Question

hw 9 # 15

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 0.05 significance level to test for a difference between the measurements from the two arms. What can be​ concluded?

Right arm	143	137	121	138	131
Left arm	171	174	191	149	136

In this​ example,

mu Subscript dμd

is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the measurement from the right arm minus the measurement from the left arm. What are the null and alternative hypotheses for the hypothesis​ test?

Answer 1

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 0.05 significance level to test for a difference between the measurements from the two arms. What can be​ concluded?

Right arm	143	137	121	138	131
Left-arm	171	174	191	149	136

μd is the mean value of the differences d for the population of all pairs of​ data,

where each individual difference d is defined as the measurement from the right arm minus the measurement from the left arm.

we want to test that there is a difference between the measurements from the two arms.

ie.e μd = 0 or not

The null and alternative hypotheses for the hypothesis​ test.

Ho:- μd = 0   

vs

Ha:- μd ≠ 0

i.e Ho:-  There is no difference between the measurements from the two arms.

Vs

Ha:-  There is a difference between the measurements from the two arms.

df = n-1 =5-1 = 4

S2 = SS⁄df = 2638.8/(5-1) = 659.7
S2M = S2/N = 659.7/5 = 131.94
SM = √S2M = √131.94 = 11.49

T-value Calculation

t = (M - μ)/SM

= (30.2 - 0)/11.49

T= 2.63

P-value =p(|t4 |> |2.63|)

= 2* P(t4 > 2.63)

= 2 * 0.291

= 0.0582

P-value = 0.0582

P-value = 0.0582 > 0.05 level of signficance

We fail to reject Ho

So we may conclude that the data do not provide sufficient evidence to support the claim that there is a difference between the measurements from the two arms.