Answer :
Complete question is;
A textile fiber manufacturer is investigating a new drapery yarn, which has a standard deviation of 0.3 kg. The company wishes to test the hypothesis H0: μ = 14 against H1: μ < 14kg using a random sample of five specimens.
(a) What is the P-value if the sample average is x¯ = 13.7kg?
(b) Find β for the case where the true mean elongation force is 13.5 kg and we assume that α = 0.05.
(c) What is the power of the test from part (b)?
Answer:
A) P-value = 0.012545
B) β = 0.018763
C) Power = 0.981237
Step-by-step explanation:
A) We are given;
Null hypothesis; H0: μ = 14
Alternative hypothesis; H1: μ < 14kg
Sample mean; x¯ = 13.7 kg
Population standard deviation; σ = 0.3 kg
Sample size; n = 5
To find the p-value, let's first find the z-score given by;
z = (x¯ - μ)/(σ/√n)
z = (13.7 - 14)/(0.3/√5)
z = -0.3/0.134164
z = -2.24
From online p-value from z-score calculator attached, using z = -2.24, significance level = 0.05, one tailed, we have:
P-value = 0.012545
B) β means the probability of getting a type II error.
Now , at a significance level of 0.05 given, the critical value of z in this case will be -1.645.
From the equation; z = (x¯ - μ)/(σ/√n)
Let's make sample mean (x¯) the subject.
Thus;
x¯ = μ + (zσ/√n)
x¯ = 14 + (-1.645 × 0.3/√5)
x¯ = 14 - 0.2207
x¯ = 13.7793
We are told the true mean is now 13.5 kg. Thus; μ_n = 13.5 kg
Thus;
z = (13.7793 - 13.5)/(0.3/√5)
z = 2.08
From online p-value from z-score calculator attached, using z = 2.08, significance level = 0.05, one tailed, we have:
P- value = β = 0.018763
C) The power of the test is simply the complement of β.
Power = 1 - β
Power = 1 - 0.018763
Power = 0.981237
