1. The 2 (chi-squared) test for independence
IB Math Studies SL
West Hall High School

2. A random sample of 200 teachers in higher education, secondary schools and primary schools gave the following numbers of men and women in each sector:
Higher Education
Secondary Education
Primary Education
Male 21 39 20
Female 13 55 52
We might want to find out whether or not there is an association between ‘age-group taught’ and ‘gender’.
One way of finding out is to perform a 2 (chi-squared) test for independence.
To set up the test:
We first set up a null hypothesis, H0, and an alternative hypothesis, H H0 always states that the data sets are independent, and H1 always states that they are related.
In this case, H0 could be “The age-group taught is independent of gender” H1 could be “There is an association between age-group taught and gender.”

3. We put the data into a table
We put the data into a table . The elements in the table are our observed data and the table is known as a contingency table.
Higher Education
Secondary Education
Primary Education
Male 21 39 20
Female 13 55 52

4. We put the data into a table
We put the data into a table . The elements in the table are our observed data and the table is known as a contingency table.
Higher Education
Secondary Education
Primary Education
TOTAL
Male 21 39 20
Female 13 55 52 80
120 34 94 72
200

5. From the observed data we can calculate the expected frequencies.
We put the data into tables. The elements in the table are our observed data and the table is known as a contingency table.
Higher Education
Secondary Education
Primary Education
TOTAL
Male 21 39 20 80
Female 13 55 52
120 34 94 72
200
From the observed data we can calculate the expected frequencies.
The expected frequency for each cell will be:
row total x column total
total sample size
Higher Education
Secondary Education
Primary Education
TOTAL
Male 80
Female
120 34 94 72
200
13.6
37.6

6. This gives us the degree of freedom for this table - it is 2
In fact for this table we only need to actually work out two of the expected values, and the rest will follow from the totals.
This gives us the degree of freedom for this table - it is 2
The expected frequency for each cell will be:
row total x column total
total sample size
Higher Education
Secondary Education
Primary Education
TOTAL
Male 80
Female
120 34 94 72
200
13.6
37.6
28.8
20.4
56.4
43.2

7. In fact for this table we only need to actually work out two of the expected values, and the rest will follow from the totals.
This tells us that the degree of freedom for this table is 2
You can always find the degree of freedom by going back to the original table (without the totals). Crossing off one column and one row, and the number of cells left is the degree of freedom. (No. of columns – 1) x (No. of rows – 1)
Higher Education
Secondary Education
Primary Education
Male 21 39 20
Female 13 55 52
df = 2
Higher Education
Secondary Education
Primary Education
TOTAL
Male 80
Female
120 34 94 72
200
13.6
37.6
28.8
20.4
56.4
43.2

8. 2calc fo is the observed value fe is the expected value 2calc
Contingency Table – Observed Data
Expected Frequencies
Higher Education
Secondary Education
Primary Education
Male 21 39 20
Female 13 55 52
Higher Education
Secondary Education
Primary Education
Male
13.6
37.6
28.8
Female
20.4
56.4
43.2
Now we are ready to calculate the 2 value using the formula:
2calc
fo is the observed value
fe is the expected value
2calc
Finally look at the critical value that you have been given.
If the 2 calc value is less than the critical value, we accept H0, the null hypothesis.
If the 2 calc value is more than the critical value, we do not accept the null hypothesis, so we accept H1
In this case the 2 calc value is 11.3, and the critical value at 5% is So we do not accept H0, the null hypothesis.
There is an association between age-group taught and gender.

9. If the 2 calc value is less than the critical value, we do accept H0, the null hypothesis. (The 2 calc value is small – there is nothing of significance going on!)
If the 2 calc value is more than the critical value, we do not accept the null hypothesis, so we accept H (The 2 calc value is large – there is something of significance going on!)
If the p-value is less than the significance level, we do not accept H0, the null hypothesis. We accept H (The probability of this happening just by chance is small – there is probably something of significance going on!)
If the p-value is more than the significance level, we do accept the null hypothesis, so we accept H (The probability of this happening just by chance is large – there is probably nothing of significance going on!)

10. 2 is given to you. p is the probability df is the degree of freedom
You can do all this on the GDC:
Enter the data into a Matrix
MATRIX
ENTER
[EDIT]
Enter the size of your matrix ; in this case 2 x 3 (2 rows, 3 columns)
Enter your data, pressing after every value.
ENTER
STAT
[TESTS]
Scroll up to find 2
ENTER
You will now see where your table of expected values will be ; change it if you wish. Otherwise scroll down to Calculate and
ENTER
2 is given to you. p is the probability df is the degree of freedom
To see the table of expected values:
ENTER
MATRIX
Finally look at the critical value that you have been given.
If the 2 calc value is less than the critical value, we accept the null hypothesis.
If the 2 calc value is more than the critical value, we do not accept the null hypothesis, so we accept H1

11. Suppose we collect data on the favourite colour of car for men and women.
Black
White
Red
Blue
Male 51 22 33 24
Female 45 36 27
We may want to find out whether favourite colour of car and gender are independent or related.
One way of finding out is to perform a 2 (chi-squared) test for independence.
To set up the test:
We first set up a null hypothesis, H0, and an alternative hypothesis, H H0 always states that the data sets are independent, and H1 always states that they are related.
In this case, H0 could be “The favourite colour of car is independent of gender” H1 could be “There is an association between favourite colour of car and gender.”

12. Black
White
Red
Blue
TOTAL
Male 51 22 33 24
130
Female 45 36 27
130 96 58 55 51
260

13. From the observed data we can calculate the expected frequencies.
Black
White
Red
Blue
TOTAL
Male 51 22 33 24
130
Female 45 36 27 96 58 55
260
From the observed data we can calculate the expected frequencies.
The expected frequency for each cell will be:
row total x column total
total sample size
Black
White
Red
Blue
TOTAL
Male
130
Female 96 58 55 51
260 48 29
27.5

14. This gives us the degree of freedom for this table - it is 3
In fact for this table we only need to actually work out three of the expected values, and the rest will follow from the totals.
This gives us the degree of freedom for this table - it is 3
The expected frequency for each cell will be:
row total x column total
total sample size
Black
White
Red
Blue
TOTAL
Male 48 29
27.5
130
Female 96 58 55 51
260
25.5 48 29
27.5
25.5

15. In fact for this table we only need to actually work out two of the expected values, and the rest will follow from the totals.
This tells us that the degree of freedom for this table is 2
You can always find the degree of freedom by going back to the original table (without the totals). Crossing off one column and one row, and the number of cells left is the degree of freedom. (No. of columns – 1) x (No. of rows – 1)
Black
White
Red
Blue
Male 51 22 33 24
Female 45 36 27
df = 3
Black
White
Red
Blue
TOTAL
Male 48 29
27.5
25.5
130
Female 96 58 55 51
260

16. 2calc fo is the observed value fe is the expected value 2calc
Contingency Table – Observed Data
Expected Frequencies
Black
White
Red
Blue
Male 51 22 33 24
Female 45 36 27
Black
White
Red
Blue
Male 48 29
27.5
25.5
Female
Now we are ready to calculate the 2 value using the formula:
fo is the observed value
fe is the expected value
2calc
2calc
Finally look at the critical value that you have been given.
If the 2 calc value is less than the critical value, we accept H0, the null hypothesis.
If the 2 calc value is more than the critical value, we do not accept the null hypothesis, so we accept H1
In this case the 2 calc value is 6.13, and the critical value at 5% is So we do accept H0, the null hypothesis.
There is no association between favourite colour of car and gender.

17. The entries in the contingency table must be frequencies.
The expected frequencies must not be less than 1, and no more than 20% of the entries can be between 1 and 5. Otherwise the test is invalid.