Question

Suppose we’re testing ?0 : ? = 80 vs. ?0 : ? > 80 . Assume throughout that the population standard deviation is ? = 10.

a) Suppose a sample of size 56 is collected and ?̅= 83.5 is observed. Compute the standardized test statistic and p-value.

b)Suppose that the significance level of the test is set at ? = 0.09, and a sample of size 56 is to be chosen. Find the rejection region, that is, find the set of values of ?̅ that would lead to the rejection of ?0 . Now find the rejection region if ? = 0.04.

c)Suppose the significance level of the test is set at ? = 0.09, and a sample of size 81 is to be chosen. What is the probability that the test will fail to reject the null hypothesis, if in reality ? = 83? If this event were to happen, would it be a type I error, type II error or a correct decision?

d)Suppose the significance level of the test is set at ? = 0.09, and a sample of size 81 is to be chosen. What is the probability that the test will fail to reject the null hypothesis, if in reality ? = 79? If this event were to happen, would it be a type I error, type II error or a correct decision?

Answer 1

a)

Ho :   µ =   80                  
Ha :   µ >   80       (Right tail test)          
                          
Level of Significance ,    α =    0.00                  
population std dev ,    σ =    10.0000                  
Sample Size ,   n =    56                  
Sample Mean,    x̅ =   83.5000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   10.0000   / √    56   =   1.3363      
Z-test statistic= (x̅ - µ )/SE = (   83.500   -   80   ) /    1.3363   =   2.619
                          
  
p-Value   =   0.0044   [ Excel formula =NORMSDIST(-z) ]              

b)hypothesis mean,   µo =    80              
significance level,   α =    0.09              
sample size,   n =   56              
std dev,   σ =    10.0000              
                      
δ=   µ - µo =    -80              
                      
std error of mean,   σx = σ/√n =    10.0000   / √    56   =   1.33631

Zα =       1.3408   (right tailed test)      
                  
We will reject the null if we get a Z statistic > 1.341  
this Z-critical value corresponds to X critical value( X critical), such that                  
                  
(x̄ - µo)/σx > Zα                  
x̄ > Zα*σx + µo                  
x̄ > 1.341   *   1.3363   +   80
       x̄ ≥ 81.7917   (rejection region)  
-----------------

for α=0.04

Zα =       1.7507   (right tailed test)

       We will reject the null if we get a Z statistic > 1.7507   
this Z-critical value corresponds to X critical value( X critical), such that                  
                  
(x̄ - µo)/σx > Zα                  
x̄ > Zα*σx + µo                  
x̄ > 1.7507 *   1.3363   +   80
       x̄ ≥ 82.3395 (rejection region)      

c)

now, type II error is ,ß =    P( x̄ ≤    81.490   given that µ =   83   )
                  
   = P ( Z < (x̄-true mean)/σx )              
=P( Z < (   81.490   -   83   ) /   1.1111
                  
   = P ( Z <    -1.359   ) =   0.0870   [excel fucntion: =normsdist(z)
Correct decision

d)

now, type II error is ,ß =    P( x̄ ≤    81.490   given that µ =   79   )  
                      
   = P ( Z < (x̄-true mean)/σx )                  
=P( Z < (   81.490   -   79   ) /   1.1111   )
                      
   = P ( Z <    2.241   ) =   0.9875   [excel fucntion: =normsdist(z)  

TyPe I error