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					Binomial Calculation of Type1 and Type 2 errors:
Ex 2. Since 1990 soil ecologists have been sounding the alarm about an invasion of
exotic earth worms in US forests. Of the more than 3500 species of earth worms the
concern is not with any of the native species but with 2 or 3 Chinese species that
consume forest floor leaf detritus at a much more rapid rate than that of native species.
Before an invasion the leaf detritus of the forest floor can be one to two feet thick and the
nutrients of this layer is slowly digested back into the soil through fungus. After an exotic
earthworm invasion the forest floor detritus layer has been reduced to a few inches. The
nutrients of the leaf detritus are released much faster to the forest floor, the soil is at risk
for erosion and the fungus ecology is diminished.
        Suppose that a soil ecologist wants to determine if the reduced detritus loam in a
local forest watershed is the result of an exotic earthworm invasion. She has drawn 8
random leaf detritus samples and has put forth the following hypothesis:
                 The proportion of exotic earth worms exist in less than 50% of the
H 0 : p  0.5  8 detritus samples and their roll in reducing the leaf detritus loam
                 can be no more than a contributory cause.
H A : p  0.5  The proportion of exotic earth worms exist in more than 50% of
                 the 8 detritus samples and they are primarily responsible for
                 reducing the leaf detritus loam.

       Consider the 2 competing views of reality offered by the null and alternate
hypothesis and for the sake of calculation assume that the true proportions for each
scenario have the assigned values:

         p = 0.4              40% of all possible samples contain exotic earth worms
         p = 0.6              60% of all possible samples contain exotic earth worms

        H 0 : p  0 .5                      H A : p  0.5




             0.3         0.4          0.5       0.6         0.7     0.8        1.0


       Find the probability of making a Type I error and a Type II error.
Answer:        a) P(Type I error) =      H O is true but we have rejected it

First translate the Type I error into English: The null hypothesis is true but our sample
data has persuaded us to reject it. The exotic earth worms do not form a majority in all
possible soil samples but in our 8 soil samples we find that a majority did contain the
exotic earth worm. We incorrectly conclude that the exotic earth worm is primarily to
blame for the reduced leaf detritus when in fact it is likely due to other causes. How many
ways can we make this rejection error? How many ways can we find a majority of exotic
earthworms in 8 samples? Let’s look at all the possible scenarios of making this Type 1
error:

No = 8 No = 7       No = 6      No = 5    No = 4    No = 3 No = 2        No = 1    No = 0
Yes = 0 Yes = 1     Yes = 2     Yes = 3   Yes = 4   Yes = 5 Yes = 6      Yes = 7   Yes = 8

Let X = Number of 8 selected         When H O is true:
       samples that contained         p = 0.4, probability of an exotic worm in a sample.
       exotic earth worms            1 – p = 0.6, probability of no exotic worm in a sample.

Then X ~ binomial(n  8, p  0.4) , PType 1 error   

            8            8             8             8
P( X  5)   0.45 0.63   0.46 0.6 2   0.47 0.61   0.48 0.60
             5            6            7             8
                                                     
P( X  5) = 0.17367             PType 1 error    = 0.17367


Answer:        b) P(Type II error) =  = H O is false but we have accepted it.

First translate the Type II error into English: The null hypothesis is false but our sample
data has persuaded us to accept it. The exotic earth worms form a majority in all possible
soil samples but in our 8 soil samples we find only that a minority contain the exotic
earth worm. We incorrectly conclude that the exotic earth worm is blameless for the
reduced leaf detritus when in fact the exotic earth worm is likely the primary cause. How
many ways can we make this acceptance error? How many ways can we find a minority
of exotic earthworms in 8 samples? Let’s look at all the possible scenarios of making this
Type 2 error:

No = 8 No = 7       No = 6      No = 5    No = 4    No = 3 No = 2        No = 1    No = 0
Yes = 0 Yes = 1     Yes = 2     Yes = 3   Yes = 4   Yes = 5 Yes = 6      Yes = 7   Yes = 8

Let Y = Number of 8 selected        When H A is true:
       samples that contained       p = 0.6, probability of an exotic worm in a sample.
       exotic earth worms           1 – p = 0.4, probability of no exotic worm in a sample.
Then Y ~ binomial(n  8, p  0.6) , PType 2 error   

           8             8            8              8
P(Y  3)   0.6 0 0.48   0.610.4 7   0.6 2 0.4 6   0.6 3 0.4 5
            0            1             2              3
                                                      
P(Y  3)  0.40591          0.40591
               So would you say the choice of the critical region was prudently made?

				
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posted:10/26/2012
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