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 0.01 significance level to test for a difference between the measurements from the two arms. What can be concluded? Right arm Left arm 148 149 172 124 133 156 137 168 176 135 In this example, Hd 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 altenative hypotheses for the hypothesis test? A. Ho: H 0 O B. Ho: Hg0 H 0 O C. Ho:Hg#0 H:0 O D Ho: Hd 0 H Ha0 Identify the test statistic. t (Round to two decimal places as needed.)

Answer 1

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
149	172	-23.000	0.040
148	168	-20.000	10.240
124	176	-52.000	829.440
133	156	-23.000	0.040
137	135	2.000	635.040

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	691	807	-116.000	1474.800

mean of difference ,    D̅ =ΣDi / n =   -23.200                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    19.2016                  
                          
std error , SE = Sd / √n =    19.2016   / √   5   =   8.5872      
                          
t-statistic = (D̅ - µd)/SE = (   -23.2   -   0   ) /    8.5872   =   -2.70

Degree of freedom, DF=   n - 1 =    4
p-value =        0.0540