Solution:
Here, we have to use paired t test.
H0: µd = 0 versus Ha: µd < 0
We assume D = Y – X for the given scenario.
We are given α = 0.05
Dbar = Ybar – Xbar = 69.70 – 74.98 = -5.28
Sd = 9
n = 9
df = n – 1 = 8
α = 0.05
Q.(1)
Critical value = -1.8595
(by using t-table)
Q.(2)
Test statistic for paired t test is given as below:
t = (Dbar - µd)/[Sd/sqrt(n)]
t = (-5.28 – 0)/[9/sqrt(9)]
t = -5.28/3
t = -1.76
Test statistic value = -1.76
P-value = 0.0582
Q.(3)
Rejection rule:
Reject H0 if test statistic t value < critical value -1.8595
We have test statistic > critical value, so we do not reject the null hypothesis
P-value > α = 0.05
Q.(4)
So, we do not reject the null hypothesis
A group of n - 9 students was selected for a comparative study that involved their Exam 1 scores ...
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