Listed below are systolic blood pressure measurements (mm Hg) taken from the rig

Question

Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data isa simple tandom sample and that the dillerences have a disibution that is approximately normal Use a 0.05 significance level to test for a diference between the measurements from the to arms What can be concluded? Rightam 144 146 140 129 130 Leam 159 186 151 136 htisexample, ,the mean valueofthe dnerences dfor the population of al pars of data, where each ideaal dierence disdehedashe measurement from the right arm minus the measurement from the left arm. What are the null and altemative hypotheses for the hypothesis test? H, "0 denitly the tesstaic Round totwe decinal places as needed ) Cick to sesect your answens 6

Explanation / Answer

Given that,
null, H0: Ud > 0
alternate, H1: Ud < 0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.776
since our test is two-tailed
reject Ho, if to < -2.776 OR if to > 2.776
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -23.2
We have d = -23.2
pooled variance = calculate value of Sd= S^2 = sqrt [ 3646-(-116^2/5 ] / 4 = 15.45
to = d/ (S/n) = -3.36
critical Value
the value of |t | with n-1 = 4 d.f is 2.776
we got |t o| = 3.36 & |t | =2.776
make Decision
hence Value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.3577 ) = 0.0284
hence value of p0.05 > 0.0284,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud > 0
alternate, H1: Ud < 0
test statistic: -3.36
critical value: reject Ho, if to < -2.776 OR if to > 2.776
decision: Reject Ho
p-value: 0.0284

X Y X-Y (X-Y)^2 144 173 -29 841 146 159 -13 169 140 186 -46 2116 129 151 -22 484 130 136 -6 36 -116 3646

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