Question

10. One of the major U.S. tire makers wishes to review its warranty for their rainmaker tire. The warranty is for 40,000 miles. The distribution of tire wear is normally distributed with a population standard deviation of 15,000 miles. The tire company believes that the tire actually lasts more than 40,000 miles. A sample of 49 tires revealed that the mean number of miles is 45,000 miles. If we test the hypothesis with a 0.05 significance level, what is the probability of a Type II error if the actual true tire mileage is 45,000 miles? Seleccione una: A. Type II error = 0.4524 B. Type II error = 0.2549 C. Type II error = 0.2451 D. Type II error = 0.4925

Answer 1

At 0.05 significance level, the critical value is z_crit = 1.645

z_crit = 1.645

or, ( - )/() = 1.645

or, ( - 40000)/(15000/) = 1.645

or, = 1.645 * 15000/ + 40000

or, = 43525

P(Type II error) = P( < 43525)

                       = P(( - )/() < (43525 - )/())

                       = P(Z < (43525 - 45000)/(15000/))

                       = P(Z < -0.69)

                       = 0.2451

Option - C) Type II error = 0.2451