Question

We have a sample of 9254 participants, 4557 men and 4697 women. The mean age of the sample is 34.33 years; men mean age is 34.12 (Variance= 663.063). The Variance in the women's group is 637.5625. Can you calculate women mean age? If so, is there a statistically significant difference between men and women mean age?  

Answer 1

total sum =9254*34.33 =317689.82

total of male =4557*34.12 =155484.84

so, mean of female = (317689.82 -155484.84)/4697

=34.53

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Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   34.12                  
standard deviation of sample 1,   s1 =    25.75                  
size of sample 1,    n1=   4557                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   34.53                  
standard deviation of sample 2,   s2 =    25.25                  
size of sample 2,    n2=   4697                  
                          
difference in sample means =    x̅1-x̅2 =    34.1200   -   34.5   =   -0.41  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    25.4974                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.5302                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -0.4100   -   0   ) /    0.53   =   -0.773
                          
Degree of freedom, DF=   n1+n2-2 =    9252                  
  
p-value =        0.439339   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis                      
                          
There is not enough evidence to say taht there a statistically significant difference between men and women mean age

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THANKS

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