Question

In a random sample of​ males, it was found that 26 write with their left hands and 217 do not. In a random sample of​ females, it was found that 61 write with their left hands and 438 do not. Use a 0.01 significance level to test the claim that the rate of​ left-handedness among males is less than that among females. Find the test statistic Z to two or more decimals and the P-value to three or more.

Answer 1

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 <   0          
                  
sample #1   ----->   Males   
first sample size,     n1=   243          
number of successes, sample 1 =     x1=   26          
proportion success of sample 1 , p̂1=   x1/n1=   0.1070          
                  
sample #2   ----->   Females   
second sample size,     n2 =    499          
number of successes, sample 2 =     x2 =    61          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.122          
                  
difference in sample proportions, p̂1 - p̂2 =     0.1070   -   0.1222   =   -0.0152
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.1173          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0252          
Z-statistic = (p̂1 - p̂2)/SE = (   -0.015   /   0.0252   ) =   -0.6059
                  

p-value =        0.2723   [Excel function =NORMSDIST(z)      
decision :    p-value>α,Don't reject null hypothesis

              

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