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 is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.01 significance level to test for a difference between the measurements from the two arms. What can be​ concluded?

Right arm: 149, 142, 136, 136, 134

Left arm: 182, 175, 192, 142, 136

Hypothesis test. Please include confidence interval.

Answer 1

Subject	Right Arm	Left Arm	d = right - left
1	149	182	-33
2	142	175	-33
3	136	192	-56
4	136	142	-6
5	134	136	-2

Using the data of difference (right - left) from the above table, we calculate
xd(bar) = -26
s(dbar) = 22.2149
sample size, n = 5, hence df = 4

H0:
Ha:

This is paired sample t-test
test statistic,

t = -2.6171

This is two tailed test, hence p-value = 2*P(t < -2.6171)
p-value = 0.0590

As p-value > 0.01, fail to reject H0
there are not significant evidence to conclude that there is a difference between the measurements from the two arms.