Question

Fowle Marketing Research, Inc., bases charges to a client on the assumption that telephone surveys can be completed within an average time of 15 minutes or less. If more time is required, a premium rate is charged.

The testable hypotheses in this situation are ?0:?=15H0:μ=15 vs. ??:?>15Ha:μ>15

1. Identify the consequences of making a Type I error.

A. The company charges the customer the premium rate when they should.
B. The company does not charge the customer the premium rate when they should.
C. The company charges the customer the premium rate when they should not.
D. The company does not charge the customer the premium rate when they should not.

2. Identify the consequences of making a Type II error.

A. The company does not charge the customer the premium rate when they should.
B. The company charges the customer the premium rate when they should.
C. The company charges the customer the premium rate when they should not.
D. The company does not charge the customer the premium rate when they should not.

To monitor the billing rate, the manager is going to take a random sample of 35 surveys each shift and calculate the average survey time in the sample. They make a decision rule that if ?¯≥17x¯≥17, they will charge the premium rate for that shift's work. Assume the population standard deviation is 7 minutes.

3. What is the probability that the company 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 complete the survey is 17.5 minutes? That is, what is the probability they will reject ?0H0 when the actual average time is 17.5 minutes? Round your answer to four decimal places.

Answer 1

1)

C. The company charges the customer the premium rate when they should not.

2)

A. The company does not charge the customer the premium rate when they should.

3)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	15
std deviation   =σ=	7.000
sample size       =n=	35
std error=σx̅=σ/√n=	1.18322

probability =P(X>17)=P(Z>(17-15)/1.183)=P(Z>1.69)=1-P(Z<1.69)=1-0.9545=0.0455

4)

probability =P(X>17)=P(Z>(17-17.5)/1.183)=P(Z>-0.42)=1-P(Z<-0.42)=1-0.3363=0.6637