Question

451 438 430 471 434 449 460 470 449 452 448 445 454 452 442 436 457 446 462 454 414 453 457 442 442 447 446 447 456 447 431 452 437 446 466 454 443 439 443 433 Assuming that the population standard deviation for the 40 arrival times above is 12. And you are performing a 1-sample t test… A) How many data points would be needed if you wanted to detect a difference of 5 minutes. Assume that alpha is 0.05. Beta is 0.10 (Power is .9). B) Given that you have already collected 40 data points, what was the Power of the test? Assume alpha is 0.05 and the difference is 5 again. C) Suppose you would like to look at several possibilities at once…Find the sample size needed if the difference is 3, 4 and 5 minutes while the power is 0.7, 0.8, and 0.9.

Answer 1

A)

Level of Significance ,    α =    0.05  
std dev =    σ =    12.0000  
power =    1-ß =    0.9  
ß=       0.1  
δ=   µ - µo =    5  
          
Z (α/2)=       1.9600   [excel function: =normsinv(α/2)
          
Z (ß) =        1.2816   [excel function: =normsinv(ß)
          
sample size needed =    n = ( σ [ Z(ß)+Z(α/2) ] / δ )² =    60.5228  
          
so, sample size =        61.000

B)

significance level,   α =    0.0500
sample size,   n =   40
std dev,   σ =    12
      
δ=   µ - µo =    5
      
std error of mean,   σx = σ/√n =    1.8974

ß =   P(Z < Zα/2 - δ√n/σ) - P(Z < -Zα/2-δ√n/σ)              
   = P ( Z <    -0.6753   ) - P ( Z <   -4.5952   )
                  
   =   0.2498   -    0.0000  
   =   0.2498          

power =1-ß = 0.7502

C)

i)

Level of Significance ,    α =    0.05      
std dev =    σ =    12.0000      
power =    1-ß =    0.7      
ß=       0.3      
δ=   µ - µo =    3      
              
Z (α/2)=       1.9600   [excel function: =normsinv(α/2)  
              
Z (ß) =        0.5244   [excel function: =normsinv(ß)  
              
sample size needed =    n = ( σ [ Z(ß)+Z(α/2) ] / δ )² =    98.7531      
              
so, sample size =        99.000      

ii)

Level of Significance ,    α =    0.05      
std dev =    σ =    12.0000      
power =    1-ß =    0.8      
ß=       0.2      
δ=   µ - µo =    4      
              
Z (α/2)=       1.9600   [excel function: =normsinv(α/2)  
              
Z (ß) =        0.8416   [excel function: =normsinv(ß)  
              
sample size needed =    n = ( σ [ Z(ß)+Z(α/2) ] / δ )² =    70.6399      
              
so, sample size =        71.000      

iii)

Level of Significance ,    α =    0.05      
std dev =    σ =    12.0000      
power =    1-ß =    0.9      
ß=       0.1      
δ=   µ - µo =    5      
              
Z (α/2)=       1.9600   [excel function: =normsinv(α/2)  
              
Z (ß) =        1.2816   [excel function: =normsinv(ß)  
              
sample size needed =    n = ( σ [ Z(ß)+Z(α/2) ] / δ )² =    60.5228      
              
so, sample size =        61.000