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
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