Question

In: Statistics and Probability

19) Your friends are at it again. This time they are testing H0 : μ =...

19) Your friends are at it again. This time they are testing H0 : μ = 14 versus H1 : μ ≠ 14. They have found that x̅ = 13.5, x̅cu = 14.7, n = 64 and they know σ = 2.5 .

a) Do your reject H0 ? Explain.

b) Find alpha.

c) Find the p-value.

d) Find beta if mu is really 14.3.

Solutions

Expert Solution

a) No, Ho is not rejected because Xbar = 13.5 < Xbar critical = 14.7

b)

µ =    14                                  
σ =    2.5                                  
n=   64                                  
                                      
X =   14.7                                  
                                      
Z =   (X - µ )/(σ/√n) = (   14.7   -   14   ) / (    2.5   / √   64   ) =   2.240
                                      
P(X ≥   14.7   ) = P(Z ≥   2.24   ) =   P ( Z <   -2.240   ) =    0.0125      

α = 2*0.0125 = 0.025

c)

Ho :   µ =   14
Ha :   µ ╪   14
      
Level of Significance ,    α =    0.025
population std dev ,    σ =    2.5000
Sample Size ,   n =    64
Sample Mean,    x̅ =   13.5000
      
'   '   '
      
Standard Error , SE = σ/√n =   2.5/√64=   0.3125
Z-test statistic= (x̅ - µ )/SE =    (13.5-14)/0.3125=   -1.6000
      
critical z value, z* =   ±   2.2400
      
p-Value   =   0.1096

d)

true mean ,    µ =    14.3  
          
hypothesis mean,   µo =    14  
significance level,   α =    0.025090923  
sample size,   n =   64  
std dev,   σ =    2.5000  
          
δ=   µ - µo =    0.3  
          
std error of mean=σx = σ/√n =    2.5/√64=   0.3125  
          
(two tailed test) Zα/2   = ±   2.240  
We will fail to reject the null (commit a Type II error) if we get a Z statistic between          
-2.240   and   2.240  
these Z-critical value corresponds to some X critical values ( X critical), such that          
-2.240   ≤(x̄ - µo)/σx≤   2.240  
13.300   ≤ x̄ ≤   14.700  
now, type II error is ,          
ß = P (   13.300   ≤ x̄ ≤   14.700
Z =    (x̄-true mean)/σx      
Z1=(13.3-14.3)/0.3125=       -3.2000  
Z2=(14.7-14.3)/0.3125=       1.2800  
          
P(Z<1.28)-P(Z<-3.2)=          
=   0.899727432   -   0.0007
          
=   0.899027432   (answer)  


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