Notes for Grade 12 Maths Studies on the 𝛘𝟐 test of independence
1)
State 𝐻0 , the null hypothesis: variable x is independent of variable y.
(State 𝐻1 , the alternative hypothesis: variable x is not independent of variable
y.)
NB the alternative hypothesis is NOT that x is dependent on y.
2
2)
State the rejection inequality 𝜒𝑐𝑎𝑙𝑐
> 𝑘, where k is the critical value of 𝜒 2 .
3)
Construct the expected frequency table.
4)
Find χ2 , use the calculator in an exam, but do the table in your project.
5)
Either accept or reject 𝐻𝑜 using the critical value or the p-value.
6)
Interpret your results!
Example 1:
There are 60 students in a year group, of whom 30 are male and 30 are female.
Data was collected about which students study which science. You may assume that
each student studies only one science. 36 study Biology, 16 study Chemistry and 8
study Physics.
The NULL HYPOTHESIS, 𝐻0 , would be that science subject studied is independent of
gender. So, the expected values would be as shown in the table below.
Male
Biology
15
Chemistry
8
Physics
7
30
This is the expected frequency table.
Female
15
8
7
30
Totals
30
16
14
60
However, the figures show that the students do not study as expected.
There are 22 females studying Biology, and 5 studying Chemistry.
Use the information, and the χ2 test of independence to determine whether the null
hypothesis 𝐻0 , is accepted or rejected.
Biology
Chemistry
Physics
Male
8
11
11
30
Female
22
5
3
30
Totals
30
16
14
60
Only two more pieces of information were necessary to fix the table of observed values.
There are 2 degrees of freedom.
Now, the expected values and the observed values are different. But are they
significantly different?
𝑓𝑜
𝑓𝑒
𝑓𝑜 − 𝑓𝑒
(𝑓0 − 𝑓𝑒 )2
8
22
11
5
11
3
15
15
8
8
7
7
-7
7
3
-3
4
-4
49
49
9
9
16
16
Total
(𝑓0 − 𝑓𝑒 )2
𝑓𝑒
3.267
3.267
1.125
1.125
2.286
2.286
13.355
2
So for this situation 𝜒𝑐𝑎𝑙𝑐
=13.355.
We look at the table of critical values in your text book and see that for 2 degrees of
freedom at 5% significance, the critical value is 5.99.
2
Since 13.355> 5.99, then 𝜒𝑐𝑎𝑙𝑐
> critical value, then we reject the null hypothesis and
accept the alternative hypothesis.
Thus we can say that at 5% significance Science studied is not independent of gender.
In an examination you would need to be able to do this on a calculator. You would not
have a table of values to look up, so you would use the p-value.
To do χ2 test on the calculator, enter your table of observed values as a matrix.
[2nd, 𝑥 −1, edit, 1, rows, columns, then go to Stat, test, c: χ2 test]
You will see χ2 , p and df.
In this case, from the calculator, the p value is 0.001259, which is less the 0.05 (5% in
decimal), so we reject the null hypothesis.
Example 2:
Of 60 grade 12 students, 30 males and 30 females, 34 play basketball, of whom 22 are
boys. Is playing basketball independent of gender?
𝐻𝑜 : playing basketball is independent of gender.
𝐻1 : playing basketball is not independent of gender
Observed Frequency table (contingency table):
Male
Female
Plays basketball
20
14
34
Does Not Play BB
10
16
26
30
30
60
One degree of freedom (r-1)(c-1)=1
Expected frequency table:
Male
Female
Plays basketball
34 × 30
= 17
60
34 × 30
= 17
60
34
Does Not Play BB
26 × 30
= 13
60
26 × 30
= 13
60
26
30
30
60
2
Since the degree of freedom is only one we have to find 𝜒𝑐𝑎𝑙𝑐
using the Yates’ continuity
correction.
𝑓𝑜
𝑓𝑒
𝑓𝑜 − 𝑓𝑒
|𝑓𝑜 − 𝑓𝑒 |
|𝑓𝑜 − 𝑓𝑒 | − 0.5
20
10
14
16
17
13
17
13
3
-3
-3
3
3
3
3
3
2.5
2.5
2.5
2.5
(|𝑓𝑜 − 𝑓𝑒 | − 0.5)2
(|𝑓𝑜 − 𝑓𝑒 | − 0.5)2
𝑓𝑒
6.25
0.8929
6.25
0.4808
6.25
0.8929
6.25
0.4808
Total
2.7474
Using the table shown in example 1, the critical value at 5% significance for 1 degree of
freedom is 3.84.
2
Since 𝜒𝑐𝑎𝑙𝑐
≈ 2.75, which is < 3.84, we do not reject 𝐻0 .
NB The calculator does not use the Yates’ continuity correction, so gives a different
2
value of 𝜒𝑐𝑎𝑙𝑐
, but in this case it still leads to the same conclusion, i.e. that we do not
reject 𝐻0 .
We conclude that at a 5% level of significance, the variables gender and playing
basketball are independent.