Transcript Inferential Statistics 3: The Chi Square Test

Advanced Higher Geography
Statistics
Inferential Statistics 3:
The Chi Square Test
Ollie Bray – Knox Academy, East Lothian
Introduction (1)
 We often have occasions to make
comparisons between two characteristics
of something to see if they are linked or
related to each other.
 One way to do this is to work out what we
would expect to find if there was no
relationship between them (the usual null
hypothesis) and what we actually observe.
Introduction (2)
 The test we use to measure the
differences between what is observed
and what is expected according to an
assumed hypothesis is called the chisquare test.
For Example
 Some null hypotheses may be:
– ‘there is no relationship between the height
of the land and the vegetation cover’.
– ‘there is no difference in the location of
superstores and small grocers shops’
– ‘there is no connection between the size of
farm and the type of farm’
Important
 The chi square test can only be used on
data that has the following
characteristics:
The data must be in the form
of frequencies
The frequency data must have a
precise numerical value and must be
organised into categories or groups.
The expected frequency in any one cell
of the table must be greater than 5.
The total number of observations must be
greater than 20.
Formula
χ 2 = ∑ (O – E)2
E
χ2 = The value of chi square
O = The observed value
E = The expected value
∑ (O – E)2 = all the values of (O – E) squared then added
together
 Write down the NULL HYPOTHESIS and
ALTERNATIVE HYPOTHESIS and set
the LEVEL OF SIGNIFICANCE.
 NH ‘ there is no difference in the distribution of old
established industries and food processing industries in
the postal district of Leicester’
 AH ‘There is a difference in the distribution of old
established industries and food processing industries in
the postal district of Leicester’
 We will set the level of significance at 0.05.
Construct a table with the information you have observed
or obtained.
Observed Frequencies (O)
Post
Codes
LE1
LE2
LE3
LE4
LE5 &
LE6
Row
Total
Old
Industry
9
13
10
10
8
50
Food
Industry
4
3
5
9
21
42
Column
Total
13
16
15
19
29
92
(Note: that although there are 3 cells in the table that are not greater than 5,
these are observed frequencies. It is only the expected frequencies that have to
be greater than 5.)
 Work out the expected frequency.
Expected frequency = row total x column total
Grand total
Eg: expected frequency for old industry in LE1 = (50 x 13) / 92 = 7.07
Post
Codes
LE1
Old
Industry
7.07
Food
Industry
Column
Total
LE2
LE3
LE4
LE5 &
LE6
Row
Total
Post
Codes
LE1
LE2
LE3
LE4
LE5 &
LE6
Row
Total
Old
Industry
7.07
8.70
8.15
10.33
15.76
50
Food
Industry
5.93
7.30
6.85
8.67
13.24
42
Column
Total
13
16
15
19
29
92
 For each of the cells calculate.
(O – E)2
E
Eg: Old industry in LE1 is (9 –
7.07)2 / 7.07 = 0.53
Post
Codes
LE1
Old
Industry
0.53
Food
Industry
Column
Total
LE2
LE3
LE4
LE5 &
LE6
Row
Total
Post
Codes
LE1
LE2
LE3
LE4
LE5 &L
E6
Old
Industry
0.53
2.13
0.42
0.01
3.82
Food
Industry
0.63
2.54
0.50
0.01
4.55
Add up all of the above numbers to obtain
the value for chi square: χ2 = 15.14.
 Look up the significance tables. These will
tell you whether to accept the null
hypothesis or reject it.
The number of degrees of freedom to use is: the number
of rows in the table minus 1, multiplied by the number of
columns minus 1. This is (2-1) x (5-1) = 1 x 4 = 4 degrees of
freedom.
We find that our answer of 15.14 is greater than the
critical value of 9.49 (for 4 degrees of freedom and a
significance level of 0.05) and so we reject the null
hypothesis.
‘The distribution of old established industry and
food processing industries in Leicester is
significantly different.’
Now you have to look for geographical factors
to explain your findings
Your Turn
 Read page 46, 47 and 48 of ‘Geographical
Measurements and Techniques: Statistical
Awareness, by LT Scotland, June 2000.
 Answer Task 1 on page 48.