# 11.3.2 - Minitab Express: Test of Independence

11.3.2 - Minitab Express: Test of IndependenceIf you have a data file with the responses for individual cases then you have "raw data" and can follow the directions below. If you have a table filled with data, then you have "summarized data." There is an example of conducting a chi-square test of independence using summarized data on a later page. After data entry the procedure is the same for both data entry methods.

## MinitabExpress – Chi-square Test Using Raw Data

**Research question**: Is there a relationship between where a student sits in class and whether they have ever cheated?

Null hypothesis: Seat location and cheating are not related in the population.

Alternative hypothesis: Seat location and cheating are related in the population.

To perform a chi-square test of independence in Minitab Express using raw data:

- Open Minitab data set
- On a
**PC**: Select**STATISTICS > Cross Tabulation and Chi-Square**

On a**Mac**: Select**Statistics**> Tables > Cross Tabulation and Chi-Square - Double-click on the variable
*Seating*to insert it into the*Rows*box - Double-click on the variable
*Ever Cheat*to insert it into the*Columns*box - Click the
*Display*tab and check the boxes*Chi-square test for association*and*Expected cell counts* - Click
*OK*

This should result in the following output:

Cell contents grouped by Back, Front, Middle, All; First row: count, Next row: expected count

No | Yes | All | |
---|---|---|---|

Back | 24 | 8 | 32 |

24.21 | 7.79 | ||

Front | 38 | 8 | 46 |

34.81 | 11.19 | ||

Middle | 109 | 39 | 148 |

111.98 | 36.02 | ||

All | 1714 | 55 | 226 |

Chi-Square | DF | P-Value | |
---|---|---|---|

Pearson | 1.54 | 2 | 0.4633 |

Likelihood Ratio | 1.63 | 2 | 0.4435 |

Select your operating system below to see a step-by-step guide for this example.

All expected values are at least 5 so we can use the Pearson chi-square test statistic. Our results are \(\chi^2 (2) = 1.54\). \(p = 0.4633\). Because our \(p\) value is greater than the standard alpha level of 0.05, we fail to reject the null hypothesis. There is not evidence of a relationship in the population between seat location and whether a student has cheated.

# 11.3.2.1 - Video Example: Dog & Cat Ownership (Raw Data)

11.3.2.1 - Video Example: Dog & Cat Ownership (Raw Data)This example uses the dataset:

\(H_0:\) There is not a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students

\(H_a:\) There is a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students

Cell contents grouped by No, Yes, Missing; First row: count, Next row: expected count

No | Yes | All | |
---|---|---|---|

BaNock | 183 | 69 | 252 |

176.02 | 75.98 | ||

Yes | 183 | 89 | 272 |

189.98 | 82.02 | ||

Missing | 1 | 0 | |

All | 366 | 158 | 524 |

Assumption: All expected counts are at least 5. The expected counts here are 176.02, 75.98, 189.98, and 82.02, so this assumption has been met.

Chi-Square | DF | P-Value | |
---|---|---|---|

Pearson | 1.77 | 1 | 0.1833 |

Likelihood Ratio | 1.77 | 1 | 0.1828 |

Since the assumption was met in step 1, we can use the Pearson chi-square test statistic.

\(Pearson\;\chi^2 = 1.77\)

\(p = 0.1833\)

Our p value is greater than the standard 0.05 alpha level, so we fail to reject the null hypothesis.

There is not evidence of a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students.