Chi-Square Tests for Independence

11.1 Chi-Square Tests for Independence

Learning Objectives

To understand what chi-square distributions are.
To understand how to use a chi-square test to judge whether two factors are independent.

Chi-Square Distributions

As you know, there is a whole family of t-distributions, each one specified by a parameter called the degrees of freedom, denoted $d f .$ Similarly, all the chi-square distributions form a family, and each of its members is also specified by a parameter $d f$ , the number of degrees of freedom. Chi is a Greek letter denoted by the symbol $χ$ and chi-square is often denoted by $χ^{2} .$ Figure 11.1 "Many " shows several chi-square distributions for different degrees of freedom. A chi-square random variableA random variable that follows a chi-square distribution. is a random variable that assumes only positive values and follows a chi-square distribution.

Figure 11.1 Many $χ^{2}$ Distributions

Definition

The value of the chi-square random variable $χ^{2}$ with $d f = k$ that cuts off a right tail of area c is denoted $χ_{c}^{2}$ and is called a critical value. See Figure 11.2.

Figure 11.2 $χ_{c}^{2}$ Illustrated

Figure 12.4 "Critical Values of Chi-Square Distributions" gives values of $χ_{c}^{2}$ for various values of c and under several chi-square distributions with various degrees of freedom.

Tests for Independence

Hypotheses tests encountered earlier in the book had to do with how the numerical values of two population parameters compared. In this subsection we will investigate hypotheses that have to do with whether or not two random variables take their values independently, or whether the value of one has a relation to the value of the other. Thus the hypotheses will be expressed in words, not mathematical symbols. We build the discussion around the following example.

There is a theory that the gender of a baby in the womb is related to the baby’s heart rate: baby girls tend to have higher heart rates. Suppose we wish to test this theory. We examine the heart rate records of 40 babies taken during their mothers’ last prenatal checkups before delivery, and to each of these 40 randomly selected records we compute the values of two random measures: 1) gender and 2) heart rate. In this context these two random measures are often called factorsA variable with several qualitative levels.. Since the burden of proof is that heart rate and gender are related, not that they are unrelated, the problem of testing the theory on baby gender and heart rate can be formulated as a test of the following hypotheses:

\begin{matrix} H_{0} & : & Baby gender and baby heart rate are independent \\ vs . H_{a} & : & Baby gender and baby heart rate are n o t independent \end{matrix}

The factor gender has two natural categories or levels: boy and girl. We divide the second factor, heart rate, into two levels, low and high, by choosing some heart rate, say 145 beats per minute, as the cutoff between them. A heart rate below 145 beats per minute will be considered low and 145 and above considered high. The 40 records give rise to a 2 × 2 contingency table. By adjoining row totals, column totals, and a grand total we obtain the table shown as Table 11.1 "Baby Gender and Heart Rate". The four entries in boldface type are counts of observations from the sample of n = 40. There were 11 girls with low heart rate, 17 boys with low heart rate, and so on. They form the core of the expanded table.

Table 11.1 Baby Gender and Heart Rate

		Heart Rate
		Low	High	Row Total
Gender	Girl	11	7	18
Gender	Boy	17	5	22
Column Total		28	12	Total = 40

In analogy with the fact that the probability of independent events is the product of the probabilities of each event, if heart rate and gender were independent then we would expect the number in each core cell to be close to the product of the row total R and column total C of the row and column containing it, divided by the sample size n. Denoting such an expected number of observations E, these four expected values are:

1st row and 1st column: $E = (R \times C) ∕ n = 18 \times 28 ∕ 40 = 12.6$
1st row and 2nd column: $E = (R \times C) ∕ n = 18 \times 12 ∕ 40 = 5.4$
2nd row and 1st column: $E = (R \times C) ∕ n = 22 \times 28 ∕ 40 = 15.4$
2nd row and 2nd column: $E = (R \times C) ∕ n = 22 \times 12 ∕ 40 = 6.6$

We update Table 11.1 "Baby Gender and Heart Rate" by placing each expected value in its corresponding core cell, right under the observed value in the cell. This gives the updated table Table 11.2 "Updated Baby Gender and Heart Rate".

Table 11.2 Updated Baby Gender and Heart Rate

		Heart Rate
		Low	High	Row Total
Gender	Girl	$\begin{matrix} O = 11 \\ E = 12.6 \end{matrix}$	$\begin{matrix} O = 7 \\ E = 5.4 \end{matrix}$	R = 18
Gender	Boy	$\begin{matrix} O = 17 \\ E = 15.4 \end{matrix}$	$\begin{matrix} O = 5 \\ E = 6.6 \end{matrix}$	R = 22
Column Total		C = 28	C = 12	n = 40

A measure of how much the data deviate from what we would expect to see if the factors really were independent is the sum of the squares of the difference of the numbers in each core cell, or, standardizing by dividing each square by the expected number in the cell, the sum $Σ {(O - E)}^{2} ∕ E .$ We would reject the null hypothesis that the factors are independent only if this number is large, so the test is right-tailed. In this example the random variable $Σ {(O - E)}^{2} ∕ E$ has the chi-square distribution with one degree of freedom. If we had decided at the outset to test at the 10% level of significance, the critical value defining the rejection region would be, reading from Figure 12.4 "Critical Values of Chi-Square Distributions", $χ_{α}^{2} = χ_{0.10}^{2} = 2.706$ , so that the rejection region would be the interval $[2.706, \infty) .$ When we compute the value of the standardized test statistic we obtain

Σ \frac{{(O - E)}^{2}}{E} = \frac{{(11 - 12.6)}^{2}}{12.6} + \frac{{(7 - 5.4)}^{2}}{5.4} + \frac{{(17 - 15.4)}^{2}}{15.4} + \frac{{(5 - 6.6)}^{2}}{6.6} = 1.231

Since 1.231 < 2.706, the decision is not to reject H₀. See Figure 11.3 "Baby Gender Prediction". The data do not provide sufficient evidence, at the 10% level of significance, to conclude that heart rate and gender are related.

Figure 11.3 Baby Gender Prediction

With this specific example in mind, now turn to the general situation. In the general setting of testing the independence of two factors, call them Factor 1 and Factor 2, the hypotheses to be tested are

\begin{matrix} H_{0} & : & The two factors are independent \\ vs . H_{a} & : & The two factors are n o t independent \end{matrix}

As in the example each factor is divided into a number of categories or levels. These could arise naturally, as in the boy-girl division of gender, or somewhat arbitrarily, as in the high-low division of heart rate. Suppose Factor 1 has I levels and Factor 2 has J levels. Then the information from a random sample gives rise to a general I × J contingency table, which with row totals, column totals, and a grand total would appear as shown in Table 11.3 "General Contingency Table". Each cell may be labeled by a pair of indices $(i, j) .$ $O_{i j}$ stands for the observed count of observations in the cell in row i and column j, R_i for the $i^{t h}$ row total and C_j for the $j^{t h}$ column total. To simplify the notation we will drop the indices so Table 11.3 "General Contingency Table" becomes Table 11.4 "Simplified General Contingency Table". Nevertheless it is important to keep in mind that the Os, the Rs and the Cs, though denoted by the same symbols, are in fact different numbers.

Table 11.3 General Contingency Table

		Factor 2 Levels
		1	$· · ·$	j	$· · ·$	J	Row Total
Factor 1 Levels	1	O₁₁	$· · ·$	$O_{1 j}$	$· · ·$	$O_{1 J}$	R₁
	⋮	⋮	⋮	⋮	⋮	⋮	⋮
	i	$O_{i 1}$	$· · ·$	$O_{i j}$	$· · ·$	$O_{i J}$	R_i
	⋮	⋮	⋮	⋮	⋮	⋮	⋮
	I	$O_{I 1}$	$· · ·$	$O_{I j}$	$· · ·$	$O_{I J}$	R_I
Column Total		C₁	$· · ·$	C_j	$· · ·$	C_J	n

Table 11.4 Simplified General Contingency Table

		Factor 2 Levels
		1	$· · ·$	j	$· · ·$	J	Row Total
Factor 1 Levels	1	O	$· · ·$	O	$· · ·$	O	R
	⋮	⋮	⋮	⋮	⋮	⋮	⋮
	i	O	$· · ·$	O	$· · ·$	O	R
	⋮	⋮	⋮	⋮	⋮	⋮	⋮
	I	O	$· · ·$	O	$· · ·$	O	R
Column Total		C	$· · ·$	C	$· · ·$	C	n

As in the example, for each core cell in the table we compute what would be the expected number E of observations if the two factors were independent. E is computed for each core cell (each cell with an O in it) of Table 11.4 "Simplified General Contingency Table" by the rule applied in the example:

E = \frac{R \times C}{n}

where R is the row total and C is the column total corresponding to the cell, and n is the sample size.

After the expected number is computed for every cell, Table 11.4 "Simplified General Contingency Table" is updated to form Table 11.5 "Updated General Contingency Table" by inserting the computed value of E into each core cell.

Table 11.5 Updated General Contingency Table

		Factor 2 Levels
		1	$· · ·$	j	$· · ·$	J	Row Total
Factor 1 Levels	1	$\begin{matrix} O \\ E \end{matrix}$	$· · ·$	$\begin{matrix} O \\ E \end{matrix}$	$· · ·$	$\begin{matrix} O \\ E \end{matrix}$	R
	⋮	⋮	⋮	⋮	⋮	⋮	⋮
	i	$\begin{matrix} O \\ E \end{matrix}$	$· · ·$	$\begin{matrix} O \\ E \end{matrix}$	$· · ·$	$\begin{matrix} O \\ E \end{matrix}$	R
	⋮	⋮	⋮	⋮	⋮	⋮	⋮
	I	$\begin{matrix} O \\ E \end{matrix}$	$· · ·$	$\begin{matrix} O \\ E \end{matrix}$	$· · ·$	$\begin{matrix} O \\ E \end{matrix}$	R
Column Total		C	$· · ·$	C	$· · ·$	C	n

Here is the test statistic for the general hypothesis based on Table 11.5 "Updated General Contingency Table", together with the conditions that it follow a chi-square distribution.

Test Statistic for Testing the Independence of Two Factors

χ^{2} = Σ \frac{{(O - E)}^{2}}{E}

where the sum is over all core cells of the table.

the two study factors are independent, and
the observed count O of each cell in Table 11.5 "Updated General Contingency Table" is at least 5,

then $χ^{2}$ approximately follows a chi-square distribution with $d f = (I − 1) \times (J − 1)$ degrees of freedom.

The same five-step procedures, either the critical value approach or the p-value approach, that were introduced in Section 8.1 "The Elements of Hypothesis Testing" and Section 8.3 "The Observed Significance of a Test" of Chapter 8 "Testing Hypotheses" are used to perform the test, which is always right-tailed.

Example 1

A researcher wishes to investigate whether students’ scores on a college entrance examination (CEE) have any indicative power for future college performance as measured by GPA. In other words, he wishes to investigate whether the factors CEE and GPA are independent or not. He randomly selects n = 100 students in a college and notes each student’s score on the entrance examination and his grade point average at the end of the sophomore year. He divides entrance exam scores into two levels and grade point averages into three levels. Sorting the data according to these divisions, he forms the contingency table shown as Table 11.6 "CEE versus GPA Contingency Table", in which the row and column totals have already been computed.

Table 11.6 CEE versus GPA Contingency Table

		GPA
		<2.7	2.7 to 3.2	>3.2	Row Total
CEE	$< 1800$	35	12	5	52
CEE	$\geq 1800$	6	24	18	48
Column Total		41	36	23	$Total = 100$

Test, at the 1% level of significance, whether these data provide sufficient evidence to conclude that CEE scores indicate future performance levels of incoming college freshmen as measured by GPA.

Solution:

We perform the test using the critical value approach, following the usual five-step method outlined at the end of Section 8.1 "The Elements of Hypothesis Testing" in Chapter 8 "Testing Hypotheses".

Step 1. The hypotheses are
$\begin{matrix} H_{0} & : & CEE and GPA are independent factors \\ vs . H_{a} & : & CEE and GPA are not independent factors \end{matrix}$
Step 2. The distribution is chi-square.

Step 3. To compute the value of the test statistic we must first computed the expected number for each of the six core cells (the ones whose entries are boldface):

1st row and 1st column: $E = (R \times C) ∕ n = 41 \times 52 ∕ 100 = 21.32$
1st row and 2nd column: $E = (R \times C) ∕ n = 36 \times 52 ∕ 100 = 18.72$
1st row and 3rd column: $E = (R \times C) ∕ n = 23 \times 52 ∕ 100 = 11.96$
2nd row and 1st column: $E = (R \times C) ∕ n = 41 \times 48 ∕ 100 = 19.68$
2nd row and 2nd column: $E = (R \times C) ∕ n = 36 \times 48 ∕ 100 = 17.28$
2nd row and 3rd column: $E = (R \times C) ∕ n = 23 \times 48 ∕ 100 = 11.04$

Table 11.6 "CEE versus GPA Contingency Table" is updated to Table 11.7 "Updated CEE versus GPA Contingency Table".

Table 11.7 Updated CEE versus GPA Contingency Table

		GPA
		<2.7	2.7 to 3.2	>3.2	Row Total
CEE	$< 1800$	$\begin{matrix} O = 35 \\ E = 21.32 \end{matrix}$	$\begin{matrix} O = 12 \\ E = 18.72 \end{matrix}$	$\begin{matrix} O = 5 \\ E = 11.96 \end{matrix}$	R = 52
CEE	$\geq 1800$	$\begin{matrix} O = 6 \\ E = 19.68 \end{matrix}$	$\begin{matrix} O = 24 \\ E = 17.28 \end{matrix}$	$\begin{matrix} O = 18 \\ E = 11.04 \end{matrix}$	R = 48
Column Total		C = 41	C = 36	C = 23	n = 100

The test statistic is

\begin{matrix} χ^{2} & = & Σ^{​} \frac{{(O - E)}^{2}}{E} \\ = & \frac{{(35 - 21.32)}^{2}}{21.32} + \frac{{(12 - 18.72)}^{2}}{18.72} + \frac{{(5 - 11.96)}^{2}}{11.96} \\ + \frac{{(6 - 19.68)}^{2}}{19.68} + \frac{{(24 - 17.28)}^{2}}{17.28} + \frac{{(18 - 11.04)}^{2}}{11.04} \\ = & 31.75 \end{matrix}

Step 4. Since the CEE factor has two levels and the GPA factor has three, I = 2 and J = 3. Thus the test statistic follows the chi-square distribution with $d f = (2 - 1) \times (3 - 1) = 2$ degrees of freedom.

Since the test is right-tailed, the critical value is $χ_{0.01}^{2} .$ Reading from Figure 12.4 "Critical Values of Chi-Square Distributions", $χ_{0.01}^{2} = 9.210$ , so the rejection region is $[9.210, \infty) .$
Step 5. Since 31.75 > 9.21 the decision is to reject the null hypothesis. See Figure 11.4. The data provide sufficient evidence, at the 1% level of significance, to conclude that CEE score and GPA are not independent: the entrance exam score has predictive power.

Figure 11.4 Note 11.9 "Example 1"

Key Takeaways

Critical values of a chi-square distribution with degrees of freedom $d f$ are found in Figure 12.4 "Critical Values of Chi-Square Distributions".
A chi-square testA test based on a chi-square statistic to check whether two factors are independent. can be used to evaluate the hypothesis that two random variables or factors are independent.

Exercises

Basic

Find $χ_{0.01}^{2}$ for each of the following number of degrees of freedom.
1. $d f = 5$
2. $d f = 11$
3. $d f = 25$
Find $χ_{0.05}^{2}$ for each of the following number of degrees of freedom.
1. $d f = 6$
2. $d f = 12$
3. $d f = 30$
Find $χ_{0.10}^{2}$ for each of the following number of degrees of freedom.
1. $d f = 6$
2. $d f = 12$
3. $d f = 30$
Find $χ_{0.01}^{2}$ for each of the following number of degrees of freedom.
1. $d f = 7$
2. $d f = 10$
3. $d f = 20$
For $d f = 7$ and $α = 0.05$ , find
1. $χ_{α}^{2}$
2. $χ_{\frac{α}{2}}^{2}$
For $d f = 17$ and $α = 0.01$ , find
1. $χ_{α}^{2}$
2. $χ_{\frac{α}{2}}^{2}$

A data sample is sorted into a 2 × 2 contingency table based on two factors, each of which has two levels.

		Factor 1
		Level 1	Level 2	Row Total
Factor 2	Level 1	$\begin{matrix} 20 \end{matrix}$	$\begin{matrix} 10 \end{matrix}$	R
Factor 2	Level 2	$\begin{matrix} 15 \end{matrix}$	$\begin{matrix} 5 \end{matrix}$	R
Column Total		C	C	n

Find the column totals, the row totals, and the grand total, n, of the table.
Find the expected number E of observations for each cell based on the assumption that the two factors are independent (that is, just use the formula $E = (R \times C) ∕ n$ ).
Find the value of the chi-square test statistic $χ^{2} .$
Find the number of degrees of freedom of the chi-square test statistic.

A data sample is sorted into a 3 × 2 contingency table based on two factors, one of which has three levels and the other of which has two levels.

		Factor 1
		Level 1	Level 2	Row Total
Factor 2	Level 1	$\begin{matrix} 20 \end{matrix}$	$\begin{matrix} 10 \end{matrix}$	R
	Level 2	$\begin{matrix} 15 \end{matrix}$	$\begin{matrix} 5 \end{matrix}$	R
	Level 3	$\begin{matrix} 10 \end{matrix}$	$\begin{matrix} 20 \end{matrix}$	R
Column Total		C	C	n

Find the column totals, the row totals, and the grand total, n, of the table.
Find the expected number E of observations for each cell based on the assumption that the two factors are independent (that is, just use the formula $E = (R \times C) ∕ n$ ).
Find the value of the chi-square test statistic $χ^{2} .$
Find the number of degrees of freedom of the chi-square test statistic.

Applications

A child psychologist believes that children perform better on tests when they are given perceived freedom of choice. To test this belief, the psychologist carried out an experiment in which 200 third graders were randomly assigned to two groups, A and B. Each child was given the same simple logic test. However in group B, each child was given the freedom to choose a text booklet from many with various drawings on the covers. The performance of each child was rated as Very Good, Good, and Fair. The results are summarized in the table provided. Test, at the 5% level of significance, whether there is sufficient evidence in the data to support the psychologist’s belief.

		Group
		A	B
Performance	Very Good	32	29
	Good	55	61
	Fair	10	13

In regard to wine tasting competitions, many experts claim that the first glass of wine served sets a reference taste and that a different reference wine may alter the relative ranking of the other wines in competition. To test this claim, three wines, A, B and C, were served at a wine tasting event. Each person was served a single glass of each wine, but in different orders for different guests. At the close, each person was asked to name the best of the three. One hundred seventy-two people were at the event and their top picks are given in the table provided. Test, at the 1% level of significance, whether there is sufficient evidence in the data to support the claim that wine experts’ preference is dependent on the first served wine.

		Top Pick
		A	B	C
First Glass	A	12	31	27
	B	15	40	21
	C	10	9	7

Is being left-handed hereditary? To answer this question, 250 adults are randomly selected and their handedness and their parents’ handedness are noted. The results are summarized in the table provided. Test, at the 1% level of significance, whether there is sufficient evidence in the data to conclude that there is a hereditary element in handedness.

		Number of Parents Left-Handed
		0	1	2
Handedness	Left	8	10	12
Handedness	Right	178	21	21

Some geneticists claim that the genes that determine left-handedness also govern development of the language centers of the brain. If this claim is true, then it would be reasonable to expect that left-handed people tend to have stronger language abilities. A study designed to text this claim randomly selected 807 students who took the Graduate Record Examination (GRE). Their scores on the language portion of the examination were classified into three categories: low, average, and high, and their handedness was also noted. The results are given in the table provided. Test, at the 5% level of significance, whether there is sufficient evidence in the data to conclude that left-handed people tend to have stronger language abilities.

		GRE English Scores
		Low	Average	High
Handedness	Left	18	40	22
Handedness	Right	201	360	166

It is generally believed that children brought up in stable families tend to do well in school. To verify such a belief, a social scientist examined 290 randomly selected students’ records in a public high school and noted each student’s family structure and academic status four years after entering high school. The data were then sorted into a 2 × 3 contingency table with two factors. Factor 1 has two levels: graduated and did not graduate. Factor 2 has three levels: no parent, one parent, and two parents. The results are given in the table provided. Test, at the 1% level of significance, whether there is sufficient evidence in the data to conclude that family structure matters in school performance of the students.

		Academic Status
		Graduated	Did Not Graduate
Family	No parent	18	31
	One parent	101	44
	Two parents	70	26

A large middle school administrator wishes to use celebrity influence to encourage students to make healthier choices in the school cafeteria. The cafeteria is situated at the center of an open space. Everyday at lunch time students get their lunch and a drink in three separate lines leading to three separate serving stations. As an experiment, the school administrator displayed a poster of a popular teen pop star drinking milk at each of the three areas where drinks are provided, except the milk in the poster is different at each location: one shows white milk, one shows strawberry-flavored pink milk, and one shows chocolate milk. After the first day of the experiment the administrator noted the students’ milk choices separately for the three lines. The data are given in the table provided. Test, at the 1% level of significance, whether there is sufficient evidence in the data to conclude that the posters had some impact on the students’ drink choices.

	Student Choice
	Regular	Strawberry	Chocolate
Poster Choice
Regular	38	28	40
Strawberry	18	51	24
Chocolate	32	32	53

Large Data Set Exercise

Large Data Set 8 records the result of a survey of 300 randomly selected adults who go to movie theaters regularly. For each person the gender and preferred type of movie were recorded. Test, at the 5% level of significance, whether there is sufficient evidence in the data to conclude that the factors “gender” and “preferred type of movie” are dependent.

http://www.flatworldknowledge.com/sites/all/files/data8.xls

Answers

1. 15.09,
2. 24.72,
3. 44.31
1. 10.64,
2. 18.55,
3. 40.26
1. 14.07,
2. 16.01
1. $C_{1} = 35$ , $C_{2} = 15$ , $R_{1} = 30$ , $R_{2} = 20$ , n = 50,
2. $E_{11} = 21$ , $E_{12} = 9$ , $E_{21} = 14$ , $E_{22} = 6$ ,
3. $χ^{2} = 0.3968$ ,
4. $d f = 1$

$χ^{2} = 0.6698$ , $χ_{0.05}^{2} = 5.99$ , do not reject H₀
$χ^{2} = 72.35$ , $χ_{0.01}^{2} = 9.21$ , reject H₀
$χ^{2} = 21.2784$ , $χ_{0.01}^{2} = 9.21$ , reject H₀

$χ^{2} = 28.4539 .$ $d f = 3 .$ Rejection Region: $[7.815, \infty) .$ Decision: Reject H₀ of independence.