Part a
We are given
α = 0.05
n = 16
df = n – 1 = 16 – 1 = 15
So,
Critical values are -2.131, 2.131.
(by using t-table)
Test statistic formula is given as below:
txbar = (Xbar - µ)/[S/sqrt(n)]
We are given Xbar = 56, S = 20, n = 16, µ = 50
txbar = (56 – 50) / [ 20/sqrt(16)]
txbar = (56 – 50) / [ 20/4]
txbar = (56 – 50) / 5
txbar = 6/5
txbar = 1.20
Absolute value of txbar = 1.20 is less than absolute critical value 2.131
So, we do not reject the null hypothesis
Correct Answer: D.
Do not reject the null hypothesis. The data do not provide sufficient evidence to conclude that the mean differs from µ = 50.
Part b
P-value = 0.249
(by using excel)
This Question: 1 pt 15 of 20 (10 complete) Consider the hypotheses below Ho μ:50 H1...
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