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?

So the p value is less than the significance level given 0.01, we can conclude that we reject the null hypothesis that the difference mean between left and right arm is significantly different from 0 at 1% of significance.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation. Assume the following data:  

x=values for left arm , y = values for the right arm

x: 173  163  182  148  149

y: 145  142  116  133  134

The system of hypothesis for this case are:

Null hypothesis:

Alternative hypothesis:

Or equivalently:

Null hypothesis:

Alternative hypothesis:

The first step is calculate the difference and we obtain this:

d: 28, 21, 66, 15, 15

The second step is calculate the mean difference  

The third step would be calculate the standard deviation for the differences, and we got:

The 4 step is calculate the statistic given by :

The next step is calculate the degrees of freedom given by:

Now we can calculate the p value, since we have a two tailed test the p value is given by:

So the p value is less than the significance level given 0.01, we can conclude that we reject the null hypothesis that the difference mean between left and right arm is significantly different from 0 at 1% of significance.

answer: y = -12x + 108

step-by-step explanation: do (y = m x+b) formula.