Question

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Answer 1

Given that,
mean(x)=78.8824
standard deviation , s.d1=14.3042
number(n1)=17
y(mean)=86.75
standard deviation, s.d2 =7.9673
number(n2)=12
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.796
since our test is two-tailed
reject Ho, if to < -1.796 OR if to > 1.796
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =78.8824-86.75/sqrt((204.61014/17)+(63.47787/12))
to =-1.89
| to | =1.89
critical value
the value of |t α| with min (n1-1, n2-1) i.e 11 d.f is 1.796
we got |to| = 1.89015 & | t α | = 1.796
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.8902 ) = 0.085
hence value of p0.1 > 0.085,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.89
critical value: -1.796 , 1.796
decision: reject Ho
p-value: 0.085
we have enough evidence to support the claim that themean time interval has changed