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: 14 against using a random sample of five specimens. (a) What is the P-value if the sample average is (b) Find for the case where the true mean elongation force is 13.5 kg and we assume that 0.05. (c) What is the power of the test from part (b)

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)?

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

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