ANSWER:
a)
Ho : µd= 0
Ha : µd ╪ 0
b)
Level of Significance , α =
0.1
sample size , n = 8
mean of sample 1, x̅1= 85.225
mean of sample 2, x̅2= 85.988
mean of difference , D̅ =ΣDi / n =
-0.7625
std dev of difference , Sd = √ [ (Di-Dbar)²/(n-1) =
2.1771
std error , SE = Sd / √n = 2.1771 /
√ 8 = 0.7697
t-statistic = (D̅ - µd)/SE = ( -0.7625
- 0 ) / 0.7697
= -0.991
c)
Degree of freedom, DF= n - 1 =
7
p-value = 0.3549 [excel
function: =t.dist.2t(t-stat,df) ]
d)
Decision: p-value>α , Do not reject null hypothesis
We can conclude that average scores are not different.
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