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 0.10 significance level to test for a difference between the measurements from the two arms. What can be​ concluded? Right arm 152 138 135 139 132 Left arm 178 176 176 142 151

Answer:1. The null hypothesisH0: Ud=0The alternate hypothesisH1:Ud≠02. Test statistic = -3.043. P value = 0.0394. Reject h0Step-by-step explanation:X = right armY = left armd = difference between both armsX. Y. d(x-y). d²145. 173. -28. 784142. 163. -21. 441116. 182. -66. 4356133. 148. -15. 225134. 149. -15. 225Totald = -145d² = 6031We have sample space n = 5d' = -145/5= -29Sd = √1/n-1(Σd²-(Σd)²/nSd = √1/4(6031-(-145)²/5= √1/4(6931-4205)= √456.5= 21.366To get t=d'/(sd/√n)= -29/(21.366/√5)= -3.035In summaryThis is a two tailed test1. The null hypothesisH0: Ud=0The alternate hypothesisH1:Ud≠02. Test statistic = -3.043. P value = 0.0394. We have pvalue to be 0.039 which is less than the level of significance 0.050.039<0.05, so we take decision to reject h0 which is the null hypothesisWe have enough evidence to support claim that there is a measurement difference between left and right arms