Question

Assume you want to test the hypothesis that the mean weight of the male student population is 160 lbs.

If we assume that the standard deviation has a known value of 12 lbs and we want to test with a confidence level of 95%

to. What sample size should we use if we want one to have a probability of making a type II error of no more than 5% in the test when the actual mean is 163 lbs?

b. If the actual mean is 160 lbs, what is the probability that you conclude that the hypothesis is false?

c. If we want to use a 90% confidence interval to estimate the mean and we determine that the estimation error should not be greater than 3 lbs, what sample size should we use?

d. If we decide to take 49 observations, what will be the probability of making a type II error in case the real mean is 163?

Answer 1

a)

true mean ,    µ =    163  
          
hypothesis mean,   µo =    160  
significance level,   α =    0.05  
sample size,   n =   208  
std dev,   σ =    12.0000  
          
δ=   µ - µo =    3  
          
std error of mean=σx = σ/√n =    12/√208=   0.8321  
          
(two tailed test) Zα/2   = ±   1.960  
We will fail to reject the null (commit a Type II error) if we get a Z statistic between          
-1.960   and   1.960  
these Z-critical value corresponds to some X critical values ( X critical), such that          
-1.960   ≤(x̄ - µo)/σx≤   1.960  
158.369   ≤ x̄ ≤   161.631  
now, type II error is ,          
ß = P (   158.369   ≤ x̄ ≤   161.631
Z =    (x̄-true mean)/σx      
Z1=(158.3692-163)/0.8321=       -5.5655  
Z2=(161.6308-163)/0.8321=       -1.6456  
          
P(Z<-1.6454)-P(Z<-5.5653)=          
=   0.049924379   -   0
          
=   0.049924379   (answer)  

sample size,   n =   208

b)

5%

c)

Standard Deviation ,   σ =    12                  
sampling error ,    E =   3                  
Confidence Level ,   CL=   90%                  
                          
alpha =   1-CL =   10%                  
Z value =    Zα/2 =    1.645   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.645   *   12   /   3   ) ² =   43.289
                          
                          
So,Sample Size needed=       44                  

d)

true mean ,    µ =    163  
          
hypothesis mean,   µo =    160  
significance level,   α =    0.05  
sample size,   n =   49  
std dev,   σ =    12.0000  
          
δ=   µ - µo =    3  
          
std error of mean=σx = σ/√n =    12/√49=   1.7143  
          
(two tailed test) Zα/2   = ±   1.960  
We will fail to reject the null (commit a Type II error) if we get a Z statistic between          
-1.960   and   1.960  
these Z-critical value corresponds to some X critical values ( X critical), such that          
-1.960   ≤(x̄ - µo)/σx≤   1.960  
156.640   ≤ x̄ ≤   163.360  
now, type II error is ,          
ß = P (   156.640   ≤ x̄ ≤   163.360
Z =    (x̄-true mean)/σx      
Z1=(156.6401-163)/1.7143=       -3.7100  
Z2=(163.3599-163)/1.7143=       0.2100  
          
P(Z<0.21)-P(Z<-3.7099)=          
=   0.583152109   -   0.0001
          
=   0.5831 (answer)  

Please let me know in case of any doubt.

Thanks in advance!

Please upvote!