# Poohsticks and hypothesis tests involving the binomial distribution

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```					 Poohsticks and hypothesis tests
involving the binomial distribution
L.O.: To carry out one-sided and
two-sided hypothesis tests
involving a binomial distribution
The House at Pooh Corner (by AA Milne)

Pooh believes that he is able to predict which of two fir-
cones will emerge first from under a bridge in a game of
poohsticks. To see whether he is right, he devises a
simple experiment in which he plays the game of
poohsticks 12 times.

He is able to correctly predict the winning fir-cone on 10 of
these occasions.

Test whether Pooh seems able to predict the winning fir
cone using a 5% significance test.
The House at Pooh Corner (by AA Milne)

H0: p = 0.5
H1: p > 0.5
5% significance test                               This
probability
is called
Let X be the number of correct predictions.       the p-
value.
Under H0, X ~ B(12, 0.5).

P(X ≥ 10) = 1 – P(X ≤ 9) = 1 – 0.9807 = 0.0193 or 1.93%
As 1.93% < 5%, we can reject H0. There is evidence that
Pooh has some ability to predict the winning fir-cone.
One-sided and two-sided hypothesis
tests
Example:
A drug is claimed to have a 40% success rate for curing
patients with a certain disease. This is suspected of being an
exaggeration. To test this claim, 16 people were treated with
one drug and it was found that 3 were cured. Carry out a test
at the 10% significance level.

Solution:
H0: p = 0.4                H1: p < 0.4
10% significance test
Let X be the number of people cured. Under Ho, X ~ B(16, 0.4)
P(X ≤ 3) = 0.0651 or 6.51% (from tables).
Since 6.51% < 10%, we can reject H0. There is weak
evidence that the claimed success rate is exaggerated.
One-sided and two-sided hypothesis
tests
Compare this example to the following slightly differently
worded question:-
One-sided and two-sided hypothesis
tests
Example:
A drug is claimed to have a 40% success rate for curing
patients with a certain disease. A doctor wishes to test
whether this success rate is correct. Note: we are not indicating
and finds success rate
The doctor gives the drug to 16 patients whether the that 3 are
cured. Carry out a hypothesis test using ais believed to be too
10% level.
high or too low.

Notice that we have the same data as before, i.e. 16 patients,
3 cured. However the hypotheses are now different:
H0: p = 0.4
H1: p ≠ 0.4
This test is now called a two-sided (or two-tailed) test. We are
essentially testing two different hypotheses, i.e. whether p <
0.4 and p > 0.4. We use a 5% significance level for each test.
One-sided and two-sided hypothesis
tests
Solution:
H0: p = 0.4               H1: p ≠ 0.4
10% significance test

Let X be the number of people cured. Under Ho, X ~ B(16, 0.4)

P(X ≤ 3) = 0.0651 or 6.51% (from tables).

Since 6.51% > 5%, we cannot reject H0. There is no evidence
that the claimed success rate is incorrect.
One-sided and two-sided hypothesis
tests
Example:
A magazine claims that 35% of all adults wear glasses.
Anna wishes to test whether this figure is correct. She takes a
sample of 18 adults and notices that exactly 13 of them wear
glasses. Test at the 5% level whether the magazine claim is
accurate.

Solution:
H0: p = 0.35               H1: p ≠ 0.35 (two-sided)
Significance level: 5%

Let X = number of people wearing glasses in sample.
One-sided and two-sided hypothesis
tests
Under Ho, X ~ B(18, 0.35).

P(X ≥ 13) = 1 – P(X ≤ 12) =
= 1 – 0.9986
= 0.0014 or 0.14%
As this test is two-sided we compare this probability with 2.5%.

As 0.14% < 2.5%, we can reject H0. We have some evidence
that the proportion of adults wearing glasses is not 35%.
Use of a normal approximation

Example:
There are many complaints from passengers about the late
running of trains on a particular route. The railway company
claims that the proportion of trains that are delayed is 10%.
The Railway Passengers’ Association conducts a survey of a
random sample of 200 trains and finds that 27 are delayed.
Test at the 5% level whether the railway company is
underestimating the proportion of trains that are delayed.
Use of a normal approximation

Solution:
H0: p = 0.1
H1: p > 0.1 (one-sided test)
Significance level: 5%

Let X be the number of trains in the sample that are delayed.
Under H0, X ~ B(200, 0.1).
But X ≈ N[200 × 0.1, 200 × 0.1 × 0.9] = N[20, 18].

     26.5  20 
P(X ≥ 27) = P(X ≥ 26.5) = P  Z            
         18 
Continuity    P ( Z  1.532)  1  0.9372  0.0628
correction
Use of a normal approximation

Conclusion:
Since 6.28% > 5%, we cannot reject H0. There is no evidence
that the train company is understating the proportion of
delayed trains.

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