Question

Using the table below, what do

M_D, df_D, and s_D²

equal? (Round your answers to three decimal places as needed.)

Participants	Groups
	A	B
1	8	4
2	3	6
3	6	2
4	6	9
5	2	7
6	6	5
7	11	6
8	4	2
9	6	6
10	9	2

M_D=

df_D=

s_D²⁼

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 1.2
We have d = 1.2
pooled variance = calculate value of Sd= √S^2 = sqrt [ 154-(12^2/10 ] / 9 = 3.938
to = d/ (S/√n) = 0.964
critical Value
the value of |t α| with n-1 = 9 d.f is 2.262
we got |t o| = 0.964 & |t α| =2.262
make Decision
hence Value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 0.9635 ) = 0.3605
hence value of p0.05 < 0.3605,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: 0.964
critical value: reject Ho, if to < -2.262 OR if to > 2.262
decision: Do not Reject Ho
mean difference = 1.2 *10 =12
d = ( Xi-Yi)/n) = 1.2
p-value: 0.3605
pooled variance =15.507
we do not have enough evidence to support the claim that difference in means .