### Beyond AP Statistics

#### Probability Basics

#### Small Samples

#### Distributions

#### Power

### Beyond AP Statistics

#### Probability Basics

#### Small Samples

#### Distributions

#### Power

# Chi-Square Test for Independence

This lesson explains how to conduct a
**chi-square test for independence**.
The test is applied when you have two
categorical variables
from a single population. It is used to determine whether
there is a significant association between the two variables.

For example, in an election survey, voters might be classified by gender (male or female) and voting preference (Democrat, Republican, or Independent). We could use a chi-square test for independence to determine whether gender is related to voting preference. The sample problem at the end of the lesson considers this example.

## When to Use Chi-Square Test for Independence

The test procedure described in this lesson is appropriate when the following conditions are met:

- The sampling method is simple random sampling.
- The variables under study are each categorical.
- If sample data are displayed in a contingency table, the expected frequency count for each cell of the table is at least 5.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

## State the Hypotheses

Suppose that Variable A has *r* levels, and
Variable B has *c* levels.
The
null hypothesis states that knowing the level of
Variable A does not help you predict the level of
Variable B. That is, the variables are independent.

H_{o}: Variable A and Variable B are independent.

H_{a}: Variable A and Variable B are not independent.

The
alternative hypothesis is that knowing the level of
Variable A *can* help you predict the level of
Variable B.

**Note:** Support for the alternative hypothesis suggests that the variables
are related; but the relationship is not necessarily causal, in
the sense that one variable "causes" the other.

## Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. The plan should specify the following elements.

- Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
- Test method. Use the chi-square test for independence to determine whether there is a significant relationship between two categorical variables.

## Analyze Sample Data

Using sample data, find the degrees of freedom, expected frequencies, test statistic, and the P-value associated with the test statistic. The approach described in this section is illustrated in the sample problem at the end of this lesson.

**Degrees of freedom.**The degrees of freedom (DF) is equal to:DF = (r - 1) * (c - 1)

where r is the number of levels for one catagorical variable, and c is the number of levels for the other categorical variable.**Expected frequencies.**The expected frequency counts are computed separately for each level of one categorical variable at each level of the other categorical variable. Compute r * c expected frequencies, according to the following formula.E

where E_{r,c}= (n_{r}* n_{c}) / n_{r,c}is the expected frequency count for level*r*of Variable A and level*c*of Variable B, n_{r}is the total number of sample observations at level r of Variable A, n_{c}is the total number of sample observations at level*c*of Variable B, and n is the total sample size.**Test statistic.**The test statistic is a chi-square random variable (Χ^{2}) defined by the following equation.Χ

where O^{2}= Σ [ (O_{r,c}- E_{r,c})^{2}/ E_{r,c}]_{r,c}is the observed frequency count at level*r*of Variable A and level*c*of Variable B, and E_{r,c}is the expected frequency count at level*r*of Variable A and level*c*of Variable B.**P-value.**The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a chi-square, use the Chi-Square Distribution Calculator to assess the probability associated with the test statistic. Use the degrees of freedom computed above.

## Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level, and rejecting the null hypothesis when the P-value is less than the significance level.

## Test Your Understanding

**Problem**

A public opinion poll surveyed a simple random sample of 1000 voters. Respondents were classified by gender (male or female) and by voting preference (Republican, Democrat, or Independent). Results are shown in the contingency table below.

Voting Preferences | Row total | |||
---|---|---|---|---|

Rep | Dem | Ind | ||

Male | 200 | 150 | 50 | 400 |

Female | 250 | 300 | 50 | 600 |

Column total | 450 | 450 | 100 | 1000 |

Is there a gender gap? Do the men's voting preferences differ significantly from the women's preferences? Use a 0.05 level of significance.

**Solution**

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

**State the hypotheses.**The first step is to state the null hypothesis and an alternative hypothesis.H

_{o}: Gender and voting preferences are independent.H

_{a}: Gender and voting preferences are not independent.**Formulate an analysis plan**. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for independence.**Analyze sample data**. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2

E

_{r,c}= (n_{r}* n_{c}) / n

E_{1,1}= (400 * 450) / 1000 = 180000/1000 = 180

E_{1,2}= (400 * 450) / 1000 = 180000/1000 = 180

E_{1,3}= (400 * 100) / 1000 = 40000/1000 = 40

E_{2,1}= (600 * 450) / 1000 = 270000/1000 = 270

E_{2,2}= (600 * 450) / 1000 = 270000/1000 = 270

E_{2,3}= (600 * 100) / 1000 = 60000/1000 = 60Χ

^{2}= Σ [ (O_{r,c}- E_{r,c})^{2}/ E_{r,c}]

Χ^{2}= (200 - 180)^{2}/180 + (150 - 180)^{2}/180 + (50 - 40)^{2}/40

+ (250 - 270)^{2}/270 + (300 - 270)^{2}/270 + (50 - 60)^{2}/60

Χ^{2}= 400/180 + 900/180 + 100/40 + 400/270 + 900/270 + 100/60

Χ^{2}= 2.22 + 5.00 + 2.50 + 1.48 + 3.33 + 1.67 = 16.2where DF is the degrees of freedom, r is the number of levels of gender, c is the number of levels of the voting preference, n

The P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 16.2._{r}is the number of observations from level*r*of gender, n_{c}is the number of observations from level*c*of voting preference, n is the number of observations in the sample, E_{r,c}is the expected frequency count when gender is level*r*and voting preference is level*c*, and O_{r,c}is the observed frequency count when gender is level*r*voting preference is level*c*.We use the Chi-Square Distribution Calculator to find P(Χ

^{2}> 16.2) = 0.0003.**Interpret results**. Since the P-value (0.0003) is less than the significance level (0.05), we cannot accept the null hypothesis. Thus, we conclude that there is a relationship between gender and voting preference.

**Note:** If you use this approach on an exam, you may also want to mention
why this approach is appropriate. Specifically, the approach is
appropriate because the sampling method was simple random sampling, the
variables under study were categorical, and the expected frequency count
was at least 5 in each cell of the contingency table.

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