Question

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. p-value?

Answer 1

Solution:

Given that,

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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