Answer:
We should reject H₀ for values of | x₁| > 5,26 |x₁| > 6,12
Explanation:
Sample mean μ₀ = 5,7
Sample size n₁ = 24
Sample standard deviation s = 1
Significance level α = 0,05 CI = 95 %
Hypothesis criteria: Valves don´t meet the specification, which means pressure could be higher or lower than nominal
Normal Distribution where n < 30 we should use a two-tail t-student test
degree of freedom df = 24 - 1 = 23
And with df = 23 and α = 0,025 we find in t-tables the value for
t(c) = - 2,0687 ( on the left tail )
Test Hypothesis:
Null Hypothesis H₀ x = μ₀
Alternative Hypothesis Hₐ x ≠ μ₀
To compute t(s)
t(s) = ( x - μ₀ ) / s/√n
t(s) = ( x - 5,7 )*4,80
t(s) = 4,80*x - 27,34
Then for values of t(s) |t(s)| > 2,0687 we have to reject H₀
If we make t(s) = t(c) we find
-2,0687 = 4,8*x - 27,34
x = ( 27,34 - 2,07 ) / 4,8 and on the other tail x = 6,12
x = 5,26
Therefore for values of 5,26 ( under ) and above 6,12 we shoud reject the Null hypothesis