Question

We use a paired-samples t test when we have two samples, and the same participants are in both samples; to conduct the test, we calculate a difference score for every individual in the study. Discuss what areas in psychology would benefit from using a paired-sample t test to better understand that phenomena. For example, when measuring intelligence, we could develop an intervention to help improve intelligence, and the compare before and after scores for each participant. I asked this question and I got a list of number with no explanation. I need to know what are the two samples.

Answer 1

assumed values

X	Y	X-Y	(X-Y)^2
150	150	0	0
200	230	-30	900
110	130	-20	400
240	220	20	400
200	190	10	100
180	150	30	900
190	200	-10	100
260	300	-40	1600
280	300	-20	400
260	240	20	400
		-40	5200

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) = -4
We have d = -4
pooled variance = calculate value of Sd= √S^2 = sqrt [ 5200-(-40^2/10 ] / 9 = 23.66
to = d/ (S/√n) = -0.53
critical Value
the value of |t α| with n-1 = 9 d.f is 2.262
we got |t o| = 0.53 & |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 != -0.5345 ) = 0.6059
hence value of p0.05 < 0.6059,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: -0.53
critical value: reject Ho, if to < -2.262 OR if to > 2.262
decision: Do not Reject Ho
p-value: 0.6059
we do not have enough evidence to support the claim that difference score for every individual in the study