Question

(4 points) Fowle Marketing Research, Inc., bases charges to a cient on the assumption that telephone surveys can be completed within an average time of 15 mminutes or less. If more time is required, a premium rate is charged The testable hypotheses in this situation are H0 : μ = 15 vs. 14: μ > 15 1. Identify the consequences of making a Type l eror A. The company charges the customer the premium rate when they should. B. The company does not charge the customer the premium rale when they should O 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 ll error. A. The company does not charge the customer the premium rate when they should not B. The company does not charge the customer the premium rale when they should O C. The company charges the customer the premium rate when they should. D. The company charges the customer the premium rate when they should not To monitor the billing rate, the manager is going to take a random sample of 25 surveys each shift and calculate the average survey time in the sample. They make a decision rule that if standard demiation is 7 minutes 16.5, they will charge the premium rate for that shift's work Assume the population 3. What is the probability that the company wil make a Type l error using this decision rule? Round your answwer 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 16.75 minutes? That is, what is the probability they will reject Ho when the actual average time is 16.75 minutes? Round your answer to four decimal places

Answer 1

1)  
type I error is reject Ho, when it is true  
answer is   
the company charge sthe customer the premium rate when they should not  
  
2)  
type II error is fail to reject false Ho  
so, answer is   
the company do not charge the customer permium rate when they should.  

3)

population std dev ,    σ =    7
Sample Size ,   n =    25
Sample Mean,    x̅ =   16.50
      
'   '   '
      
Standard Error , SE =   σ/√n =   1.4000
      
Z-test statistic=   (x̅ - µ )/SE =    1.071

p-Value   =   0.1420   [excel function =NORMSDIST(z)]
so, type I error = 0.1420

4)

true mean ,    µ =    15
      
hypothesis mean,   µo =    16.5
significance level,   α =    0.1420
sample size,   n =   25
std dev,   σ =    7
      
δ=   µ - µo =    -1.5
      
std error of mean,   σx = σ/√n =    1.4000
      

Zα =       1.0714   (right tailed test)  
              
We will fail to reject the null (commit a Type II error) if we get a Z statistic <               1.0714
this Z-critical value corresponds to X critical value( X critical), such that              
              
       (x̄ - µo)/σx ≤ Zα      
       x̄ ≤ Zα*σx + µo      
       x̄ ≤    18.000   (acceptance region)
              
now, type II error is ,ß =    P( x̄ ≤    18.000   given that µ =   15
              
   = P ( Z < (x̄-true mean)/σx )          
   = P ( Z <    2.143   )  
   =   0.98394      
              
              
              
power =    1 - ß =   0.0161