Question

16. The boiling point of each of the 20 samples of a certain brand of hydrocarbon was determined, resulting in ¯x = 93.98. Assume that the distribution of boiling point of this brand of hydrocarbon is normal with σ = 1.30. a) Formulate and test your hypotheses with µ0 = 95 at α = 0.02. b) Find the Type II error when µ 0 = 94 c) What value of n is necessary to ensure that β(94) = 0.1 when α = 0.02?

Answer 1

a)

Ho :   µ =   95  
Ha :   µ ╪   95   (Two tail test)
          
Level of Significance ,    α =    0.020  
population std dev ,    σ =    1.3000  
Sample Size ,   n =    20  
Sample Mean,    x̅ =   93.9800  
          
'   '   '  
          
Standard Error , SE = σ/√n =   1.3/√20=   0.2907  
Z-test statistic= (x̅ - µ )/SE =    (93.98-95)/0.2907=   -3.5089  
          
          
p-Value   =   0.0004   [ Excel formula =NORMSDIST(z) ]
Decision:   p-value≤α, Reject null hypothesis

.........

b)

true mean ,    µ =    94
      
hypothesis mean,   µo =    95
significance level,   α =    0.02
sample size,   n =   20
std dev,   σ =    1.3000
      
δ=   µ - µo =    -1
      
std error of mean=σx = σ/√n =    1.3/√20=   0.2907

(two tailed test) Zα/2   = ±   2.326      
We will fail to reject the null (commit a Type II error) if we get a Z statistic between              
-2.326   and   2.326      
these Z-critical value corresponds to some X critical values ( X critical), such that              
-2.326   ≤(x̄ - µo)/σx≤   2.326      
94.324   ≤ x̄ ≤   95.676      
now, type II error is ,              
ß = P (   94.324   ≤ x̄ ≤   95.676   )
Z =    (x̄-true mean)/σx          
Z1=(94.3238-94)/0.2907=       1.1138      
Z2=(95.6762-94)/0.2907=       5.7665      
              
P(Z<5.7663)-P(Z<1.1136)=              
=   0.999999996   -   0.8673  
              
=   0.1327 (answer)      
.............

c)

true mean= µ =   94  
hypothesized mean=µo =    95  
α=   0.02  
std dev,σ=   1.3000  
power= 1- ß =   0.9  
ß =    0.1  
δ=µ - µo =    1  
Zα/2=   2.3263  
Z (ß ) =    2.3263  
n = ( ( Z(ß)+Z(α) )*σ / δ )² =    ((2.3263+2.3263)*1.3/1)^2=   36.58
so, sample size=   37  

.........

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