O StatListed 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 asimple random sample and that the differences have a distribution that is approximately normal. Use a 0.01 significance level to test for a difference between themeasurements from the two arms. What can be concluded?ionRight armLeft armk145150143129135 D178162181136150ncementsIn this example, Ha 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 fromthe right arm minus the measurement from the left arm. What are the null and alternative hypotheses for the hypothesis test?mentsO A. Ho: Ho #0O B. Ho: Hd #0PlanH1: Hd = 0Hq: Hg>0O C. Ho: Hd = 0O D. Ho: Hd =0H1: Hd #0ebookH:Hd

Name: O StatListed below are systolic blood pressure measurements (mm Hg) t
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: O StatListed 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 asimple random sample and that the differences have a distribution that is approximately

Given information: Significance level α=0.01 The systolic blood pressure measurement data for the right and left arms are summarized in the following table: Right arm X1 Left arm X2 DX1-X2 145 178 -33 150 162 -12 143 181 -38 129 136 -7 135 150 -15The hypotheses is defined as follows: H0:μD=0H1:μD≠0 This corresponds to a two-tailed test, for which a t-test for two dependent samples will be used. Therefore, the correct answer is option D. The mean difference is calculated as follows: D¯=∑i=1nDin=-33-12-38-7-155=-21 The standard deviation is calculated as follows: sD=∑i=1nDi-D¯2n-1=-33+212+....+(-15+21)25-1=13.6565 The degrees of freedom is calculated as follows: df=n-1=5-1=4 The critical value, at 0.01 significance level, for a two-tailed test is: tc=±4.604 The test statistic is calculated as follows: t=D¯sD/n=-2113.6565/5=-3.4385 Therefore, t=-3.44 The P-value, 0.01 significance level, for 4 degrees of freedom is: P-value = 0.026 Based on the hypothesis test, the conclusion is: Since the P-value is greater than the significance level, accept the null hypothesis. There is not sufficient evidence to support the claim of a difference in measurements between the two arms.