Question

A new treatment has been developed for a certain type of cement, resulting in a mean pressure resistance of 5000kg/cm² and a standard deviation of 130kg/cm². To verify the null hypothesis of p = 5000 against the alternative of u <5000, a random sample of 50 pieces of cement are examined. Determine the probability of committing a Type II error if = 4960 and a = 0.05. A. 0.298 B. 0.954 C. 0.082 D. 0.105 E. none of the preceding

Answer 1

For 0.05 level of significance, the critical mean value here is computed from standard normal tables here:
P(Z < -1.645) = 0.05

Therefore, we have here:

$\bar X_{crit} = \mu_0 - 1.645*\frac{\sigma}{\sqrt{n}} = 5000 - 1.645*\frac{130}{\sqrt{50}} = 4969.7570$

The probability of type II is computed as the probability of not rejecting the null hypothesis when it is false given that the true mean is 4960. Therefore the probability here is computed as:

$P(\bar X> 4969.7570)$

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

$P(Z > \frac{4969.7570 - 4960}{\frac{130}{\sqrt{50}}})$

P(Z > 0.5307)

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

Therefore A) 0.298 is the required probability here.