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

Given that,
mean(x)=34.12
standard deviation , σ1 =25.75
number(n1)=4557
y(mean)=34.33
standard deviation, σ2 =25.25
number(n2)=4697
null, Ho: u1 = u2
alternate, H1: μ1 != u2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=34.12-34.33/sqrt((663.0625/4557)+(637.5625/4697))
zo =-0.396
| zo | =0.396
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo | =0.396 & | z α | =1.96
make decision
hence value of | zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.396 ) = 0.69211
hence value of p0.05 < 0.69211,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: μ1 != u2
test statistic: -0.396
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.69211
we donot have enough evidence to support the claim that there is a statistically significant difference between men and women mean age