Question

Please write the full derivation of the answers. Thank you!

Q2) 1 Point for part a, 1 point for part b; total 2 points A company 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 kilogram. To test the hypothesis that u = 15 kilograms against the alternative that u < 15 kilograms, a random sample of 50 lines will be tested. The critical region is defined to be x < 14.9. (a) Find the probability of committing a type I error when He is true. (b) Evaluate ß for the alternatives u = 14.8 and u = 14.9 kilograms.

Answer 1

a) Given that the null hypothesis is true that is the mean is 15, the probability of type I error is computed here as:

$\alpha = P(\bar X < 14.9)$

Converting it to a standard normal variable, we have here:

$\alpha = P(Z < \frac{14.9 - 15}{\frac{0.5}{\sqrt{50}}})$

$\alpha = P(Z < - 1.4142)$

Getting it from the standard normal tables, we have here:

$\alpha = P(Z < - 1.4142) = 0.0786$

Therefore 0.0786 is the required proability of type I error here.

b) For a true mean of 14.8, the probability of type II error that is beta is computed here as:

= Probability that the sample mean is greater than 14.9 given that the true mean is 14.8.

$\beta_{\mu = 14.8} = P(\bar X > 14.9)$

Converting it to a standard normal variable we have here:

$\beta_{\mu = 14.8} = P(Z > \frac{14.9 - 14.8}{\frac{0.5}{\sqrt{50}}})$

$\beta_{\mu = 14.8} = P(Z > 1.4142)$

Getting it from the standard normal tables, we have here:

$\beta_{\mu = 14.8} = 0.0786$

Therefore 0.0786 is the required type II error probability here.

For a true mean value of 14.9, the probability of type II error is computed here as:

$\beta_{\mu = 14.8} = P(Z > \frac{14.9 - 14.9}{\frac{0.5}{\sqrt{50}}})$

$\beta_{\mu = 14.8} = P(Z > 0) = 0.5$

Therefore 0.5 is the required type II error probability here.