Question

Use a t-distribution to find a confidence interval for the difference in means μd=μ1-μ2 using the relevant sample results from paired data. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using d=x1-x2.

A 99% confidence interval for μd using the paired data in the following table:

Case	1	2	3	4	5
Treatment 1	21	29	30	24	27
Treatment 2	19	32	26	21	20

Give the best estimate for μd, the margin of error, and the confidence interval.

Enter the exact answer for the best estimate, and round your answers for the margin of error and the confidence interval to two decimal places.

best estimate =

margin of error =

The 99% confidence interval is:

Answer 1

S. No	treatment 1	treatment 2	diff:(d)=x1-x2	d2
1	21	19	2	4.00
2	29	32	-3	9.00
3	30	26	4	16.00
4	24	21	3	9.00
5	27	20	7	49.00
	total	=	Σd=13	Σd2=87
	mean dbar=	d̅     =	2.6000
	degree of freedom =n-1                            =		4
	Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) =		3.646917

from above

best estimate = 2.6

for 99% CI; and 4 degree of freedom, value of t=

4.604

margin of error          =t*std error=

7.51

lower confidence limit                     =		-4.91
upper confidence limit                    =		10.11
from above 99% confidence interval for population mean =(-4.91 ,10.11)