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
0.050.05
significance level to test for a difference between the measurements from the two arms. What can be concluded?
Right arm |
152152 |
148148 |
115115 |
129129 |
136136 |
|
---|---|---|---|---|---|---|
Left arm |
172172 |
168168 |
177177 |
148148 |
149149 |
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 dμdequals=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 dμdgreater than>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 not
is
sufficient evidence to support the claim of a difference in measurements between the two arms.
Click to select your answer(s).
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