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“Teach A Level Maths”
Statistics 1
Hypothesis Testing
© Christine Crisp
Hypothesis Testing
Statistics 1
MEI/OCR
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Hypothesis Testing
Suppose there is a new drug treatment which we hope will
be better than the existing one.
To test whether it is better, we could set up a trial
involving a certain number of people. We could then see how
long it takes people to get well with the new drug and
compare it with results for the old one.
However, sampling involves random effects and we need to
know whether the apparently good results are really due to
improvements in treatment.
Statistics has a big part to play in making decisions of this
type and this presentation introduces an important theory
that is widely used.
Hypothesis Testing
e.g. 1. In a trial of 20 patients with a new drug, the
condition of 15 was greatly improved. On the older drug,
60% reported the same improvement. Is there evidence
that the new drug is more effective than the old one?
We can’t just say that the drug is clearly better because
75% of patients improved compared with 60% before.
We need to find out what the probability is that 15
patients improve even if the new drug is no more effective
than the old one.
The situation can be modelled with the Binomial distribution.
We let p, the probability of success, be the probability
that a patient improved with the new drug.
Hypothesis Testing
We set up a hypothesis ( a theory ) that the new drug is
not more effective and only reject this if we have
significant evidence against it.
We write
H0 :
p  06
H0 is called the null hypothesis. ( I think of it as “no
change”. )
The null hypothesis is using the value of p from data for
the old drug.
We want to test if the new drug is better than the old
one so we also have
H1 :
p 06
H1 is the alternative hypothesis.
Hypothesis Testing
We now want to test the hypothesis.
Let X be the random variable ”the number of patients
that improve”
There were 20 patients in the trial so n = 20.
So,
Also,
H0 :
H1 :
X ~ B( 20, p)
p  06
p 06
We want to find out how likely, or unlikely it is that, with
the old drug, 15 or more patients will improve.
We will only reject H0 if this probability is less than
0·05.
We say that we reject the null hypothesis at the 5% level
of significance.
Hypothesis Testing
X ~ B( 20, p)
H0 : p  0  6
H1 : p  0  6
Test at 5% level
of significance.
Assuming the null hypothesis we have n = 20 and p = 0·6:
P ( X  15)  1  P ( X  14)
 1  0  8744
 0  1256
There is a probability of 0·1256 ( more than 12% ) that
15 or more patients will improve even with the old drug.
We do not reject the null hypothesis.
If you have Autograph you can illustrate this and see the
probabilities for other values of x. I’ve copied an example
on the next slide.
Hypothesis Testing
P( X  15)  0  1256  12  56%
We have to consider P ( X  15) because we are comparing
with the end 5% of the distribution.
Let’s see how many patients would need to improve on the
new drug before we could reject H0 and accept that the
new drug is better than the old.
Hypothesis Testing
P ( X  16)
 0  0510  5  10%
( Near, but still
greater than 5% )
P ( X  17)
 0  0160  1  60%
So, if 17 patients improved we would have our evidence.
H0 would be rejected.
Hypothesis Testing
Critical Region
The critical region for the test consists of the values of
the random variable, X, that would cause us to reject the
null hypothesis.
In this example, the critical region is 17, 18, 19 and 20.
5%
I’ll write out the example again with the complete
solution so you see how we set it out.
Hypothesis Testing
e.g. 1. In a trial of 20 patients with a new drug, the
condition of 15 was greatly improved. On the older drug,
60% reported the same improvement. Is there evidence
that the new drug is more effective than the old one?
Solution:
Let X be the random variable ”the number of patients
that improve”
X ~ B( 20, p)
Test at 5% level
H0 : p  0  6
of significance.
H1 : p  0  6
P ( X  15)  1  P ( X  14)
 1  0  8744
 0  1256  0  05
There is a probability of 0·1256 ( more than 12% ) that
15 or more patients will improve even with the old drug.
We do not reject the null hypothesis. There is
insufficient evidence to suggest the new drug is better.
Hypothesis Testing
1-tailed and 2-tailed tests
In our example the alternative hypothesis, H1, was given
by
H1 :
p 06
This is called a 1-tailed test because we are only
considering values of p greater than 0·6.
We would also have a 1-tailed test if we had H 1 : p  0  6.
If our alternative hypothesis were
H1 :
p  06
we would have a 2-tailed test.
For a 2-tailed test, the percentage for the significance
is split into 2, one half at each end of the distribution.
Hypothesis Testing
e.g. 2. In a trial, 16 seeds are sown and only 11 germinate.
Use a 10% significance level to test the supplier’s claim
that 85% germinate. Find the critical region for the test.
Solution:
Let X be the random variable ”the number of seeds
that germinate”
X ~ B(16, p)
Test at 10% level
H 0 : p  0  85
of significance.
H
:
p

0

85
To1 test the supplier’s claim, the alternative hypothesis
is that
fewerathan
85% test
This
is again
1-tailed
but 
this
P( X
 11
)  germinate.
0  0791
0  1time we need to
test the bottom end of the distribution.
There is a probability of 0·0791 ( less than 10% ) that 11
or fewer seeds will germinate.
We reject the null hypothesis at the 10% level of
significance and conclude that the germination rate
is below 85% .
Hypothesis Testing
The Autograph illustration is as follows:
X ~ B(16, 0  85)
P ( X  11)  0  0791 7  91%
The probability of 12 or fewer germinating is 0·2101
( 21·01% ), so the critical region for the test is
0, 1, 2, . . . 10, 11.
Hypothesis Testing
e.g. 3. A team captain thinks the coin tossed at the start
of a match is biased. Find the critical region for the
number of heads in 30 tosses using a 5% significance level.
Solution:
Let X be the random variable “ number of heads ”.
So,
X ~ B( 30, p)
The null hypothesis is that the coin is fair, so
H0 :
p  05
H1 :
p  05
We don’t know whether there might be more or fewer
heads than on a fair coin so the alternative hypothesis is
We have a 2-tailed test, so the 5% must be split into
2·5% at each end of the distribution.
Hypothesis Testing
We want to find out how many heads would convince us that
the coin is biased. We refer to tables to get an idea of
the number we want and then check by doing the
calculation.
1  P ( X  19)
P ( X  20) 
 1  0  9506
 0  0494
We would not reject H0 with 20 heads because, although
we are using a 5% significance level, we are looking for
values in the top or bottom 2·5% of the distribution.
2  5%
2  5%
Hypothesis Testing
P ( X  20)  0  0494
We now try 21: P ( X  21)  1  P ( X  20)
 1  0  9786
 0  0214
This gives 2·14% which is less than 2·5% so with 21 or
more heads we reject H0 in favour of H1.
However, we are doing a 2-tailed test so we must also
consider the bottom end of the distribution.
With a value of p other than 0·5, the Binomial distribution
is not symmetric so we would need to do the calculation.
However, with p = 0·5, the Binomial is symmetric, so
without further calculations, we know that X  9 would
also cause us to reject H0.
Hypothesis Testing
2  5%
2  5%
The critical region has 2 parts:
0, 1, 2, . . . 9,
and 21, 22, . . . 30.
The values of X where we would accept H0 form the
acceptance region, so in this example they are 10, 11, . . .
20.
Hypothesis Testing
SUMMARY
 To carry out a hypothesis test on a Binomial model we
•
define a random variable
•
•
write down the distribution with n equal to the
number of trials and using p for the probability of
success
state the null (H0) and alternative (H1) hypotheses
using < or > for a 1-tailed test and  for a 2-tailed
test
write down the significance level of the test
•
calculate the cumulative probability
•
reject H0 if the probability is less than the
significance level ( or, for a 2-tailed test, less than
half the total significance level ).
continued
•
Hypothesis Testing
SUMMARY
 The critical region consists of the values of the random
variable that cause H0 to be rejected.
 A 1-tailed test is used if we are testing a claim that
the probability is higher or lower than defined in H0.
If there is no indication, we use a 2-tailed test.
 For a 2-tailed test, the significance level is split into 2
halves so, for example, 5% becomes 2·5% at each
end.
Hypothesis Testing
Exercise
1. In a survey it was found that 35% of members of a
political party supported candidate D to be leader. A
month later, D claimed his support had increased. A
2nd survey of 20 members showed that 11 supported
him. Would you accept D’s claim at the 5%
significance level?
2.
Hyacinth bulbs are sold to a retailer in packs of 100
which claim to have equal numbers of bulbs producing
blue and pink flowers. A random sample of 25 produces
17 blue flowers. Test at the 10% level of significance
whether the retailer has a right to complain that there
are not equal numbers of each colour.
Find the critical region for the test.
Hypothesis Testing
Solution:
1. Let X be the r.v.” number supporting D”
X ~ B( 20, p)
H0 :
H1 :
p  0  35
Test at 5%
significance level.
p  0  35
P ( X  11)  1  P ( X  10)
 1  0  9468
 0  0532  0  05
We do not reject H0 and so the candidate cannot claim
greater support.
At a 10% level of significance we would reject H0 and
conclude that D’s support had increased.
The Autograph diagram on the next slide illustrates the
solution.
Hypothesis Testing
10%
5%
P ( X  11)  0  0532
11 is outside the critical region at 5% significance but
inside it at 10%.
Hypothesis Testing
2.
Let X be the r.v. “ number of blue flowers”
X ~ B( 25, p)
H0 :
H1 :
p  05
p  05
Test at 10%
significance level.
( This is a 2-tailed test so we check against 5% each
end of the distribution. )
P ( X  17)  1  P ( X  16)
 1  0  9461
 0  0539  0  05
We do not reject H0. There is insufficient evidence at
the 10% level of significance to claim that there are
more of one colour than the other.
Hypothesis Testing
The value 0·0539 is so close to 0·05 that we are sure the
critical region starts at 18. However, we should check.
P ( X  18)  1  P ( X  17)
 1  0  9784
 0  0216  0  05
18 blue flowers would fall within the critical region.
Since this is a 2-tailed test the critical region is in 2
parts.
With p = 0·5, the
Binomial distribution
is symmetrical, so
the critical region is
0, 1, . . . 7 and
18, 19, . . . 25
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
Hypothesis Testing
SUMMARY
 To carry out a hypothesis test on a Binomial model we
•
define a random variable
•
•
write down the distribution with n equal to the
number of trials and using p for the probability of
success
state the null (H0) and alternative (H1) hypotheses
using < or > for a 1-tailed test and  for a 2-tailed
test
write down the significance level of the test
•
calculate the cumulative probability
•
reject H0 if the probability is less than the
significance level ( or, for a 2-tailed test, less than
half the total significance level ).
continued
•
Hypothesis Testing
 The critical region consists of the values of the random
variable that cause H0 to be rejected.
 A 1-tailed test is used if we are testing a claim that
the probability is higher or lower than defined in H0.
If there is no indication, we use a 2-tailed test.
 For a 2-tailed test, the significance level is split into 2
halves so, for example, 5% becomes 2·5% at each
end.
Hypothesis Testing
e.g. 1. In a trial of 20 patients with a new drug, the
condition of 15 was greatly improved. On the older drug,
60% reported the same improvement. Is there evidence
that the new drug is more effective than the old one?
Solution:
Let X be the random variable ”the number of patients
that improve”
X ~ B( 20, p)
Test at 5% level
H0 : p  0  6
of significance.
H1 : p  0  6
P ( X  15)  1  P ( X  14)
 1  0  8744
 0  1256  0  05
There is a probability of 0·1256 ( more than 12% ) that
15 or more patients will improve even with the old drug.
We do not reject the null hypothesis.
Hypothesis Testing
e.g. 2. In a trial, 16 seeds are sown and only 11 germinate.
Use a 10% significance level to test the supplier’s claim
that 85% germinate. Find the critical region for the test.
Solution:
Let X be the random variable ”the number of seeds
that germinate”
H0 :
H1 :
p  0  85
p  0  85
X ~ B(16, p)
Test at 10% level of significance.
To test the supplier’s claim, the alternative hypothesis is
that fewer than 85% germinate. This is again a 1-tailed
test but this time we need to test the bottom end of
the distribution.
P ( X  11)  0  0791  0  1
There is a probability of 0·0791 ( less than 10% ) that 11
or fewer seeds will germinate.
Hypothesis Testing
We reject the null hypothesis at the 10% level of
significance and conclude that the germination rate
is below 85% .
The Autograph illustration is as follows:
P ( X  11)  0  0791 7  91%
The probability of 12 or fewer germinating is 0·2101
( 21·01% ), so the critical region for the test is
0, 1, 2, . . . 10, 11.
Hypothesis Testing
e.g. 3. A team captain thinks the coin tossed at the
start of a match is biased. Find the critical region for
the number of heads in 30 tosses using a 5% significance
level.
Solution:
Let X be the random variable “ number of heads ”.
So,
X ~ B( 30, p)
The null hypothesis is that the coin is fair, so
H0 :
p  05
H1 :
p  05
We don’t know whether there might be more or fewer
heads than on a fair coin so the alternative hypothesis is
We have a 2-tail test, so the 5% must be split into 2·5%
at each end of the distribution.
Hypothesis Testing
We want to find out how many heads would convince us that
the coin is biased. A reasonable guess would be 20 out of
30.
P ( X  20)  1  P ( X  19)
 1  0  9506
 0  0494  0  025
2  5%
2  5%
We would not reject H0 with 20 heads because we are
looking for values in the top or bottom 2·5% of the
distribution.
Hypothesis Testing
P ( X  20)  0  0494
We now try 21: P ( X  21)  1  P ( X  20)
 1  0  9786
 0  0214
This gives 2·14% which is less than 2·5% so with 21 or
more heads we reject H0 in favour of H1.
However, we are doing a 2-tailed test so we must also
consider the bottom end of the distribution.
With a value of p other than 0·5, the Binomial distribution
is not symmetric so we would need to do the calculation.
However, with p = 0·5, the Binomial is symmetric, so
without further calculations, we know that X  9 would
also cause us to reject H0.
Hypothesis Testing
2  5%
2  5%
The critical region has 2 parts:
0, 1, 2, . . . 9,
and 21, 22, . . . 30.
The values of X where we would accept H0 form the
acceptance region, so in this example they are 10, 11, . . .
20.