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

0.010.01

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

Right arm	144	134	116	132	130	Open in StatCrunch + Copy to Clipboard + Open in Excel +
Left arm	184	177	172	154	134

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?

A.

Upper H 0H0​:

mu Subscript dμdnot equals≠0

Upper H 1H1​:

mu Subscript dμdgreater than>0

B.

Upper H 0H0​:

mu Subscript dμdequals=0

Upper H 1H1​:

mu Subscript dμdless than<0

C.

Upper H 0H0​:

mu Subscript dμdnot equals≠0

Upper H 1H1​:

mu Subscript dμdequals=0

D.

Upper H 0H0​:

mu Subscript dμdequals=0

Upper H 1H1​:

mu Subscript dμdnot equals≠0

Identify the test statistic.

tequals=nothing

​(Round to two decimal places as​ needed.)

Identify the​ P-value.

​P-valueequals=nothing

​(Round to three decimal places as​ needed.)

What is the conclusion based on the hypothesis​ test?

Since the​ P-value is

▼

greater

less

than the significance​ level,

▼

fail to reject

reject

the null hypothesis. There

▼

is

is not

sufficient evidence to support the claim of a difference in measurements between the two arms.

Answer 1

null and alternative hypotheses:

null Hypothesis:		μd	=	0
alternate Hypothesis:		μd	≠	0

  test statistic t =-3.64

​ P-value =0.022

Since the​ P-value is greater than the significance​ level , fail to reject the null hypothesis. There is not sufficient evidence to support the claim of a difference in measurements between the two arms