Question

In: Statistics and Probability

Listed below is a simple random sample of systolic blood pressure measurements (mm Hg) taken from...

Listed below is a simple random sample of systolic blood pressure measurements (mm Hg) taken from the right and left arms of seven people

Right arm 138 128 154 144 137 157 126

Left arm 179 173 167 125 173 146 158

Use a 0.04 significance level to test the claim that the blood pressure in the right arm is less than the blood pressure in the left arm.

Is there sufficient evidence to support the claim that the blood pressure in the right arm is less than the blood pressure in the left arm?

Solutions

Expert Solution

null Hypothesis:μd = 0
alternate Hypothesis: μd < 0
for 0.04 level with left tailed test and n-1= 6 df, critical value of t= -2.104
Decision rule: reject Ho if test statistic t<-2.104
S. No Right left diff:(d)=x1-x2 d2
1 138 179 -41 1681.00
2 128 173 -45 2025.00
3 154 167 -13 169.00
4 144 125 19 361.00
5 137 173 -36 1296.00
6 157 146 11 121.00
7 126 158 -32 1024.00
total = Σd=-137 Σd2=6677
mean dbar= d̅                           = -19.5714
degree of freedom =n-1                            = 6.000
Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) = 25.806
std error=Se=SD/√n= 9.7538
test statistic            =     (d̅-μd)/Se         = -2.007~ -2.01 (if required to 2 decimals)
p value = 0.046

since p value >0.04 , we fail to reject the null we do not have sufficient evidence to conclude that  the blood pressure in the right arm is less than the blood pressure in the left arm


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