Question

The following table contains information on matched sample values whose differences are normally distributed.

Number - Sample 1 - Sample 2

1 17 21

2 12 13

3   21 22

4 23   19

5 19   19

6 13 17

7   19 16

8 17 21

a. Construct the 95% confidence interval for the mean difference μD.

Answer 1

SOLUTION:

From given data,

The following table contains information on matched sample values whose differences are normally distributed.

Number	Sample 1 ()	Sample 2 ()	Difference = -	( - )2
1	17	21	-4	(-4+0.875)2 = 9.765625
2	12	13	-1	(-1+0.875)2 = 0.015625
3	21	22	-1	(-1+0.875)2 = 0.015625
4	23	19	4	(4+0.875)2​​​​​​​ = 23.765625
5	19	19	0	(0+0.875)2​​​​​​​ = 0.765625
6	13	17	-4	(-4+0.875)2​​​​​​​ = 9.765625
7	19	16	3	(3+0.875)2​​​​​​​ = 15.015625
8	17	21	-4	(-4+0.875)2​​​​​​​ = 9.765625
			= -7	( - )2 = 68.875

= 8

  ​​​​​​​ = -4-1-1+4+0-4+3-4 = -7

The mean of difference = =   ​​​​​​​ / = -7 / 8 = -0.875

( - )²

= 9.765625+0.015625+0.015625+23.765625+0.765625+9.765625+15.015625+9.765625

= 68.875

= * ( - )²

= *  68.875

= 9.83928

(a). Construct the 95% confidence interval for the mean difference

Sample standard deviation , S = = = 3.1367626

95% CI for using t-dist

Confidence interval is 95%

95% = 95/100 = 0.95

= 1 - Confidence interval = 1-0.95 = 0.05

/2 = 0.05 / 2 = 0.025

Degree of freedom = df = n-1 = 8-1 = 7

Critical value,

t_/2,df = t_0.025,7 = 2.3646 ( from t- table , two -tails , df =7 )

Margin of Error = E = t_/2,df * ( S  / )

= 2.3646 * ( 3.1367626 / )

= 2.3646 *1.109013

= 2.62237

Limits of 95% confidence interval are given by:

Lower limit = - E =  -0.875 - 2.62237 = -3.49737   -3.49

Upper limit = + E =  -0.875+ 2.62237 = 1.74737 1.74

Therefore, Confidence interval for the mean difference μD

(-3.49 , 1.74 )

-3.49 < D < 1.74

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