Question

You survey 20 men and 40 women regarding their excitement level for a DIY channel marathon. The average excitement level for a man is 1.6 with SD of .4; the average for a woman is 1.76 with SD of .8. You do not believe there is a statistical difference between genders.

Which row of the t-table should you use for the critical values?

Answer 1

Given that,
mean(x)=1.6
standard deviation , s.d1=0.4
number(n1)=20
y(mean)=1.76
standard deviation, s.d2 =0.8
number(n2)=40
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.093
since our test is two-tailed
reject Ho, if to < -2.093 OR if to > 2.093
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.6-1.76/sqrt((0.16/20)+(0.64/40))
to =-1.0328
| to | =1.0328
critical value
the value of |t α| with min (n1-1, n2-1) i.e 19 d.f is 2.093
we got |to| = 1.0328 & | t α | = 2.093
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.0328 ) = 0.315
hence value of p0.05 < 0.315,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.0328
critical value: -2.093 , 2.093
decision: do not reject Ho
p-value: 0.315
we do not have enough evidence to support the claim that believe there is a statistical difference between genders.