Question

(1 point) A manufacturer of electronic kits has found that the mean time required for novices to assemble its new circuit tester is 3 hours, with a standard deviation of 0.9 hours. A consultant has developed a new instructional booklet intended to reduce the time an inexperienced kit builder will need to assemble the device and the manufacturer needs to decide whether or not to send out the new booklet.

The testable hypotheses in this situation are ?0:?=3H0:μ=3 vs ??:?<3HA:μ<3.

1. Identify the consequences of making a Type I error.
A. The manufacturer does not send out an unhelpful instructional booklet.
B. The manufacturer sends out a helpful instructional booklet.
C. The manufacturer does not send out a helpful instructional booklet.
D. The manufacturer sends out an unhelpful instructional booklet.

2. Identify the consequences of making a Type II error.
A. The manufacturer does not send out a helpful instructional booklet.
B. The manufacturer does not send out an unhelpful instructional booklet.
C. The manufacturer sends out an unhelpful instructional booklet.
D. The manufacturer sends out a helpful instructional booklet.

To monitor the assembly time of inexperienced kit builders using the booklet, the manufacturer is going to take a random sample of 12 novices and calculate the mean time to assemble the circuit tester. If it is less than 2.75, they will send out the new instructional booklet. Assume the population standard deviation is 0.9 hours.

3. What is the probability that the manufacturer 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 time to assemble the circuit tester is 2.25 hours? That is, what is the probability they will reject ?0H0 when the actual average time is 2.25 hours? Round your answer to four decimal places.

Answer 1

1)

D. The manufacturer sends out an unhelpful instructional booklet.

2)

A. The manufacturer does not send out a helpful instructional booklet.

3)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	3
std deviation   =σ=	0.900
sample size       =n=	12
std error=σx̅=σ/√n=	0.25981

probability =P(X<2.75)=(Z<(2.75-3)/0.26)=P(Z<-0.96)=0.1685

(please try 0.1680 if this comes wrong and reply)

4)

probability =P(X<2.75)=(Z<(2.75-2.25)/0.26)=P(Z<1.92)=0.9726

(please try 0.9729 if this comes wrong and reply)