Question

A production line operation is tested for filling weight accuracy using the following hypotheses.
Hypothesis Conclusion and Action
H0: μ = 16
Filling okay; keep running.
Ha: μ ≠ 16
Filling off standard; stop and adjust machine.
The sample size is 30 and the population standard deviation is
σ = 0.9.
Use
α = 0.05.
(a)
What would a type II error mean in this situation?
Accepting H0 and letting the process continue to run when actually over-filling or under-filling exists.
Accepting H0 and stopping the process when actually no over-filling or under-filling exists.
Rejecting H0 and stopping the process when actually no over-filling or under-filling exists.
Rejecting H0 and letting the process continue to run when actually over-filling or under-filling exists.
Correct: Your answer is correct.
(b)
What is the probability of making a type II error when the machine is overfilling by 0.5 ounces? (Round your answer to four decimal places.)
  

(c)
What is the power of the statistical test when the machine is overfilling by 0.5 ounces? (Round your answer to four decimal places.)

Answer 1

From the information, observe that a production line operation is tested for filling weight accuracy using the following hypothesis. Null hypothesis, Ho:u=16 Alternative hypothesis, H.: 4616 The level of significance is, a = 0.05 a) The type-II error is the accepting null hypothesis when it is false. That is, accepting H, and letting the process continue to run when actually over filling or under filling exists. Hence, the correct option is (A). b) Calculate the probability of Type-II error when the machine is overfilling by 0.5 ounce. The level of significance is, a = 0.05 From the standard normal table values, observe that the critical value for the two-tail test at the 5% level of significance is £1.96 Now, calculate the lower and upper values of x. y = 4+242 = 16+(1.96) = 16 +0.29 =16.29

=16-(196) =16 -0.29 =15.71 Therefore, the required probability is, P(Type – Ilerror) = P(15.71 <ă <16.29| = 16.5) p(15.71-4 1- u 16.29 - M oln olun olan ) (15.71-16.5 16.29 - 16.5 0.91 130 0.91130) = P(-4.81<Z<-1.28) = P(Z <-1.28) - P(Z<-4.81) = 0.1003 -0.0000 = 0.1003 The power of the test is, Power =1- P(Type - II error) =1 -0.1003 = 0.8997