Given that,
mean(x)=69.937
standard deviation , s.d1=11.132
number(n1)=32
y(mean)=70.031
standard deviation, s.d2 =11.262
number(n2)=32
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.04
since our test is two-tailed
reject Ho, if to < -2.04 OR if to > 2.04
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =69.937-70.031/sqrt((123.92142/32)+(126.83264/32))
to =-0.0336
| to | =0.0336
critical value
the value of |t α| with min (n1-1, n2-1) i.e 31 d.f is 2.04
we got |to| = 0.03358 & | t α | = 2.04
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 != -0.0336 )
= 0.973
hence value of p0.05 < 0.973,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.0336
critical value: -2.04 , 2.04
decision: do not reject Ho
p-value: 0.973
we do not have enough evidence to support the claim that women and
men have the same mean diastolic
blood pressure.
Assume that the two samples are independent simple random samples selected from normally distributed populations. Do...
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