Answer:
A.
Null:
H0: u= 5.65
Alternative:
H1: u > 5.65
B.
The test statistic
Xbar = 6.24
u = 5.65
S = 1.24
N = 57
T = 6.24-5.65/(1.24/√57)
= 0.59/(1.24/7.5498)
= 0.59/0.1642
Test statistic = 3.593
C.
With alpha = 0.05
Df = n-1 = 57-1 = 56
T critical = 1.673
D.
If t test > 1.673 reject the null hypothesis
3.593>1.673, reject the null hypothesis
We conclude that there is enough evidence that the technique helped the participants to learn more.