Question

A tire manufacturer produces tires that have a mean life of at least 35000 miles when the production process is working properly. The operations manager stops the production process if there is evidence that the mean tire life is below 35000 miles.

The testable hypotheses in this situation are ?0:?=35000H0:μ=35000 vs ??:?<35000HA:μ<35000.

1. Identify the consequences of making a Type I error.
A. The manager stops production when it is necessary.
B. The manager does not stop production when it is not necessary.  
C. The manager stops production when it is not necessary.
D. The manager does not stop production when it is necessary.

2. Identify the consequences of making a Type II error.
A. The manager stops production when it is not necessary.  
B. The manager does not stop production when it is necessary.
C. The manager does not stop production when it is not necessary.  
D. The manager stops production when it is necessary.

To monitor the production process, the operations manager takes a random sample of 15 tires each week and subjects them to destructive testing. They calculate the mean life of the tires in the sample, and if it is less than 34750, they will stop production and recalibrate the machines. They know based on past experience that the standard deviation of the tire life is 2000 miles.

3. What is the probability that the manager will make a Type I error using this decision rule? Round your answer to four decimal places.

4. Using this decision rule, what is the power of the test if the actual mean life of the tires is 34650 miles? That is, what is the probability they will reject ?0H0 when the actual average life of the tires is 34650 miles? Round your answer to four decimal places.

Answer 1

1)

C. The manager stops production when it is not necessary.

2(

B. The manager does not stop production when it is necessary.

3)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	35000
std deviation   =σ=	2000.000
sample size       =n=	15
std error=σx̅=σ/√n=	516.39778

probability =P(X<34750)=(Z<(34750-35000)/516.398)=P(Z<-0.48)=0.3156

(please try 0.3141 if this comes incorrect)

4)

probability =P(X<34750)=(Z<(34750-34650)/516.398)=P(Z<0.19)=0.5753

(please try 0.5768 if this comes incorrect)