Tests for Independence (College Board AP® Statistics)

Tests for Independence (College Board AP® Statistics)

Revision Note

Test for independence

Revision Note

Test for independence

What is a test for independence?

What are the null and alternative hypotheses for an independence test?

What are the conditions for an independence test?

How do I calculate a chi-square value?

What are degrees of freedom?

How do I use the chi-square distribution table?

How do I conclude a hypothesis test?

How can I perform an independence test on the calculator?

Unlock more, it's free!

What is a test for independence?

What are the null and alternative hypotheses for an independence test?

What are the conditions for an independence test?

How do I calculate a chi-square value?

What are degrees of freedom?

How do I use the chi-square distribution table?

How do I conclude a hypothesis test?

How can I perform an independence test on the calculator?

Unlock more, it's free!

Author: Naomi C

Exam Tip

Exam Tip

Exam Tip

Exam Tip

Worked Example

You've read 0 of your 10 free revision notes

Join the 100,000+ Students that ❤️ Save My Exams

Exam Tip

Exam Tip

Exam Tip

Exam Tip

Worked Example

You've read 0 of your 10 free revision notes

Join the 100,000+ Students that ❤️ Save My Exams

Author

Naomi C

Expertise

Maths

A chi-square $(χ^{2})$ test for independence is used to determine whether there is a significant relationship between two categorical variables
- i.e. if two variables are independent of each other or if they are related (dependent)
For example, you may have collected data on the grade level and school subject preference from a group of students
- A test for independence could indicate whether the grade level of a student has an impact on their preferred school subject
A chi-square test for independence is a specific goodness of fit test
- The observed values are compared with values that you would expect if the two variables are independent
Observed values can be shown in a two-way table
- This is also known as a contingency table
E.g. a contingency table for the grade level and school subject preference of a group of students is shown below

The null hypothesis, $H_{0}$ , is the assumption that the two categorical variables are independent
- e.g. $H_{0} :$ The grade level of a student is independent of their school subject preference (there is no association)
  - It is assumed to be correct, unless evidence proves otherwise
The alternative hypothesis, $H_{a}$ , is the assumption that the two categorical variables are not independent
- e.g. $H_{a} :$ The grade level of a student is not independent of their school subject preference (there is an association)

In an exam, a test for independence may also be referred to as a test for an association between two variables, but be careful with the wording: if two variables are independent, then there is no association between them.

When performing a chi-square independence test:
- Observed values must come from a random sample
- Observed values must be independent
  - They are sampled with replacement
  - or the sample size is less than 10% of the population size
- Expected values must meet the large counts condition
  - Each expected value must be greater than or equal to 5
  - or at least 80% of the expected values are greater than 5 and all are greater than or equal to 1

In the exam, either condition is accepted for the large counts condition.

The chi-square value for the test of independence, $X^{2}$ , can be calculated from the formula given to you in the exam
- $χ^{2} = \sum_{}^{} \frac{{(observed - expected)}^{2}}{expected}$
The larger $X^{2}$ is, the more different the observed values are from the expected values
To be able to calculate the chi-square value, you therefore need to find the expected values first
To calculate the expected value for a particular cell, multiply together:
- the probability of being in that particular row
- by the probability of being in that particular column
- by the total number in the sample
- This is equivalent to simply multiplying the row total by the column total and dividing by the grand total
E.g. the expected value for the number of 10th graders who prefer languages in the example above is
- $\frac{31}{100} \cdot \frac{19}{100} \cdot 100 = 5.89$
- or just $\frac{31 \cdot 19}{100} = 5.89$

The table below shows all the expected values for the example above

		Grade level
		9th	10th	11th	12th	Total
Preferred subject	Math/Science	10.5	7.98	12.6	10.92	42
	Humanities	6.75	5.13	8.1	7.02	27
	Languages	7.75	5.89	9.3	8.06	31
	Total	25	19	30	26	100

Expected values do not need to be integer values, so leave them unrounded to avoid calculation errors!

The number of degrees of freedom, 'dof', is equal to
- the $(number of rows - 1) (number of columns - 1)$
- e.g. dof for the contingency table above is $(3 - 1) (4 - 1) = 2 \cdot 3 = 6$

You can use the chi-square tables given to you in the exam to find the critical value
- This is the threshold value that determines whether you reject the null hypothesis or not
To find the critical value from the tables, you need the significance level, $α %$ and the dof
- The critical value is located in the cell where the relevant row and column intersect

Conclusions to a hypothesis test need to show two things:
- a decision about the null hypothesis
- an interpretation of this decision in the context of the question
To make the decision, compare the calculated chi-square value, $X^{2}$ , to the critical value from the table
- If $X^{2} > critical value$ then the null hypothesis should be rejected
  - The two categorical variables are not independent
- If $X^{2} < critical value$ then the null hypothesis should not be rejected
  - There is not enough evidence to say that the two categorical variables are not independent

To complete an independence test on your calculator:
- Create a matrix of the observed values
- Perform a chi-square test
  - This is often called a $χ^{2}$ two-way test on a calculator
- Compare your calculated $X^{2}$ , with the critical value from the chi-square tables
Alternatively, you can compare the given significance level, $α$ , with the calculator's $p$ -value
- If using the $p$ -value, remember
  - $p < α$ , reject the null hypothesis
  - $p > α$ , do not reject the null hypothesis

Even if you perform the independence test on your calculator, it is still important to show all of your working to demonstrate full understanding. Depending on the question, you may need to show how the chi-square statistic is calculated in full or just how an expected value and the degrees of freedom are calculated.

If you compare the $p$ -value with $α$ , don't forget that the inequalities are the opposite to when you are comparing the $X^{2}$ value to the critical value when you are determining whether or not to reject the null hypothesis!

A coffee company wanted to understand more about who their customers were. They took a random sample of 200 individuals to see if there was an association between an individual's relationship status and whether they were a coffee drinker or not.

The outcomes of their research are shown in the table below.

	Single	Married / Cohabiting	Other	Total
Coffee drinker	48	25	37	110
Non-coffee drinker	23	32	35	90
Total	71	57	72	200

Determine, at the 5% significance level, if there is an association between an individual's relationship status and whether they are a coffee drinker or not.

Write the null and alternative hypotheses

$H_{0} :$ There is no association between an individual's relationship status and whether they drink coffee or not, they are independent

$H_{a} :$ There is an association between an individual's relationship status and whether they drink coffee or not, they are not independent

State the type of test being used

The correct inference procedure is a chi-square test of independence at $α = 0.05$

Calculate the expected values (by multiplying the row total by the column total and dividing by the grand total)

	Single	Married / Cohabiting	Other	Total
Coffee drinker	39.05	31.35	39.6	110
Non-coffee drinker	31.95	25.65	32.4	90
Total	71	57	72	200

Verify the conditions for the test

All conditions for inference have been met:

The observed values are independent as the sample of individuals is randomly selected
All expected values are greater than 5

Calculate the chi-square value, $X^{2}$ , $\sum \frac{{(observed - expected)}^{2}}{expected}$

$\begin{array}{rcl} X^{2} & = & \frac{{(48 - 39.05)}^{2}}{39.05} + \frac{{(25 - 31.35)}^{2}}{31.35} + \frac{{(37 - 39.6)}^{2}}{39.6} + \frac{{(23 - 31.95)}^{2}}{31.95} + \frac{{(32 - 25.65)}^{2}}{25.65} + \frac{{(35 - 32.4)}^{2}}{32.4} \\ = & 7.795 . . . \end{array}$

State the number of degrees of freedom

degrees of freedom = $(2 - 1) (3 - 1) = 2$

Find the critical value from the chi-square tables

Find the row corresponding to 2 degrees of freedom and the column corresponding to $α = 0.05$

$critical value = 5.99$

Compare the calculated $X^{2}$ value to the critical value and state the conclusion of the test

$\begin{array}{rcl} 7.795 . . . & > & 5.99 \\ X^{2} & > & critical value \end{array}$

$H_{0}$ is rejected

Interpret the result in the context of the question

There is sufficient evidence to suggest that there is an association between an individual's relationship status and whether they are a coffee drinker or not, i.e. sufficient evidence that they are not independent

the (exam) results speak for themselves:

Next topic

Did this page help you?

Grade level

Preferred subject

Math/Science

Humanities