Question

Suppose that a forester wants to see if the average height of lodgepole pines in Yellowstone is different from the national average of 70 ft. The standard deviation lodgepole pine height is known to be 8.2 ft. The forester decides to measure the height of 22 trees in Yellowstone and use a one-sample ?‑test with a significance level of 0.01. She constructs the following null and alternative hypotheses, where ? is the mean height of lodgepole pines in Yellowstone.

?0:?=70?1:?≠70

Use software to determine the power of the hypothesis test if the true mean height of lodgepole pines in Yellowstone is 65 ft. You may find one of these software manuals useful. Write your answer in decimal form and round to three decimal places.

power =

Answer 1

Answer:

true mean ,    µ =    65                          
                                  
hypothesis mean,   µo =    70                          
significance level,   α =    0.01                          
sample size,   n =   22                          
std dev,   σ =    8.2                          
                                  
  
std error of mean,   σx = σ/√n =    8.2000   / √    22   =   1.74825          
                                  
Zα/2   = ±   2.576   (two tailed test)                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic between                           -2.576   and   2.576
these Z-critical value corresponds to some X critical values ( X critical), such that                                  
                                  
-2.576   ≤(x̄ - µo)/σx≤   2.576                          
65.497   ≤ x̄ ≤   74.503                          
                                  
now, type II error is ,ß =        P (   65.497   ≤ x̄ ≤   74.503   )          
       Z =    (x̄-true mean)/σx                      
       Z1 = (   65.497   -   65   ) /   1.74825   =   0.284
       Z2 = (   74.503   -   65   ) /   1.74825   =   5.436
                                  
   so, P(   0.284   ≤ Z ≤   5.436   ) = P ( Z ≤   5.436   ) - P ( Z ≤   0.284   )
                                  
       =   1.000   -   0.612   =   0.3881   [ Excel function: =NORMSDIST(z) ]  
                                  
power =    1 - ß = 1-0.388 = 0.612   

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