Chi-square test is a non-parametric test. The assumption of normal distribution in the population is not required for this test. The statistical technique chi-square can be used to find the association (dependencies) between sets of two or more categorical variables by comparing how close the observed frequencies are to the expected frequencies. In other words, a chi-square ($\chi^2$) statistic is used to investigate whether the distributions of categorical variables are different from one another. Note that the responses of categorical variables should be independent of each other. We use the chi-square test for a relationship between two nominal scaled variables.

The chi-square test of independence is used as a test of goodness of fit and as a test of independence. In a test of goodness of fit, we check whether or not the observed frequency distribution is different from the theoretical distribution, while in a test of independence, we assess, whether paired observations on two variables are independent from each other (from the contingency table).

**Example:** A social scientist sampled 140 people and classified them according to income level and whether or not they played a state lottery in the last month. The sample information is reported below. Is it reasonable to conclude that playing the lottery is related to income level? Use the 0.05 significance level.

Income | ||||

Low | Middle | High | Total | |

Played | 46 | 28 | 21 | 95 |

Did not play | 14 | 12 | 19 | 45 |

Total | 60 | 40 | 40 | 140 |

A step-by-step procedure for testing the hypothesis about the association between these two variables is described, below.

**Step1:**

$H_0$: There is no relationship between income and whether the person played the lottery.

$H_1$: There is a relationship between income and whether the person played the lottery.

**Step2:** Level of Significance 0.05

**Step 3:** Test statistics (calculations)

Observed Frequencies ($f_o$) | Expected Frequencies ($f_e$) | $\frac{(f_o – f_e)^2}{f_e}$ |
---|---|---|

46 | 95*60/140= 40.71 | $\frac{(46-40.71)^2}{40.71}$ |

28 | 95*40/140= 27.14 | $\frac{(28-27.14)^2}{27.14}$ |

21 | 95*40/140= 27.14 | $\frac{(21-27.14)^2}{27.14}$ |

14 | 45*60/140= 19.29 | $\frac{(14-19.29)^2}{19.29}$ |

12 | 45*40/140= 12.86 | $\frac{(12-12.6)^2}{12.86}$ |

19 | 45*40/140= 12.86 | $\frac{(19-12.86)^2}{12.86}$ |

$ \chi^2=\sum[\frac{(f_0-f_e)^2}{f_e}]=$ | 6.544 |

**Step 4:** Critical Region:

Tabular Chi-Square value at 0.05 level of significance and $(r-1) \times (c-1)=(2-1)\times(3-1)=2$ is 5.991.

**Step 5:** Decision

As the calculated Chi-Square value is greater than the tabular Chi-Square value, we reject $H_0$, which means that there is a relationship between income level and playing the lottery.

Note that there are several types of chi-square tests (such as Yates, Likelihood ratio, test in time series) available which depend on the way data was collected and also the hypothesis being tested.

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