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 0.05 significance level to test for a difference between the measurements from the two arms. What can be​ concluded?
Right arm Left Arm
145 173
142 163
116 182
133 148
134 149
In this​ example,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. What are the null and alternative hypotheses for the hypothesis​ test?
A. H0​:μd≠0
H1​:μd=0
B. H0​:μd≠0
H1​:μd>0
C. H0​:μd=0
H1​:μd≠0
D. Upper H0​:μd=0
H1​:μd<0
Identify the test statistic.
t=
​(Round to two decimal places as​ needed.)
Identify the​ P-value.
​P-value=
​(Round to three decimal places as​ needed.)
What is the conclusion based on the hypothesis​ test?
Since the​ P-value is less/greaterthan the significance​ level,rejectfail to 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

Answer:

1.

The null hypothesis

H0: Ud=0

The alternate hypothesis

H1:Ud≠0

2. Test statistic = -3.04

3. P value = 0.039

4. Reject h0

Step-by-step explanation:

X = right arm

Y = left arm

d = difference between both arms

X. Y. d(x-y). d²

145. 173. -28. 784

142. 163. -21. 441

116. 182. -66. 4356

133. 148. -15. 225

134. 149. -15. 225

Total

d = -145

d² = 6031

We have sample space n = 5

d' = -145/5

= -29

Sd = √1/n-1(Σd²-(Σd)²/n

Sd = √1/4(6031-(-145)²/5

= √1/4(6931-4205)

= √456.5

= 21.366

To get t

=d'/(sd/√n)

= -29/(21.366/√5)

= -3.035

In summary

This is a two tailed test

1.

The null hypothesis

H0: Ud=0

The alternate hypothesis

H1:Ud≠0

2. Test statistic = -3.04

3. P value = 0.039

4. We have pvalue to be 0.039 which is less than the level of significance 0.05

0.039<0.05, so we take decision to reject h0 which is the null hypothesis

We have enough evidence to support claim that there is a measurement difference between left and right arms