Question

Stress between males and females

*Note: alpha = .001

1 t-Test: Two-Sample Assuming Unequal Variances Female Male 4 Mean 5 Variance 6 Observations 7 Hypothesized Mean Difference 3.655737705 3.52857143 1.296174863 1.12236025 70 61 8 df 9 t Stat 10 P(T-t) one-tail 11 t Critical one-tail 12 P(T<-t) two-tail 13 t Critical two-tail 124 0.658596658 0.255687918 3.157259054 0.511375836 3.370720124

Answer 1

Given that,
mean(x)=3.655
standard deviation , s.d1=1.1384
number(n1)=61
y(mean)=3.5285
standard deviation, s.d2 =1.059
number(n2)=70
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.001
from standard normal table, two tailed t α/2 =3.46
since our test is two-tailed
reject Ho, if to < -3.46 OR if to > 3.46
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =3.655-3.5285/sqrt((1.29595/61)+(1.12148/70))
to =0.655
| to | =0.655
critical value
the value of |t α| with min (n1-1, n2-1) i.e 60 d.f is 3.46
we got |to| = 0.65529 & | t α | = 3.46
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 != 0.6553 ) = 0.515
hence value of p 0.001 < 0.515,here we do not reject Ho
ANSWERS
---------------
1.

null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.655
critical value: -3.46 , 3.46
2.

decision: do not reject Ho
p-value: 0.515
we do not have enough evidence to support the claim that difference between males and females.