Download The Proportion of success in a large sample

Discrete Uniform
Probability distributions - Examples
A number is chosen at random from the natural numbers between 1 and 50.
What is the probability that a number over 30 is chosen?
What is the expected value of the number chosen and what is the variance of
numbers chosen?
X is the number chosen
X~DU (50)
p(X>30) = 20/50 = 0.4
E[X] = 51/2 =25.5
Var(X) = 2499/12 = 208.25
Binomial
Probability distributions - Examples
A marksman fires at a target 10 times, with probability of hitting the target
9/10.
What is the probability that he hits the target exactly 8 times?
What is the probability that he hits the target more than 6 times?
What is the expected value of the number of hits and what is the variance of
the number of hits?
X is the number of hits
X~B(10, 0.9)
p(X=8) = 0.194
p(X>6) = 1 – p(X≤6) = 0.987
E[X] = 10x0.9 = 9
Var(X) = 10x0.9x0.1 = 0.9
Geometric
Probability distributions - Examples
A marksman fires at a target with probability of hitting the target 9/10.
What is the probability that he hits the target for the first time on his
second shot?
What is the probability that he takes at least 5 shots to hit the target for
the first time?
What is the expected value of the number of shots before he hits the target
and what is the variance of the number of shots before hitting the target?
X is the number of shots before he hits the target
X~Geo(0.9)
p(X=2) = 0.09
p(X≥5) = 1 – p(X≤4) = 0.0001
E[X] = 10/9 = 1.11
Var(X) = 0.123
Poisson
Probability distributions - Examples
It is known that on average the number of people visiting a shop in an hour is
17.
What is the probability that exactly 30 people visit the shop in a two hour
period?
What is the probability that less than 50 people visit the shop in the first
three hours of opening?
What is the expected number of people visiting the shop in an 8 hour day, and
what is the variance of the number of people visiting?
X is the number of people visiting in 2 hours
p(X=30) = 0.0568
Y is the number of people visiting in 3 hours
p(Y<50) = p(Y≤49) = 0.426
Z is the number of people visiting in 8 hours
E[Z] = 136
Var(Z) = 136
X~Po(34)
Y~Po(51)
Z~Po(136)
Hypergeometric
Probability distributions - Examples
There are 20 students studying HL maths, of which 14 are girls.
If 8 students from the HL classes are chosen at random to visit the maths
department of a particular university, what is the probability that exactly 3
girls are chosen?
What is the expected number of girls chosen in the group and what is the
variance of the number of girls?
X is the number of girls in the group
X~Hyp(8,14,20)
p(X=3) = 364x6/125970 = 0.0173
E[X] = 5.6
Var(X) = 504/475
Negative Binomial
Probability distributions - Examples
A marksman fires at a target with probability of hitting the target 9/10.
What is the probability that he hits the target for the third time on his
eighth shot?
What is the expected value of the number of shots before he hits the target
for the fifth time and what is the variance?
X is the number of shots before he hits the target for the third time
X~NB(3,0.9)
p(X=8) = 0.000153
Y is the number of shots before he hits the target for the fifth time
Y~NB(5,0.9)
E[Y] = 50/9
Var(Y) = 50/81
Continuous Uniform
Probability distributions - Examples
A number is chosen at random from the real numbers between 0 and 50.
What is the probability that a number over 30 is chosen?
What is the expected value of the number chosen and what is the variance of
numbers chosen?
X is the number chosen
X~U(0,50)
p(X>30) = 20/50 = 0.4
E[X] = 50/2 =25
Var(X) = 2500/12 = 208
Normal
Probability distributions - Examples
The lengths of worms are normally distributed with mean 7.2cm and standard
deviation 2.3cm
Find the probability that a worm chosen at random is greater than 8cm
X is the length of the worm
X~N(7.2, 2.32)
p(X>8) = 0.364
Another species of worm also have their lengths normally distributed and
40% of worms are longer than 8cm and 20% less than 5cm.
Find the mean and standard deviation of the lengths of these worms
Y is the length of the worm
p(Y<8) = 0.6, p(Y<5) = 0.2
8 – m = 0.253347s
5 – m = -0.841621s
m = 7.31
and
s = 2.74
Y~N(m,s2)
Remember Z = (Y – m)/s
So (Y – m)/s = F (0.6)
-1
Exponential
Probability distributions - Examples
It is known that on average the number of people visiting a shop in an hour is
18.
If there is no-one in the shop at present, find the probability that it is more
than 10 minutes before someone comes into the shop.
Find the expected amount of time before someone comes into the shop and
the variance of that time.
X is the amount of time before someone comes into the shop
X~Exp(0.3)

p(X>10) = 0.0498
e - x dx
10
E[X] = 10/3
- x 
Var(X) = 100/9
  -e 


 10
 0  e -3
 0.498
Expectation Algebra
Three coins are tossed and the number of heads is recorded
X is the number of heads
X ~ B(3, 0.5)
x
P(X = x)
0
1/8
1
3/8
2
3/8
3
1/8
Two dice are thrown and the number of sixes recorded
Y is the number of sixes
Y ~ B(2, 1/6)
y
0
1
2
P(Y = y)
25/36
10/36
1/36
x
0
1
2
3
P(X = x)
1/8
3/8
3/8
1/8
y
P(Y = y)
0
25/36
E[X] = 3/2 and Var(X) = 3/4
E[Y] = 1/3 and Var(Y) = 5/18
1
10/36
2
1/36
Remember E[X] = np
And Var(X) = npq
Now what if I want to consider the total number of heads
and sixes in the two experiments
What will be the expectation and variance for the total?
x
0
1
2
3
P(X = x)
1/8
3/8
3/8
1/8
y
P(Y = y)
0
25/36
1
10/36
2
1/36
z=x+y
0
1
2
3
4
5
P(Z = z)
25/288
85/288
106/288
58/288
13/288
1/288
P(Z = 5)
0) == (3/8)(25/36)+(1/8)(10/36)=85/288
1)
3)
2)
4)
(1/8)(25/36)= =1/288
(1/8)(25/36)+(3/8)(10/36)+(3/8)(1/36)
(3/8)(25/36)+(3/8)(10/36)+(1/8)(1/36)
(1/8)(10/36)+(3/8)(1/36)
(1/8)(1/36)
25/288 = 13/288
=58/288
=106/288
So E[X + Y] = 11/6
and Var(X + Y) = 37/36
= E[X] + E[Y]
= Var(X) + Var(Y)
In general:
E[X + Y] = E[X] + E[Y]
Var(X + Y) = Var(X) + Var(Y)
However more generally
E[aX ± bY] = aE[X] ± bE[Y]
Var(aX ± bY) = a2Var(X) + b2Var(Y)
Note only +
Example 1:
The weights of apples are normally distributed with mean 380g and
standard deviation 45g and the weights of bananas are normally
distributed with mean 240g and standard deviation 25g.
A bag contains three apples and 5 bananas. What is the probability that the
bag weighs more than 2.5kg?
Solution:
W = A 1 + A2 + A 3 + B 1 + B 2 + B 3 + B 4 + B 5
and Ak~N(380,452), Bk~N(240,252)
E[W] = E[A1 + A2 + A3 + B1 + B2 + B3 + B4 + B5]
= E[A1] + E[A2] + E[A3] + E[B1] + E[B2] + E[B3] + E[B4] + E[B5]
=3 x 380 + 5 x 240 = 2340
Var(W) = Var(A1 + A2 + A3 + B1 + B2 + B3 + B4 + B5)
= Var(A1) + Var(A2) + Var(A3) + Var(B1) + Var(B2) + Var(B3) + Var(B4) + Var(B5)
=3 x 452 + 5 x 252 = 9200
So W~N(2340, 9200)
P(W>2500) = 0.0476
Example 2:
The weights of apples are normally distributed with mean 380g and
standard deviation 45g and the weights of bananas are normally
distributed with mean 240g and standard deviation 25g.
Apples cost 3 pesos per gram and bananas cost 5 pesos per gram
If I buy an apple and a banana at random, what is the probability I spend
more than 2500 pesos?
Solution:
C = 3A + 5B
and A~N(380,452), B~N(240,252)
E[C] = E[3A + 5B]
= 3E[A] + 5E[B]
=3 x 380 + 5 x 240 = 2340
Var(C) = Var(3A + 5B)
= 9Var(A) + 25Var(B)
=9 x 452 + 25 x 252 = 33850
So W~N(2340, 33850)
P(W>2500) = 0.192
Compare this to example 1
Example 3:
The weights of apples are normally distributed with mean 380g and
standard deviation 45g and the weights of bananas are normally
distributed with mean 240g and standard deviation 25g.
If I buy an apple and a banana at random, what is the probability that the
apple is more than twice as heavy as the banana?
Solution:
D = A - 2B
and A~N(380,452), B~N(240,252)
E[D] = E[A - 2B]
= E[A] - 2E[B]
=380 – 2 x 240 = -100
Var(D) = Var(A - 2B)
= Var(A) + 4Var(B)
=452 + 4 x 252 = 4525
So D~N(-100, 4525)
P(D>0) = 0.0686
The distribution of sample means
When making inferences about a POPULATION we usually have to do so from the data
provided by a SAMPLE.
For example if I want to estimate the number of children per family in Switzerland, I may
take a sample of 50 families to make my estimate.
Suppose the mean of my sample is 2.6 children per family
I can ESTIMATE the mean of the population (all families in Switzerland) to be 2.6
children per family
If someone else did the same thing, they would very likely find a different estimate for the
mean number of children per family
If many people did the same thing, we would have many different sample means.
The CENTRAL LIMIT THEOREM states that
the distribution of sample means of size n is approximately Normal
with mean m and standard deviation
s
n
where m and s are the mean and standard deviation of the population
The distribution of sample means
 s2 
So X ~ N  m,

n


The Central Limit Theorem is exact if the sample is from a Normally distributed
population, and if it not a Normally distributed population, the Central Limit Theorem still
applies as an approximation – the larger the value of n, the better the approximation.
We can see that as n increases, so the standard deviation decreases.
Using the Central Limit theorem
Example 1
On a certain beach, turtles go each year to lay their eggs.
The number of eggs in a nest is known to be normally distributed with mean 114.78 and
standard deviation 22.4
(a) Calculate the probability that the mean number of eggs in a sample of 10 nests is
greater than 120
(b) If a sample of 5 nests is taken, calculate the value of a, such that the probability of the
mean of the sample is within a of the mean is 90%
Solution
2
(a) X ~ N  114.78, 22.4 


10 

22.42 

(b) X ~ N  114.78, 5 



a
p Z 
22.4 /

a  16.6
p  X  120   0.231
p 114.78 - a  X  114.78  a   0.9
p  X  114.78  a   0.95

  0.95
5
a
22.4 /
5
 1.64
The Proportion of success in a large sample
An egg manufacturer claims that eggs delivered to a supermarket have no more than 5%
that are broken. On a certain day, of 1000 eggs delivered to the supermarket, 80 are found to
be broken. What is the probability of this happening? Briefly comment on the
manufacturer’s claim
Method 1 – exact
X are the number of broken eggs in the delivery
X ~ B(1000, 0.05)
p(X ≥ 80) = 3.49 × 10-5
It seems very unlikely that the manufacturer’s claim is correct
The Proportion of success in a large sample
Method 2 – Using the central limit theorem
X are the number of broken eggs in the delivery
X ~ B(1000, 0.05) so E[X] = 50 and Var(X) = 47.5
Now if p 
X
n
 
then E  p   E    E  X   np  p
n
n n
X
 
1
X
and Var p  Var 
n
 pq 
p
~
N
So
 p,

n 

1
1
pq
 1

Var
X

npq




2
n2
n
 n
if n is large (by the Central Limit Theorem)
The Proportion of success in a large sample
So in this example
So

p ~ N  0.05, 0.0000475

p p  0.08  6.72 10-6
It still seems very unlikely that the manufacturer’s claim is correct.
Note that the answer is a little inaccurate due to the approximation of the CLT
The Proportion of success in a large sample
This can also be found on the GDC using the
1-proportion Z test
On the STAT menu find 1-PropZTest
Set
and Calculate
The Proportion of success in a large sample
Another Example
A drug company claims that a new drug cures 75 % of patients suffering from a certain
disease. However, a medical committee believes that less than 75 % are cured. To
test the drug company’s claim, a trial is carried out in which 100 patients suffering
from the disease are given the new drug. It is found that 68 of these patients are
cured. Does the medical committee have the evidence it requires to refute the drug
company’s claim?
Solution
We assume 75% of patients are cured by the drug
Method 1
Number of patients cured is X
X ~ B(100, 0.75)
p(X ≤ 68) = 0.0693
The Proportion of success in a large sample
Another Example
A drug company claims that a new drug cures 75 % of patients suffering from a certain
disease. However, a medical committee believes that less than 75 % are cured. To
test the drug company’s claim, a trial is carried out in which 100 patients suffering
from the disease are given the new drug. It is found that 68 of these patients are
cured. Does the medical committee have the evidence it requires to refute the drug
company’s claim?
Solution
Assuming 75% of patients are cured by the drug
Method 2
p ~ N  0.75, 0.001875 


p p  0.68  0.0530
The Proportion of success in a large sample
Another Example
A drug company claims that a new drug cures 75 % of patients suffering from a certain
disease. However, a medical committee believes that less than 75 % are cured. To
test the drug company’s claim, a trial is carried out in which 100 patients suffering
from the disease are given the new drug. It is found that 68 of these patients are
cured. Does the medical committee have the evidence it requires to refute the drug
company’s claim?
Solution
Assuming 75% of patients are cured by the drug
Method 3
From GDC using 1-PropZTest
p = 0.0530
Hypothesis Testing
In the previous example, what could you conclude?
Was the drug company’s claim valid?
Would the medical committee have sufficient evidence for a lawsuit?
We need to develop a more formal approach to testing the hypothesis
First we write the hypothesis in a formal way.
We say that the NULL HYPOTHESIS is that the company’s claim is
correct – i.e. that 75% of patients are cured
Also the ALTERNATIVE HYPOTHESIS is that LESS than 75% are
cured
Note that we are not interested in this case if MORE than 75% are
cured
This is called a 1-TAIL TEST
Hypothesis Testing
We write:
H0: 75% of patients cured by the drug
H1: Less than 75% of patients cured by the drug
OR
H0: p=0.75
H1: p<0.75
Hypothesis Testing
H0: p=0.75
H1: p<0.75
Now we can find the probability of the result found as before (by any
of the methods)
p ~ N  0.75, 0.001875 


p p  0.68  0.0530
Next we decide on a SIGNIFICANCE LEVEL
If we need to be very sure (e.g. to take out a lawsuit) then we need a
low significance level. A higher level if it is not so important
Note that you will usually be given the significance level
Usually it is 10%, 5%, or 1%
Hypothesis Testing
H0: p=0.75
H1: p<0.75


p p  0.68  0.0530
So if we test at the 10% level
(probably not sure enough for a lawsuit)
Then as 5.3% < 10%, we would reject H0 in favour of H1. The
company’s claim does not seem reasonable
On the other hand if we test at the 1% level
Then as 5.3% > 1%, there is insufficient evidence to reject H0 and
we accept the company’s claim that 75% of patients are cured
by the drug
Hypothesis Testing
In the previous example the standard deviation was known, as the
distribution of number of patients etc. was Binomial. In other cases,
the standard deviation may be known (by other sources) or
estimated from the sample.
These cases give rise to different types of hypothesis testing
Hypothesis Testing
For example
A sample of 10 crates of items is taken from a factory which states
that the mean number of defective items per crate is 12. The sample
mean of the ten crates is found to be 13. Knowing that the standard
deviation of the number of defective items is 1.6, test the
hypothesis that the mean is more than 12:
(a) at the 10% level of significance
(b) at the 1% level of significance
Hypothesis Testing
H0: m = 12
H1: m > 12
 1.62 
X ~ N 12,

10


by the central limit theorem


p X  13  0.0241
(a) so testing at the 10% level of significance we see that 2.5%<10%
so we reject H0 in favour of H1, the mean number of defective
items is greater than 12
(b) testing at the 1% level of significance we see that 2.5%>1% so
there is insufficient evidence to reject H0. We accept that the
manufacturer’s claim that the mean number of defective items is
12
Hypothesis Testing
Or!
H0: m = 12
H1: m > 12


p X  13  0.0241
(a) so testing at the 10% level of significance we see that 2.5%<10%
so we reject H0 in favour of H1, the mean number of defective
items is greater than 12
(b) testing at the 1% level of significance we see that 2.5%>1% so
there is insufficient evidence to reject H0. We accept that the
manufacturer’s claim that the mean number of defective items is
12
Hypothesis Testing
Another example
A sample of 5 crates of items is taken from a factory which states
that the mean number of defective items per crate is 12. The sample
is 12, 13, 11, 14, 13. Knowing that the standard deviation of the
number of defective items is 1.6, test the hypothesis at the 5% level
of significance
Solution
H0: m = 12
H1: m > 12
Sample mean is 12.6 (from GDC)


 1.62 
X ~ N 12,
 p X  12.6  0.201
5 

20%>5% so there is insufficient evidence to reject H0. We accept that
the mean number of defective items is 12
Or from GDC using Z-Test
Hypothesis Testing
Now if the standard deviation of the population is NOT known, and
has to be estimated from the sample, then the distribution is altered.
Rather than being Normal, the distribution follows that of a
Student-t distribution.
The standardised Student-t distribution has parameter u, known as
the degrees of freedom
The degrees of freedom is generally n - 1
Hypothesis Testing
For example
A sample of 10 crates of items is taken from a factory which states
that the mean number of defective items per crate is 12. The sample
mean of the ten crates is found to be 13 and the estimate of the
standard deviation of the number of defective items is 1.6, test the
hypothesis that the mean is more than 12:
(a) at the 10% level of significance
(b) at the 1% level of significance
Hypothesis Testing
H0: m = 12
Note the degrees of freedom is 9
and S is the estimate of the
population standard deviation
H1: m > 12
X -m
T
~ t 9
S
10

10 
p X  13  p  T 
  0.0398
1.6 



13 - 12
10

 1.6  1.6


 10 
(a) so testing at the 10% level of significance we see that
3.98%<10% so we reject H0 in favour of H1, the mean number of
defective items is greater than 12
(b) testing at the 1% level of significance we see that 3.98%>1% so
there is insufficient evidence to reject H0. We accept that the
manufacturer’s claim that the mean number of defective items is
12
Or use T-Test on the GDC
Hypothesis Testing
Error Types
Falsely rejecting H0, is known as a Type I error
Falsely accepting H0, is known as a Type II error
So in the example used previously:
..testing at the 10% level of significance we see that 2.5%<10% so we reject H0 in favour
of H1, the mean number of defective items is greater than 12.
There is a 2.5% risk of committing a type I error
Confidence intervals
A confidence interval is an interval in which we are certain to a
given probability of the mean lying in that interval.
For example if a 95% confidence interval is given, then there would
be a 95% probability that the mean is in that interval.
Example 1
On a given busy day, 1000 eggs are delivered to a supermarket and
80 are broken. Find the 95% confidence interval for the mean
percentage of broken eggs.
Confidence intervals
Example 1
On a given busy day, 1000 eggs are delivered to a supermarket and
80 are broken. Find the 95% confidence interval for the mean
percentage of broken eggs.
Method 1
 pq 
As before p ~ N  p,

n 

p ~ N  0.08, 0.0000736 
We can estimate p as 0.08 so
Now if p 0.08 - a  p  0.08  a  0.95
Then p p  0.08  a  0.975




so
a  0.0168
and the confidence interval is [6.32%, 9.68 %]
Confidence intervals
Example 1
On a given busy day, 1000 eggs are delivered to a supermarket and
7% are broken. Find the 95% confidence interval for the mean
percentage of broken eggs.
Method 2
From formula book
For a 95% confidence interval we take z=InvNorm(0.975)=1.96
So applying the formula we find the confidence interval to be
[6.32%, 9.68%]
Method 3
From GDC using 1-PropZInt, the interval is [6.32%, 9.68%]
Confidence intervals
Example 2
To test a drug company’s claim, a trial is carried out in which
100 patients suffering from a disease are given a new drug. It is
found that 68 of these patients are cured. Find a 90% confidence
interval for the proportion of patients cured by the drug.
Method 1
p ~ N  0.68, 0.002176 




p 0.68 - a  p  0.68  a  0.90  p p  0.68  a  0.95
a = 0.0767
So the confidence interval is [0.603, 0.757]
Or either of the other two methods!
Example 3
A sample of 10 crates of items is taken from a factory and the mean
of the ten crates is found to be 13. Knowing that the standard
deviation of the number of defective items is 1.6, find a 95%
confidence interval for the mean number of defective items in a crate.
Solution
 1.62 
X ~ N 13,

10 



p X  13  a  0.975
 

a 10 
a = 0.99167
 p  Z 
  0.975
1.6 
 

So the confidence interval is [12.01, 13.99]
•Or use the formula in the formula book
•Or use ZInterval on the GDC
Confidence intervals
Again if the standard deviation of the population is NOT known,
and has to be estimated from the sample, then the distribution is
altered to that of a Student-t distribution.
Example 4
A sample of 10 crates of items is taken from a factory and the
sample mean of the ten crates is found to be 13 and the estimate
of the standard deviation of the number of defective items is 1.6.
Find a 90% confidence interval for the mean number of
defective items.
Solution
X - 13
~ t (9) p (X < 13 + a ) = 0.95
As before T = 1.6
10
æ
ö
a 10 ÷
ç
÷
p çT <
= 0.95
÷
çè
1.6 ÷
ø
so a = 0.927
(1.833 × 1.6 / 10)
(from tables or solver)
So the confidence interval is [12.07, 13.93]
• Or use the formula in the formula book
• Or use TInterval on the GDC
The 2– Goodness of fit test
A dice is thrown 60 times and the following frequencies are obtained
Score
1
2
3
4
5
6
Frequency
12
13
10
11
11
3
Is there evidence to suggest that the dice is biased?
How can we calculate the probability of results as extreme as these in
order to test a hypothesis?
2
We use the fact that the quantity
 Observed Frequency - Expected frequency 
Expected Frequency
approximately follows a 2 probabilitydistribution
Note that if any of the expected frequencies are less than
5, then the chi-squared distribution may not be a
reasonable approximation
The 2– Goodness of fit test
So following standard hypothesis testing procedures
H0: the dice is fair – i.e. X~UD(6) where X is the score
H1: the dice is biased – i.e. the distribution is not uniform
Observed
frequency
Expected
Frequency
1
12
10
2
13
10
3
10
10
4
11
10
5
11
10
6
3
10
Also this can be done on a TI84
So

2
calc

 fo - fe 
2
fe
 6.4
taking u = 5
2
p   calc
 6.4   0.269
So testing at the 5% significance
level, 27%>5% so there is insuficient
evidence to reject H0. We accept
that the dice is not biased.
More on Degrees of freedom
u = no of classes – no of restrictions – no of estimated parameters
So often u = n – 1 as there is a restriction on the total frequency (as in
the previous example)
Another example
Data is collected as below:
x
0
1
frequency
10
16
2
3
4
5
12
7
3
2
Test at the 5% significance level to see if the data is from a Poisson
distribution..
H0: X~Po(m)
H1: X is not Poisson
Estimate m from the data to be 1.66
Note that last expected
frequency is that of ≥5
u =6–1–1=4
Expected frequencies are 9.51, 15.78, 13.10, 7.25, 3.01, 1.36
Now we notice that the last two expected frequencies are
less than 5 and so will have to be grouped together
Regrouping the table
x
0
1
2
≥3
Observed
frequency
10
16
12
12
Expected
frequency
9.51
15.78
13.10
11.61
So we change u = 4 – 1 – 1 = 2
2
 calc
 0.134
2
p (c calc
> 0.134) = 0.935
So 94%>5% so there is insufficient evidence to reject H0. We accept that the
data is distributed by a Poisson distribution.
The 2– Independence test
100 Tennis players are studies to see if their style is dependent on
their nationality
The results were as follows
Eastern Europe
Western Europe
Americas
Serve/Volley
8
15
5
Baseline
11
38
23
Is there evidence to suggest that tennis style is dependent on
nationality?
Again we use the fact that the quantity
2
 Observed Frequency - Expected frequency 

Expected Frequency
approximately follows a 2 probability distribution
The 2– Independence test
The expected frequencies can be calculated
Eastern Europe
Western Europe
Americas
Total
Serve/Volley
28*19/100=5.32
8
28*53/100=14.84
15
28*28/100=7.84
5
28
Baseline
72*19/100=13.68
11
72*53/100=38.16
38
72*28/100=20.16
23
72
19
53
28
100
Now, as before following procedure for significance testing
H0: Style is independent of nationality
H1: Style depends on nationality
Or we can use 2-Test
u = (row – 1)(col – 1) = 2
on the GDC
2
 calc
 3.306
2
p   calc  3.306   0.191
Testing at the 10% significance level, 19%>10% so there is
insufficient evidence to reject H0. We accept that tennis style is
independent of nationality.
The 2– Independence test
Put in the observed frequencies into matrix A
H0: Style is independent of nationality
H1: Style depends on nationality

2
calc
 3.306
2
p   calc
 3.306   0.191 (with df=2)
(and expected frequencies are in matrix B if required)
Testing at the 10% significance level, 19%>10% so there is
insufficient evidence to reject H0. We accept that tennis style is
independent of nationality.