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

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).