Question

Use the following information to answer questions (35) - (39), researcher wants to know if mood is affected by music. She conducts a test on a sample of 4 randomly selected adults and measures mood rating before and after being exposed to dassic rock music. Test the twypothesis that mood rating decreased after being exposed to classic rock music. Following are the mood ratings for the four participants. Participant #1 Participant #2 Participant #3 Participant #4 Before exposure After exposure Differences (Before - After) 5.0 4.0 6.0 5.2 4.0 3.2 4.5 6.2 ssume that all conditions for testing have been met. Use the fact that difference - 1.2868 for your computations.

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 =3.182
since our test is two-tailed
reject Ho, if to < -3.182 OR if to > 3.182
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 0.225
We have d = 0.225
pooled variance = calculate value of Sd= √S^2 = sqrt [ 5.17-(0.9^2/4 ] / 3 = 1.287
to = d/ (S/√n) = 0.35
critical Value
the value of |t α| with n-1 = 3 d.f is 3.182
we got |t o| = 0.35 & |t α| =3.182
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 != 0.3497 ) = 0.7497
hence value of p0.05 < 0.7497,here we do not reject Ho
ANSWERS
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null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: 0.35
critical value: reject Ho, if to < -3.182 OR if to > 3.182
decision: Do not Reject Ho
p-value: 0.7497
we do not have enough evidence to support the claim that difference of means between before exposure and after exposure.
35.
mean difference of before and after exposure is 0.225*4 =0.9
36.
Confidence Interval
CI = d ± t a/2 * (Sd/ Sqrt(n))
Where,
d = ∑ di/n
Sd = Sqrt( ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
d = ( ∑ di/n ) =0.899999999999999/4=0.225
Pooled Sd( Sd )= Sqrt [ 5.17- (0.9^2/4 ] / 3 = 1.287
Confidence Interval = [ 0.225 ± t a/2 ( 0.743/ Sqrt ( 4) ) ]
= [ 0.225 - 3.182 * (0.643) , 0.225 + 3.182 * (0.643) ]
= [ -1.822 , 2.272 ]
standard error = 0.643
38.
p-value: 0.7497