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

0.05

significance level to test for a difference between the measurements from the two arms. What can be​ concluded?

Identify the test statistic.

Right_arm(mm_Hg)   Left_arm(mm_Hg)
149   169
132   170
117   187
139   155
134   154

Answer 1

Here we will apply paired t test to check if there is an difference between right arm and left arm

SO here Hypothesis will be :

where is the difference between right arm and left arm

i.e., di = xi - yi

where xi's are Right arm BP and yi's are left arm BP

SO here test statistic is given by:

Now we will calculate the test statistic:

Right_arm(mm_Hg)	Left_arm(mm_Hg)	di	di^2
149	169.00	-20.00	400.00
132	170.00	-38.00	1444.00
117	187.00	-70.00	4900.00
139	155.00	-16.00	256.00
134	154.00	-20.00	400.00
	sum	-164.00	7400.00
	mean	-32.80

So here

= -32.80

So here

t statistic is given by above formula:

t = -3.263077 ~ t4  N

Now we will find critical value to find the rejection region.

since it is two tailed test ;

So rejection region is given by:

is the critcal region.

where

2.776445 is the critical value which we have calculated in R

code:

Since here we can see that t = -3.263077 < tc = -2.776445

which implies that here we have enough evidence to reject null hypotheisis

i.e., there is difference between the blood pressure of right arm and left arm at level of significance.

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