Question

Consider the following hypothesis test.
H0: μ ≥ 10
Ha: μ < 10
The sample size is 155 and the population standard deviation is assumed known with
σ = 5.
Use
α = 0.05.
(a)
If the population mean is 9, what is the probability that the sample mean leads to the conclusion do not reject
H0?
(Round your answer to four decimal places.)

(b)
What type of error would be made if the actual population mean is 9 and we conclude that
H0: μ ≥ 10
is true?
Type I
Type II
Correct: Your answer is correct.
(c)
What is the probability of making a type II error if the actual population mean is 8? (Round your answer to four decimal places. If it is not possible to commit a type II error enter NOT POSSIBLE.)

Answer 1

a)

true mean ,    µ =    9                  
                          
hypothesis mean,   µo =    10                  
significance level,   α =    0.05                  
sample size,   n =   155                  
std dev,   σ =    5                  
                          
δ=   µ - µo =    -1                  
                          
std error of mean,   σx = σ/√n =    5.0000   / √    155   =   0.40161  

Zα =       -1.6449   (left tail test)          
                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic >               -1.645      
this Z-critical value corresponds to X critical value( X critical), such that                      
                      
(x̄ - µo)/σx ≥ Zα                      
x̄ ≥ Zα*σx + µo                      
x̄ ≥    -1.645   *   0.4016   +   10  
       x̄ ≥    9.3394   (acceptance region)      
                      
now, type II error is ,ß =    P( x̄ ≥    9.339   given that µ =   9   )  
                      
   = P ( Z > (x̄-true mean)/σx )                   
=P(Z > (   9.339   -   9   ) /   0.4016   )
                      
   = P ( Z >    0.845   ) =   0.1990 (answer)

b) Type II error

c)

now, type II error is ,ß =    P( x̄ ≥    9.339   given that µ =   8   )  
                      
   = P ( Z > (x̄-true mean)/σx )                   
=P(Z > (   9.339   -   8   ) /   0.4016   )
                      
   = P ( Z >    3.335   ) =   0.0004 (answer) [ Excel function: =1-normsdist(z) ]