A manufacturer has developed a new fishing line, which the company claims has a mean breaking strength of 15 kilograms with a standard deviation of 0.5 kilograms. To test the hypothesis that μ=15 kilograms against the alternative hypothesis that μ<15 kilograms, a random sample of 50 lines will be tested. The critical region is defined to be when the sample mean is less than 14.9 kilograms.

a) Find the probability of committing a type 1 error when H˳ is true
b) Evaluate β (Beta) when μₐ =14.8 and μₐ =14.9 kilograms

a) α = 0.7927, b) at u=14.8, β = 0.99767, at u = 14.9, β= 0.2073

Step-by-step explanation: a) from the question, u= population mean = 15 and x= sample mean = 14.9

σ = population standard deviation = 0.5, n = sample size = 50.

We get the probability of committing a type 1 error by using the z score.

Z = x - u/(σ/√n)

Z = 14.9 - 15/(0.5/√50)

Z = - 0.1/0.0707

Z = - 1.41.

By checking the the probabilistic value attached to this z score using a standard normal distribution table whose area is to the left of the distribution, we have that

P(z=-1.41) = 0.7927.

Hence the probability of committing a type 1 error is 0.7927

b)

at x = 14.8 ( I let ua=x=14.8)

Z = x - u/(σ/√n)

Z = 14.8 - 15/(0.5/√50)

Z = - 0.2/ 0.0707

Z = - 2.83

Using the standard normal distribution table, we have that

P(z=-2.38) = 0.00233.

But α + β = 1

Where α= probability of committing a type 1 error

β = probability of committing a type 2 error.

β = 1 - α

β = 1 - 0.00233

β = 0.99767

At x = 14.9

Z = x - u/(σ/√n)

Z = 14.9 - 15/(0.5/√50)

Z = - 0.1/0.0707

Z = - 1.41.

P(z=-1.41) = 0.7927.

α = 0.7927.

But α + β = 1

β = 1 - 0.7927

β = 0.2073

the radius is 2 times the diameter.

step-by-step explanation: