Question

Assume that 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%.

a) 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

Sol:

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

σx =   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

Z1 = -5.5655  

Z2=(161.6308-163)/0.8321

Z2 =   -1.6456  

P(Z<-1.6454)-P(Z<-5.5653) =  0.049924379   -   0
          
P(Z<-1.6454)-P(Z<-5.5653) =   0.049924379   

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

n=   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

Z1 =   -3.7100  

Z2=(163.3599-163)/1.7143

Z2 =   0.2100  

P(Z<0.21)-P(Z<-3.7099) =   0.583152109   -   0.0001
          
P(Z<0.21)-P(Z<-3.7099)  =   0.5831  

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