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Hypothesis Testing
------------------

Hypothesis t esting asks the question is the sample statistic
different to the population parameter.

Null hypothesis      H0: =
Alternate hypothesis H1: ≠ <= 2 tailed
<  <= 1 tailed
>  <= 1 tailed

So what are the 4 possible outcomes?

1.  Correctly accept H0 and reject H1
2.  Correctly accept H1 and reject H0
3.  Accept H1 when H0 is true - TYPE 1 error = α

Where α is the SIGNIFICANCE LEVEL of the test.

4.  Accept H0 when H1 is true - TYPE 2 error = β

Where 1 - β is the POWER of the test.

p-value
-------

The p-value for any hypothesis test is the α level at which
we would be indifferent between accepting or rejecting H0.
That is, the p-value is the α level at which the given value
of the sample statistic is on the borderline between the
acceptance and rejection regions.

The p-value corresponds to the shaded areas.  The 1-tailed
test is twice the 2-tailed value.  Consider a Z-distribution
and α = 0.05:

1-tailed:  p = 0.05 for a critical value of Z = -1.645 or +1.645

Therefore, if the computed Z score was < -1.645 we would reject
_
H0 and accept H1: x < μ

If the computed Z score was < +1.645 we would reject
_
H0 and accept H1: x > μ

2-tailed:  p = 0.025 for a critical value of Z = -1.96 or +1.96

Therefore, if the computed Z score was < -1.96 or > +1.96 we
_
would reject H0 and accept H1: x ≠ μ

The χ2 and F distributions are not symmetric
and so the left and right tails are different.

The p-value can also be thought of as the probability of
obtaining a test statistic as extreme as or more extreme
than the actual test statistic obtained, given that H0 is
true.

The null hypothesis is rejected if the p-value is less than
or equal to α (i.e. the test statistic falls with in the
shaded areas).  In the language of statistics we say:

"At the α level of significance we can accept/reject H0,
there is not/is a difference (≠, < or >) between the sample
and the population".

F Test example:

Suppose you randomly select 7 marbles from company As
production line and 12 marbles from company Bs production
line and measure their diameters. Assume you are given:

sA = 1.0 and sB = 1.1

F = sA2/sB2 = 1/1.21 = 0.83

H0:  σA = σB
H1:  σA ≠ σB

From tables F0.05 for v1 = n1 - 1 = 6 and v2 = n2 - 1 = 11 is equal
to 3.0946.  Since 0.83 < 3.0946 the result is not significant and
there is no reason to reject H0.