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  
hypothesized mean=µo =    160  
=   0.05  
std dev,σ=   12.0000  

ß =    0.05  
δ=µ - µo =    3  
Zα/2=   1.9600  
Z (ß ) =    1.6449  
n = ( ( Z(ß)+Z(α) )*σ / δ )² = ((1.645+1.96)*12/3)^2=   207.92
so, sample size=   208  

b) P( hypothesis is false given µ =160) =type I error =0.05 =  5%

c) Standard Deviation ,   σ =    12
sampling error ,    E =   3
Confidence Level ,   CL=   90%
      

=   1-CL =   10%
Z value = Z_/2=    1.645
      
Sample Size,n = (Z*σ/E)² =   (1.645*12/3)² =    43.289
So, Sample Size needed=       44

d) std error of mean=σ_x = σ/√n =    12/√49=   1.7143
(two tailed test) Z_/2 = ±   1.9600  
type II error is          
ß = P(Z < Zα/2 - δ/σx) - P(Z < -Zα/2-δ/σx)          
P(Z<1.96-3/1.7143) - P(Z<-1.96-3/1.7143)=          
P(Z<0.21)-P(Z<-3.7099)=          
=   0.5832   -   0.0001
          
=   0.5831 (answer)  

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