Download null hypothesis

Testing a hypothesis
(Session 09)
SADC Course in Statistics
Learning Objectives
By the end of this session, you will be able to
• set up and carry out a test of hypothesis
concerning a population mean
• interpret the results of such a t-test
• explain what is meant by a p-value and
how it may be interpreted
• state conclusions resulting from a
hypothesis test according to the size of the
p-value
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An illustrative example
• Farmers growing maize in upland areas of
Chiradzulu EPA in southern Malawi were getting
average yields of 2900 kg/ha with s.d.  = 780
kg/ha.
• A “new” Integrated Pest Management (IPM)
approach was attempted with 16 farmers.
• Objective: To determine if the new approach
results in an increase in maize yields.
• Yields from these 16 farmers (after using IPM)
gave = 3454 kg/ha with standard error 168.
• Can we determine whether IPM has really
increased maize yields?
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Is the yield increase real?
• In above example, clearly the sample mean
of 3454 kg/ha is greater than 2990 kg/ha
• But the question of interest is “does this
result indicate a significant increase in
the yield or is it just a result of the usual
random variation of yield”
• Hypothesis testing seeks to answer such
questions by looking at the observed
change relative to the “noise”, i.e. the
standard error in the sample estimate
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Review of Session 8 ideas
• A hypothesis is a statement or a claim
concerning a population or populations.
• The main hypothesis of interest is called
the null hypothesis and usually denoted
H0.
• The alternative hypothesis, denoted H1,
is one that is considered to be true if the
null hypothesis is false.
• It is thus always the case that either H0 is
true or H1 is true, both cannot be true at
the same time.
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H0 and H1 for example
Null hypothesis H0:  = 2900
where  is the true mean yield of farmers in
the area using the new approach
The promoters of the new approach are
confident that yields with the new approach
cannot possibly decrease.
Hence the above null hypothesis needs to be
tested against the alternative hypothesis
H1:  > 2900
Note: H0 is very specific. H1 is more general
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Classical approach to testing
We use the distribution of the sample mean
( x ) to test H0, i.e.
x ~ N(, 2/n)
Assume the null hypothesis is true, i.e. IPM
has no effect on yields.


Then
x
3454  2900 

Pr  x  2900  = Pr 

 Pr  z  2.84 

780


n
16 

From tables of the standard normal distn, we see
that the above probability is about 0.0022
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Interpretation
This probability is very small.
Unlikely to get this probability value, based
on the observed mean of 3454 kg/ha, if the
null hypothesis is really true.
In other words, the chance of getting a zvalue as large as that observed (i.e. 2.84)
is very small at 0.22%.
We conclude that H0 cannot be true.
IPM has really had an effect.
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Test statistic based approach
(a) Set up the null and alternative hypotheses
(b) Compute the test statistic
z = ( x - )/(/n) = (3454 – 2900)/(780/4)
= 2.84
(c) Find the probability of getting a result,
which is as extreme, or more extreme than
the one observed, given H0 is true.
The smaller this probability value, the greater
is the evidence against the null hypothesis.
This probability is called the p-value or
significance level of the test
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Test for mean when  is unknown
It is usually the case that  is unknown.
Instead, we could use the sample standard
deviation s.
The test statistic is now
t = ( x - )/(s/n) = (3454 – 2900)/(168)
= 3.30
which follows a t-distribution with n-1=15
degrees of freedom.
Exercise: Look up tables of the t-distribution
and assess roughly what the p-value may be…
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Interpretation and conclusions
It is clear from t-tables that the p-value is
smaller than 0.01.
Using statistical software, we can get the
exact p-value as 0.0024. (Can also use Excel’s
function tdist(3.30,15,1) to get this)
This p-value is so small, there is sufficient
evidence to reject H0.
Conclusion:
Use of the new IPM technology has led to an
increase in maize yields (p-value=0.0024)
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Summary of actions to take
• If get a small p-value, i.e. close to zero,
– Conclude null-hypothesis is unlikely to be true
and select the alternative
– Reject the null hypothesis and declare that the
result is “statistically significant”
• Note that “Statistical significance” does not
necessarily imply that the effect observed is
practically important (more on this later)
• If the p-value is not small,
– There is insufficient evidence to reject H0.
– The result is not significant
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More on interpretation of p-value
• Traditionally, 0.05 has been regarded as a
cut-off point for a result to be declared as
statistically significant
• Note that the p-value is essentially the
chance (probability) that the null hypothesis
is rejected when it is actually true
• Thus the p-value =  = Probability of the
Type I error of test
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Reporting the results
Take care when reporting results. It is not a
good idea to regard a 5% level of significance,
i.e. a p-value of 0.05, as an absolute cut off
value. Report results according to size of pvalue.
For example, evidence of an effect is :
• almost conclusive if p-value < 0.001 and
could be said to be strong if p-value < 0.010
• If 0.01< p-value < 0.05, results indicate some
evidence of an effect.
• If p-value > 0.05, but close to 0.05, it may
indicate something is going on, but further
confirmatory study is needed.
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References
Altman, D.G., Machin, D., Bryant, T.N., and
Gardner, M.J. (2000) Statistics with
confidence. (2nd Edition). BMJ Books,
Bristol, UK. pp 240. (See next session for
more references)
Some practical work follows…
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