Answer:
Kindly check explanation
Explanation:
Given the data:
May August
95 100
72 80
78 95
50 65
89 85
92 98
75 70
90 96
65 87
The appropriate test is a paired t test :
d = difference between May and August
d = (-5, -8, -17, -15, 4, -6, 5, -6, -22)
The hypothesis :
H0 : μd = 0
H1 : μd ≠ 0
The test statistic :
T = dbar / (Sdbar/√n)
Where, dbar and Sdbar are the mean and standard deviation of 'd' respectively.
Using calculator :
dbar = - 7.777 ; Sdbar = 9.052
Test statistic = - 7.777 / (9.052 /√9)
Test statistic = - 2.577
The Pvalue, df = n - 1 = 9 - 1 = 8
Pvalue(-2.577, 8) = 0.0327
At α = 0.05
Pvalue < α ; WE reject the H0 ; and conclude that there has been a change in score