Question

(1 point) A manufacturer of electronic kits has found that the mean time required for novices to assemble its new circuit tester is 2.9 hours, with a standard deviation of 0.7 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

H0:μ=2.9

vs HA:μ<2.9

.

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

To monitor the assembly time of inexperienced kit builders using the booklet, the manufacturer is going to take a random sample of 14 novices and calculate the mean time to assemble the circuit tester. If it is less than 2.7, they will send out the new instructional booklet. Assume the population standard deviation is 0.7 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 2.75 hours? That is, what is the probability they will reject H0 when the actual average time is 2.75 hours? Round your answer to four decimal places.

Answer 1

1)   type I error is rejecting Ho when it is true   
   so, answer is   
   The manufacturer sends out a helpful instructional booklet.  
      
2)   type II error is fail to reject false Ho  
   so, answer is   
     The manufacturer does not send out a helpful instructional booklet  
      
3)      
type I error =P(X<2.7|) =

µ =    2.9              
σ/√n= 0.7/√14 =   0.187   

                  
Z =   (X - µ ) / σ/√n = (2.7-2.9)/0.187 = -1.07          
                  
P(X ≤   2.7   ) = P(Z ≤   -1.07   ) =   0.1425(answer)

4)

true mean ,    µ =    2.75
      
hypothesis mean,   µo =    2.9
significance level,   α =    0.1425
sample size,   n =   14
std dev,   σ =    0.7
      
δ=   µ - µo =    -0.15
      
std error of mean,   σx = σ/√n =    0.1871

Zα =       -1.0690   (left tail test)  
              
We will fail to reject the null (commit a Type II error) if we get a Z statistic >               -1.0690
this Z-critical value corresponds to X critical value( X critical), such that              
              
       (x̄ - µo)/σx ≥ Zα      
       x̄ ≥ Zα*σx + µo      
       x̄ ≥    2.700   (acceptance region)
              
now, type II error is ,ß =    P( x̄ ≥    2.700   given that µ =   2.75
              
   = P ( Z > (x̄-true mean)/σx )           
   = P ( Z >    -0.267   )  
   =   0.60537      
              
              
              
power =    1 - ß = 1-0.6054 = 0.3946