Question

Let x be a random variable that represents the batting average of a professional baseball player. Let y be a random variable that represents the percentage of strikeouts of a professional baseball player. A random sample of n = 6 professional baseball players gave the following information. x 0.318 0.290 0.340 0.248 0.367 0.269 y 3.1 7.6 4. 0 8 .6 3.1 11.1 (a) Find Ex, 2y, Ex?, y2, Exy, and r. (Round r to three decimal places.) (b) Use a 5% level of significance to test the claim that p +0. (Round your answers to two decimal places.) critical t = Conclusion Reject the null hypothesis, there is sufficient evidence that p differs from 0. Reject the null hypothesis, there is insufficient evidence that p differs from 0. Fall to reject the null hypothesis, there is insufficient evidence that p differs from 0. Fail to reject the null hypothesis, there is sufficient evidence that p differs from 0. (c) Find Sera, and b. (Round your answers to four decimal places.) Se- (d) Find the predicted percentage y of strikeouts for a player with an x = 0.3 batting average. (Round your answer to two decimal places.) (e) Find an 80% confidence interval for y when x = 0.3. (Round your answers to two decimal places.) lower limit upper limit (f) Use a 5% level of significance to test the claim that = 0. (Round your answers to two decimal places.) critical t =

Answer 1

X	Y	XY	X²	Y²
0.318	3.1	0.9858	0.101124	9.61
0.29	7.6	2.204	0.0841	57.76
0.34	4	1.36	0.1156	16
0.248	8.6	2.1328	0.061504	73.96
0.367	3.1	1.1377	0.134689	9.61
0.269	11.1	2.9859	0.072361	123.21

sample size ,   n =   6          
here, x̅ =Σx/n =   0.3053   ,   ȳ = Σy/n =   6.25  
                  
SSxx =    Σx² - (Σx)²/n =   0.01          
SSxy=   Σxy - (Σx*Σy)/n =   -0.64          
SSyy =    Σy²-(Σy)²/n =   55.78          

.a)

	X	Y	XY	X²	Y²
total sum	1.8320000	37.5000000	10.8062000	0.5693780	290.1500000

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   -0.862

b)

correlation hypothesis test      
Ho:   ρ = 0  
Ha:   ρ ╪ 0  
n=   6  
alpha,α =    0.05  
correlation , r=   -0.8617  
t-test statistic = r*√(n-2)/√(1-r²) =        -3.397

DF=n-2 =   4  
p-value =    0.0274  
Decison:   p value < α , So, Reject Ho  
critical t-value =    2.7764  

t = -3.40

critical t = 2.78

c)

std error ,Se =    √(SSE/(n-2)) =    1.8946
a=estimated slope , ß1 = SSxy/SSxx =   -0.644   /   0.010   =   -64.3328
                  
b = intercept,   ß0 = y̅-ß1* x̄ =   25.8930

d)

Predicted Y at X=   0.3   is                  
Ŷ =   25.893   +   -64.333   *   0.3   =   6.59%   

e)

Confidence Level=   80%  
      
      
Sample Size , n=   6  
Degrees of Freedom,df=n-2 =   4  
critical t Value=tα/2 =   1.533   [excel function: =t.inv.2t(α/2,df) ]
      
X̅ =    0.31  
Σ(x-x̅)² =Sxx   0.010007  
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    0.780              
margin of error,E=t*Std error=t* S(ŷ) =   1.5332   *   0.7800   =   1.1959
                  
Confidence Lower Limit=Ŷ +E =    6.593   -   1.1959   =   5.40
Confidence Upper Limit=Ŷ +E =   6.593   +   1.1959   =   7.79

f)

t = -3.40

critical t = 2.78