2. OBJECTIVES
• Recognise a suitable distribution to apply chi
square test to
• Conduct the goodness-of-fit test of hpothesis
• Conduct the test of independence
• Conduct a test of homgeneity
2
3. Chi square distribution
• Positively skewed
• Test done on right tail only
• Therefore all chi square tests are positive with
one critical value only
• Basic steps of hypothesis test are the same, only
the test statistic and distribution have changed
3
4. • Techniques used to analyse data up to now was
measured on quantitative scale.
• Results of tests can often be classified into categories
where there is no natural order:
– Categorical variable
– Categories
– Categorical data
• Categorical data can be analysed with Chi-squared
tests:
– Simple random sample
– Sample size reasonable large
4
5. Example:
• Survey of job satisfaction
• Employed persons classified as satisfied, neutral,
dissatisfied
CATEGORICAL VARIABLE – is employee satisfaction
CATEGORIES – satisfied, neutral, dissatisfied
CATEGORICAL DATA – no. of employees satisfied,
neutral or dissatisfied (also referred to as frequency of
category)
5
6. Examples
1. A persons income can be categorised as high,
medium or low. Define the categorical variable,
the categories and the categorical data
2. We want to investigate different types of
industries, e.g. information technology,
financial and transformation. Define the
categorical variable, the categories and the
categorical data
6
7. Example answers
1. Categorical variable is income. Categories are
high, medium and low. Categorical data are the
no. of people who have high, medium or low
income
2. Categorical variable is type of industry,
categories are information technology, financial
and transformation. Categorical data are the no
of industries that are information tech, financial
or transformation
7
8. • Chi-squared goodness-of-fit test
– This test describes a single population of categorical data.
– The multinomial experiment studied is an extension of the
binomial experiment.
• There are n independent trials.
• The outcome of each trial can be classified into one of
k categories.
• The probability pi of cell i remains constant for each
trial. Moreover, p1 + p2 + … +pk = 1.
– Experiment records the observed trails for each category.
– Denoted by f1, f2, …, fk and f1 + f2 + … + fk = n
8
9. EXAMPLE
In a box of smarties you will find 6 different colours:
brown, red,yellow,blue,orange and green. A
random sample of smarties (6918 in total) was
taken and the frequesncy of each colour was
counted. The distribution of colours is given below
Colour Brown Red Yellow Blue Orange Green
f 1611 1172 1308 904 921 1002
Determine whether the smarties survey fits the
description of a multinomial experiment
9
11. To use the Χ2-tests
• The goodness-of-fit test all expected
frequencies must
– Used to determine if the observed counts ofbe at least 5
the
categories agree with the probabilities specified for
each category.
– Observed frequencies (f ) compared with the expected
frequencies (e).
Testing H0: Proportions agree with specified probabilities
Alternative Decision rule:
Test statistic
hypothesis Reject H0 if …
( fi ei )2
H1: H0 is not true Χ2 > Χ2k – 1;1 – α 2
ei 11
12. • Example
– A household detergent is marketed in three sizes:
• 1 000 ml, 750 ml and 250 ml
– The distributers belief that the market share of the different sizes is
as follow:
• 1 000 ml = 40%
• 750 ml = 45%
• 250 ml = 15%.
– To study the effect of the economic climate on the sales of the
products, 200 customers were ask to state which size they will
prefer.
• Survey results:
– 82 customers preferred the 1 000 ml
– 102 customers preferred the 750 ml
– 16 customers preferred the 250 ml
12
13. • Solution
– The population investigated is the size preferences.
– The data are in categories.
– This is a multinomial experiment (three categories).
– The question of interest: Are p1, p2, and p3 different
from the expected 40%, 45% and 15%?
13
14. • The hypotheses are: Expected
frequencies are
– H0: p1 = 0,40, p2 = 0,45, p3 = 0,15 all ≥ 5
– H1: At least one pi is not equal to its specified value.
Are the observed and the expected frequencies the same?
Expected
Expected frequencies Observed values frequencies
16
15%
ei = npi
40% 1000ml 82 1000ml 40% of 200 = 80
750ml 750ml
250ml 250ml
45% of 200 = 90
102
45% 15% of 200 = 30
14
15. • The hypotheses are:
– H0: p1 = 0,40, p2 = 0,45, p3 = 0,15
– H1: At least one pi is not equal to its specified value.
Are the observed and the expected frequencies the same?
Expected
frequencies
Expected frequencies Observed values
30
15%
16 ei = npi
40% of 200 = 80
80
40% 1000ml 82 1000ml
750ml 750ml 45% of 200 = 90
250ml 250ml
15% of 200 = 30
102
90
45% 15
16. • The hypotheses are:
– H0: p1 = 0,40, p2 = 0,45, p3 = 0,15
– H1: At least one pi is not equal to its specified value.
( fi ei ) 2
2 α = 0,05
ei
0 5,9917
Χ2k – 1;1 – α
(82 80) (102 90) (16 30)
2 2 2
80 90 30 Accept H0 Reject H0
8,18 Conclusion: At 5% significance level there is
– Reject H0. sufficient evidence to reject the null hypothesis.
At least one of the probabilities pi is different.
Thus, at least two market shares have changed. 16
17. Two friends were playing a board game in which a die played a
big role. One of the players believed that the die was not fair.
60 tosses of the die produced the results below. Test at 5%
significance level whether the die was fair.
Number of dots 1 2 3 4 5 6
Number of tosses 7 6 7 18 15 7
17
18. ei = npi
= 60(1/6)
= 10
Expected values for the six categories are: 10 10 10 10 10 10
H0: p1 = … = p6 = 1/6
H1: At least one pi ≠ 1/6
= 0,05
f e
2
=
2
e
7 10 + … +
2
(7 10) 2
=
10 10
= 13,2 Accept H0 Reject H0
2
k 1;1
2
= 5; 0,95
= 11,07
Therefore, reject H0. The probabilities of the dots are not equal and the die was not
fair.
18
19. • Chi-squared test for independence
– Cross classify two categories using a contingency table.
– Rows representing one category and columns
representing the other category.
– Each value in cell indicates the frequency in the cross
classification.
– Table can be any number of rows and columns:
• r×c number of cells
19
21. • Chi-squared test for independence
– H0: the two variables are independent – no relationship.
– H1: the two variables are dependent – is a relationship.
For a 2×2 contingency table:
B
A B1 B2 Total
A1 f11 f12 r1
Observed
A2 f21 f22 r2 frequencies
Total c1 c2 n
21
22. • Chi-squared test for independence
– Contingency tables describe the relationship between two
categorical variables.
– H0: the two variables are independent – no relationship.
– H1: the two variables are dependent – is a relationship.
For a 2×2 contingency table: For each observed
B
frequency an expected
frequency must be
A B1 B2 Total calculated
A1 f11 f12 r1
A2 f21 f22 r2 row total × column total
e=
Total c1 c2 n n 22
23. • Chi-squared test for independence
– Contingency tables describe the relationship between two
categorical variables.
– H0: the two variables are independent – no relationship.
– H1: the two variables are dependent – is a relationship.
For a 2×2 contingency tabel:
B
row total × column total
A B1 B2 Total e=
n
A1 f11 f12 r1
A2 f21 f22 r2
e11 (r1 c1 ) / n ; e12 (r1 c2 ) / n
Total c1 c2 n e21 (r2 c1 ) / n ; e22 (r2 23 2 ) / n
c
24. • Chi-squared test for independence
– H0: the two variables are independent – no relationship.
– H1: the two variables are dependent – is a relationship.
Testing H0: Variables are independent
Alternative Decision rule:
Test statistic
hypothesis Reject H0 if …
( f e) 2
H1: Variables are
Χ2 > Χ2(r – 1)(c – 1);1 – α 2
dependent
e
24
25. • Example
– A household detergent is marketed in three sizes:
• 1 000 ml, 750 ml and 250 ml
– The market for potential buyers is divided into three
age groups:
• < 30 years old
• 30–50 years old
• > 50 years old
– Market researcher believe that there is a relationship
between the age of a buyer and the size of the
packaging. 25
26. • Solution
– The data is summarised in a 3×3 contingency table.
– H0: Size and age are independent.
Observed
– H1: Size and age are dependent. frequencies
Age groups
Size < 30 30–50 > 50 Total
1 000 ml 27 41 14 82
750 ml 39 18 45 102
250 ml 8 2 6 16
Total 74 61 65 200 26
27. • Solution
– Calculate the expected frequency
– (Row total×column total)/n Expected frequency:
(74×82)/200 = 30,34
Age groups
Size < 30 30–50 > 50 Total
1 000 ml 27 30,34 41 25,01 14 26,65 82
750 ml 39 37,74 18 31,11 45 33,15 102
250 ml 8 5,92 2 4,88 6 5,20 16
Total 74 61 65 200
27
28. Χ2(r – 1)(c – 1);1 – α =
Χ2(3-1)(3-1);0.95 = 9.49
• The hypotheses are:
– H0: Size and age are independent
α = 0,05
– H1: Size and age are dependent
0 9,49
( f e) 2
2
e
Accept H0 Reject H0
(27 30,34) 2 (41 25, 01) 2 (6 5, 20) 2
.....
30,34 25, 01 5, 20
28,95
– Reject H0. Conclusion: At 5% significance level there is
sufficient evidence to reject the null hypothesis.
There is a relationship between the size of detergent
that people prefer and their age. 28
29. A recent survey of marketing managers in four different industries provided
the data in the table below, which gives managers attitudes to market
research and its value in marketing decision making:-
INDUSTRY TYPE
Perceived value Consumer Industrial Retail & Finance &
of M Research businesses organisations wholesale insurance
Little value 9 22 13 9
Moderate value 29 41 6 17
Great value 26 28 6 27
TOTAL 64 91 25 53
Test at 1% level of significance whether manager’s perception of the value
of the market research is dependent on the type of industry in which a
marketing manager is employed. 29
30. Industry type
Perceived value of market Consumer Industrial Retail and Finance Total
and
research businesses organisations wholesale insurance
Little value 9 (14,56) 22 (20,7) 13 (5,69) 9 (12,06) 53
Moderate value 29 (25,55) 41 (36,32) 6 (9,98) 17 (21,15) 93
Great value 26 (23,9) 28 (33,98) 6 (9,33) 27 (19,79) 87
Total 64 91 25 53 233
H0: Manager’s perception is independent of industry type.
H1: Manager’s perception is dependent of industry type.
= 0,01
f e
2
=
2
e
9 14,56 + … + 27 19,79
2 2
=
14,56 19,79
= 20,895
2
r 1c 1;1
Accept H0 Reject H0
= 2
6; 0,99
= 16,81
Therefore, reject H0. Manager’s perception is dependent on the industry type. 30
32. • Chi-squared Test of Homogeneity
– Test if two or more populations are homogeneous (similar)
with regard to a certain characteristic.
– H0: The proportion of elements with certain characteristic in
two or more different populations are the same.
– H1: The proportion of elements with certain characteristic in
two or more different populations are not the same.
– The rest of the test is the same as the test for
independence.
32
33. An immigration attorney was investigating which industries to
target for obtaining new clients who might have problems with
change in the immigration laws. The lawyer selected five
industries and twenty workers were randomly selected in each
industry and their visa statuses were verified.
VISA STATUS INDUSTRY
A B C D E
Illegal resident 8 10 5 10 1
Legal resident 4 2 6 4 9
SA citizen 8 8 9 6 10
Test at a 1% level of significance whether the 5 industries are
homogeneous with respect to the visa status of their workers
33
34. Industry
Visa status Total
A B C D E
Illegal resident 8 (6,8) 10 (6,8) 5 (6,8) 10 (6,8) 1 (6,8) 34
Legal resident 4 (5) 2 (5) 6 (5) 4 (5) 9 (5) 25
SA citizen 8 (8,2) 8 (8,2) 9 (8,2) 6 (8,2) 10 (8,2) 41
Total 20 20 20 20 20 100
H0: Five industries are homogeneous with respect to the visa status of their
workers.
H1: Five industries are heterogeneous with respect to the visa status of their
workers.
= 0,01
f e
2
2 = e
8 6,8 + … + 10 8,2
2 2
=
6,8 8,2
= 15,32
2
r 1c 1;1
= 2
8; 0,99
= 20,09 34
Therefore, do not reject H0. The five industries are homogeneous with respect to
the visa status of their workers.
