Question

3. The table below lists pitches in a baseball game for two pitchers and the breakdown of “strikes” and “balls” (good pitches and bad pitches). The summary statistics are also listed.

Pitcher	Strikes	Balls	Total
Bumgarner	76	32	108
Cueto	76	41	117

  Bumgarner                  Cueto

            Sample Mean (pitches per inning) 12.0                              13.0

            Sample Variance (pitches per inning) 3.5 8

            Sample Size (# of innings) 9 9

             (an “inning” is one of the rounds in a game)

(5)        a. At the 99% level, test the alternate hypothesis that Cueto throws more pitches per inning than Bumgarner for the entire season (population of pitches):

(5)        b. Set up a 90% confidence interval for the true difference in the proportion of Balls thrown by each pitcher for the entire season:

(5)        c. Perform a hypothesis test to see if the proportion of throwing strikes for Bumgarner and Cueto is not equal for the entire season. Set the Type I error equal to 10%.

Answer 1

A)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
                          
Level of Significance ,    α =    0.01                  
                          
Sample #1   ---->  BumgarneR                
mean of sample 1,    x̅1=   12.00                  
standard deviation of sample 1,   s1 =    3.50                  
size of sample 1,    n1=   9                  
                          
Sample #2   ---->   Cueto                
mean of sample 2,    x̅2=   13.00                  
standard deviation of sample 2,   s2 =    8.00                  
size of sample 2,    n2=   9                  
                          
difference in sample means =    x̅1-x̅2 =    12.0000   -   13.0   =   -1.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    6.1745                  
std error , SE =    Sp*√(1/n1+1/n2) =    2.9107                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -1.0000   -   0   ) /    2.91   =   -0.344
                          
Degree of freedom, DF=   n1+n2-2 =    16                  

p-value =        0.367826   [ excel function: =T.DIST(t stat,df) ]               
Conclusion:     p-value>α , Do not reject null hypothesis             

.................

B)

sample #1   ----->     Bumgarner        
first sample size,     n1=   108          
number of successes, sample 1 =     x1=   32          
proportion success of sample 1 , p̂1=   x1/n1=   0.2963          
                  
sample #2   ----->   Cueto        
second sample size,     n2 =    117          
number of successes, sample 2 =     x2 =    41          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.350          
                  
difference in sample proportions, p̂1 - p̂2 =     0.2963   -   0.3504   =   -0.0541
level of significance, α =   0.10              
Z critical value =   Z α/2 =    1.645   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0623          
margin of error , E = Z*SE =    1.645   *   0.0623   =   0.1024
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    -0.054   -   0.1024   =   -0.1565
upper limit = (p̂1 - p̂2) + E =    -0.054   +   0.1024   =   0.0483
                  
so, confidence interval is (   -0.1565   < p1 - p2 <   0.0483   )  

.............

C)

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 ╪   0          
                  
sample #1   ----->   experimental          
first sample size,     n1=   108          
number of successes, sample 1 =     x1=   76          
proportion success of sample 1 , p̂1=   x1/n1=   0.7037          
                  
sample #2   ----->   standard          
second sample size,     n2 =    117          
number of successes, sample 2 =     x2 =    76          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.650          
                  
difference in sample proportions, p̂1 - p̂2 =     0.7037   -   0.6496   =   0.0541
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.6756          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0625          
Z-statistic = (p̂1 - p̂2)/SE = (   0.054   /   0.0625   ) =   0.8665
                      
p-value =        0.3862   [excel formula =2*NORMSDIST(z)]      
decision :    p-value>α,Don't reject null hypothesis               
                  
Conclusion:   There is not enough evidence to  the proportion of throwing strikes for Bumgarner and Cueto is not equal for the entire season

.....................

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