Question

Assume that the paired data came from a population that is normally distributed. Using a

0.05

significance level and

d=x−​y,

find

_
d,

sd​,

the t test​ statistic, and the critical values to test the claim that

μd=0.

x

13

7

8

6

6

14

8

9

 

y

13

7

5

5

6

9

9

12

_
d=

 

​(Round to three decimal places as​ needed.)

sd=

 

​(Round to three decimal places as​ needed.)

t=

 

​(Round to three decimal places as​ needed.)

tα/2=±

 

​(Round to three decimal places as​ needed.)

Answer 1

SOLUTION:

From given data,

Test hypothesis

Null hypothesis:

Ho :μd=0

Alternative Hypothesis

Ha: μd≠0

From given values,

X	Y	d= X-Y	$\bar{d}$	d-$\bar{d}$	(d-$\bar{d}$)2
13	13	13-13=0	0.625	0-0.625 = -0.625	(-0.625)2 = 0.390625
7	7	7-7 = 0	0.625	0-0.625 = -0.625	(-0.625)2 = 0.390625
8	5	8-5 = 3	0.625	3-0.625 = 2.375	(2.375)2 = 5.640625
6	5	6-5 = 1	0.625	1-0.625 = 0.375	( 0.375)2 = 0.140625
6	6	6-6=0	0.625	0-0.625 = -0.625	(-0.625)2 = 0.390625
14	9	14-9= 5	0.625	5-0.625 = 4.375	(4.375)2 = 19.140625
8	9	8-9=-1	0.625	-1-0.625 =-1.625	(-1.625)2 = 0.390625
9	12	9-12 = -3	0.625	-3-0.625 = -3.625	(-3.625)2 = 13.140625
		$\Sigma$ d = 5			$\Sigma$ (d-$\bar{d}$)2 = 39.234375

$\bar{d}$ = $\Sigma$ d / n = 5/8 = 0.625

Standard deviation.

Compute the standard deviation (sd) of the differences computed from n matched pairs.

sd = sqrt [ (Σ(d - $\bar{d}$)² / (n - 1) ]

sd = sqrt [ 39.234375/ (8 - 1) ]

sd = sqrt [ 39.234375/ 7 ]

sd = sqrt [ 35.6049107]

sd = 2.367

Test statistic.

The test statistic is a t statistic (t) defined by the following equation.

t = ($\bar{d}$ - $\mu$) / SE

the standard error can be approximated by:

SE = sd / sqrt( n )

=2.367/sqrt(8)

=0.83686

t = (0.625 - 0) / 0.83686

t = 0.625 / 0.83686

t = 0.7468

Critical value :

Degrees of freedom.

The degrees of freedom (DF) is: DF = n - 1 =8-1=7

significance level = $\alpha$ = 0.05

t_{$\alpha$/2 ,df} = t_{0.05/2 ,7} = t_{0.025 ,7} = 2.365

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