
Degree of freedom ,df = n-2=   4              
p-value =    0.0326              
decision :    p-value<α , reject Ho             

g)

confidence interval for slope                  
α=   0.1              
t critical value=   t α/2 =    2.132   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    2.69641   /√   272.83   =   0.163
                  
margin of error ,E= t*std error =    2.132   *   0.163   =   0.348
estimated slope , ß^ =    0.4117              
                  
                  
lower confidence limit = estimated slope - margin of error =   0.4117   -   0.348   =   0.064
upper confidence limit=estimated slope + margin of error =   0.4117   +   0.348   =   0.760