The questions below refer to the following data: Dr. Chung, a renowned nutrition
ID: 3337487 • Letter: T
Question
The questions below refer to the following data: Dr. Chung, a renowned nutritionist, has consistently proclaimed the benefits of a balanced breakfast. To substantiate her claim, she asks her participants to go without breakfast for one week. The following week he asks the same participants to make sure they eat a complete breakfast. Following each week, Dr. Chung asks the supervisors of each participant to rate their productivity for that week. Participant Performance with Breakfast_ Performance Without Breakfast #1 #2 #3 #4 8 6 8 8 6 Using a-.05, do a one-tailed test of the doctor's hypothesis. What is your T=Explanation / Answer
Given that,
null, H0: Ud = 0
alternate, H1: Ud < 0
level of significance, = 0.05
from standard normal table, two tailed t /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) = 2
We have d = 2
pooled variance = calculate value of Sd= S^2 = sqrt [ 22-(8^2/4 ] / 3 = 1.414
to = d/ (S/n) = 2.828
critical Value
the value of |t | with n-1 = 3 d.f is 3.182
we got |t o| = 2.828 & |t | =3.182
make Decision
hence Value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.8284 ) = 0.0663
hence value of p0.05 < 0.0663,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud =0
alternate, H1: Ud < 0
test statistic: 2.828
critical value: reject Ho, if to < -3.182 OR if to > 3.182
decision: Do not Reject Ho
p-value: 0.0663
T = 3.182 for right tailed
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