A **Chi-Square**** Test for Independence **is used when we want to formally test whether or not there is a significant association between two categorical variables from a single population.

The following example illustrates how to conduct a Chi-Square Test for Independence.

**Example: Chi-Square ****Test for Independence**

We want to know whether or not gender is associated with political party preference. We take a simple random sample of 500 voters and survey them on their political party preference. Here are the results:

Republican | Democrat | Independent | Total | |
---|---|---|---|---|

Male | 120 | 90 | 40 | 250 |

Female | 110 | 95 | 45 | 250 |

Total | 230 | 185 | 85 | 500 |

Does gender seem to be associated with political party preference? Use a 0.05 level of significance.

**Step 1. State the hypotheses. **

The null hypothesis (H_{0}): Gender and political party preference is independent.

The alternative hypothesis: (Ha): Gender and political party preference is not independent.

**Step 2. Determine a significance level to use.**

The problem tells us that we are to use a .05 level of significance.

**Step 3. Find the test statistic.**

The test statistic is X^{2} = Σ [ (O_{i} – E_{i})^{2} / E_{i} ]

Where Σ is just a fancy symbol that means “sum”, O_{i} is the observed frequency at level *i* of the variable, and E_{i} is the expected frequency at level *i* of the variable.

Notice that we surveyed an equal amount of males and females. This means that if there is no association between gender and political party preference, we can *expect* that each party is split 50/50 between males and females.

For example, we would *expect* that 50% of all the people who said they were republican to be females. That is, .50 * 230 = 115. We would also expect .50 * 230 = 115 males. Let’s find the expected number and observed number of people for each political party:

Republican | Democrat | Independent | Total | |
---|---|---|---|---|

Male | 115 | 92.5 | 42.5 | 250 |

Female | 115 | 92.5 | 42.5 | 250 |

Total | 230 | 185 | 85 | 500 |

Republican | Democrat | Independent | Total | |
---|---|---|---|---|

Male | 120 | 90 | 40 | 250 |

Female | 110 | 95 | 45 | 250 |

Total | 230 | 185 | 85 | 500 |

Lastly, calculate the Chi-Square test statistic X^{2}: (120 – 115)^{2} / 115 + (110 – 115)^{2} / 115 + (90 – 92.5)^{2} / 92.5 + (95 – 92.5)^{2} / 92.5 + (40 – 42.5)^{2} / 42.5 + (45 – 42.5)^{2} / 42.5 =** .864**

Use the Chi-Square Calculator with a degrees of freedom = (r-1)*(c-1) (where r = # rows, c = # columns) = (2-1)*(3-1) = 2, Chi-square critical value = .864, and click “Calculate p-value” to find that the p-value = .35079. Then 1 – .35079= **.649**.

**Step 4. Reject or fail to reject the null hypothesis.**

Since the p-value (.649) is not less than our significance level of .05, we fail to reject the null hypothesis.

**Step 5. Interpret the results. **

Since we failed to reject the null hypothesis, we do not have sufficient evidence to state that there is an association between gender and political party preference.