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

There is a significant difference between the two means based on this samples at the 0.10 level of significance.

Step-by-step explanation:

Let's call

mean of the systolic pressure from the right hand

mean of the systolic pressure from the left hand

and construct a confidence interval for the difference  

based on the sample of size 5.

The confidence interval whose endpoints are

where

= mean of the sample from the right hand

= mean of the sample from the left hand

= standard deviation of the sample from the right hand

= standard deviation of the sample from the left hand

= t-score corresponding to a level of significance 0.10 or a confidence level 90%

Since the sample is too small we have better use the Student's t-distribution with 4 (sample size -1) degrees of freedom, which is the approximation of the Normal distribution for small samples.

For a 90% confidence level equals 2.132

Let's compute now the means and standard deviations of the samples

From the right hand we have

= 139.2

= 7.66

From the left hand we have

= 164.6

= 16.85

Then our confidence interval would be

finally, the interval is

[-43.05, -7.75]

Since our confidence interval does not contain the zero, we can say there is a significant difference between the two means based on this samples at the 0.10 level of significance.

38.5

step-by-step explanation:

lets say, first number = r

second number = 3r+7

total = 49

(r)+(3r+7)= 49

4r+7 = 49

4r = 42

r = 42/4

r= 10.5

3r+7 = 3(10.5)+7 = 38.5

hope this

x= 149. 24+7=31. 180-31=149

your welcome