Question

In the 1996 regular baseball season, the World Series Champion New York Yankees played 80 games at home and 82 games away. They won 49 of their home games and 43 of the games played away. We can consider these games as samples from potentially large populations of games played at home and away.

a. Combining all of the games played, what proportion did the Yankees win?

b. Find the standard error needed for testing that the probability of winning is the same at home and away.

c. Most people think that it is easier to win at home than away. Formulate null and alternative hypotheses to examine this idea.

d. Compute the z statistic and its P-value. What conclusion do you draw?

Answer 1

a)

Number of Items of Interest,   x = 43+49 = 92
Sample Size,   n = 80+82 = 162
      
Sample Proportion ,    p̂ = x/n =    0.5679

b)

sample #1   ----->   home          
first sample size,     n1=   80          
number of successes, sample 1 =     x1=   49          
proportion success of sample 1 , p̂1=   x1/n1=   0.6125          
                  
sample #2   ----->   away          
second sample size,     n2 =    82          
number of successes, sample 2 =     x2 =    43          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.5244          
                  
difference in sample proportions, p̂1 - p̂2 =     0.6125   -   0.5244   =   0.0881
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.5679          
               
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.07785          

c)

Ho:   p1 - p2 =   0
Ha:   p1 - p2 >   0

d)

Z-statistic = (p̂1 - p̂2)/SE = (   0.088   /   0.0778   ) =   1.1319
                  
  
p-value =        0.1288   [excel function =NORMSDIST(-z)]