In: Statistics and Probability
Assumed values
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.447
since our test is two-tailed
reject Ho, if to < -2.447 OR if to > 2.447
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -8.143
We have d = -8.143
pooled standard deviation = calculate value of Sd= √S^2 = sqrt [
2755-(-57^2/7 ] / 6 = 19.54
to = d/ (S/√n) = -1.103
critical Value
the value of |t α| with n-1 = 6 d.f is 2.447
we got |t o| = 1.103 & |t α| =2.447
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 != -1.1026 )
= 0.3125
hence value of p0.05 < 0.3125,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: -1.103
critical value: reject Ho, if to < -2.447 OR if to >
2.447
decision: Do not Reject Ho
p-value: 0.3125
we do not have enough evidence to support the claim